CH 9 Hypothesis Testing

For which of the following is a t-test appropriate for performing a hypothesis test? A sample mean of 19 minutes and a population standard deviation of 3 minutes are obtained using a simple random sample of the commute times of 10 individuals from the residents of a city. A sample mean of 21 miles per gallon and a sample standard deviation of 2 miles per gallon are obtained using a simple random sample of 22 vehicles from a large group of vehicles, which have approximately normally distributed fuel consumption rates. A simple random sample of 12 two-inch candles is taken from the candles offered at a candle store with approximately normally distributed burning times. The sample mean is 7 hours, the sample standard deviation is 1.5 hours, and the population standard deviation is 1.1 hours. A population is approximately normal. The sample mean is 89, and the population standard deviation is 5 based on a sample of size 12.

A sample mean of 21 miles per gallon and a sample standard deviation of 2 miles per gallon are obtained using a simple random sample of 22 vehicles from a large group of vehicles, which have approximately normally distributed fuel consumption rates. In order to perform a hypothesis test of a single population mean μ using a Student's t-distribution, the data should be a simple random sample that comes from a population that is approximately normally distributed with an unknown population standard deviation. The only option meeting these criteria is the context in B.

Hypothesis testing: a quantitative method of evaluating hypotheses consists of two contradictory hypotheses or statements, a decision based on the data, and a conclusion Null hypothesis: denoted by the variable H0, a statement about the population that is believed to be true (or used to put forth an argument) unless it can be shown to be incorrect beyond a reasonable doubt Alternative hypothesis: denoted by the variable Ha, a claim about the population that is contradictory to H0; this is the conclusion if we reject H0

Hypothesis testing: a quantitative method of evaluating hypotheses consists of two contradictory hypotheses or statements, a decision based on the data, and a conclusion Null hypothesis: denoted by the variable H0, a statement about the population that is believed to be true (or used to put forth an argument) unless it can be shown to be incorrect beyond a reasonable doubt Alternative hypothesis: denoted by the variable Ha, a claim about the population that is contradictory to H0; this is the conclusion if we reject H0

Understand the Process of Hypothesis Testing

Hypothesis testing: formal process to test accepted or "believed" claims about the populationHypothesis testing is also referred to as Significance testing Claim: hypothesized statement about a parameter Significance level: Allows us to make a conclusion for the hypothesis testSignificance level is also referred to as alpha Test Statistic: Compares the randomly sampled data to the population parameter in the hypotheses Critical Value: Computed from the significance level and test type Reject the null hypothesis: When a test statistic falls within a rejection region

An office manager, who works at a finance company, claims that employees spend less than 5 hours on the phone per day, on average. If the office manager wants to conduct a hypothesis test, should they use a left-, right-, or two-tailed hypothesis test to analyze whether employees spend less than 5 hours on the phone per day, on average? Left-tailed test Right-tailed test Two-tailed test

Left-tailed test To identify the null and alternative hypothesis, first determine the purpose of the hypothesis test. In this scenario, the office manager is analyzing whether employees spend less than 5 hours on the phone per day, on average. Since the primary concern of this test is dealing with whether a population parameter, p, is less than a specified value, p0, then the office manager should conduct a left-tailed test.

Another trader would like to carry out a hypothesis test about stocks that offer dividends. Why is this hypothesis test right-tailed? This is a right-tailed test because a direction is not specified. This is a right-tailed test because a direction is specified. The population parameter is greater than the specified value. This is a right-tailed test because a direction is specified. The population parameter is less than the specified value. More information is needed.

More information is needed. There is not enough information to determine why this is a right-tailed test. We need to know how the population parameter and specified value are being compared.

Heather, a sociology major is interested in studying mass media topics. She is particularly interested in the percentage of mass media topics that relate to entertainment. Based on previous research, 72% of mass media topics relate to entertainment. She suspects this percent is different. During the process of hypothesis testing, she calculates a probability using the test statistic. What is the probability associated with the test statistic called? Critical value Alternative hypothesis P-value Significance level

P-value The p-value is a probability associated with the test-statistic and is compared to the significance level.

Joanna, a machine learning engineer is studying what programming languages are the most effective to help systems learn from data. Based on previous research, more than 50% of machine learning involves the use of Python, a programming language. She suspects this percent has increased in recent years. In the process of hypothesis testing, what set value will allow Joanna to make a conclusion about whether her sample proportion is "different enough" than the claimed population proportion?' Null hypothesis P-value Test statistic Significance level

Significance level Before calculating a test statistic, the significance level is usually determined. Deciding on a significance level allows us to make a conclusion about whether our sample parameter is "different enough" than the claimed population parameter.

A finance blogger is writing an article about credit card use. As they write their article, the finance blogger is interested in the average number of credit cards per adult. Based on previous research, an adult has 4 credit cards, on average. The finance blogger believes this average is different. If the finance blogger chooses a p-value approach, what does the p-value need to be compared to in order to determine if there's enough evidence to support their claim? Critical value Null hypothesis Test statistic Significance level

Significance level The p-value is compared to the significance level in order to make a conclusion. If the p-value is smaller than the significance level, the finance blogger can reject the null hypothesis.

A doctor is measuring body temperature for patients visiting the office. The doctor believes the average body temperature is less than 98.6 degrees Fahrenheit and would like to test this claim. During the process of hypothesis testing, the doctor computes a value from the sample data, which will be used to compare the sample data to the population parameter. What value did the doctor compute? Critical value Test Statistic P-value Significance level

Test Statistic The test statistic is computed using sample data. It helps us to compare the sample data to the population parameter in the hypotheses.

A finance journal, which publishes research on current financial topics, states that the maturity term for a certificate of deposit is, on average, 10 years. A banker believes the average maturity term at their bank is different than the amount quoted in the finance journal. After completing a study, the banker found that the average maturity term for a certificate of deposit is 8 years, on average. As the banker sets up a hypothesis test to determine if their belief is correct, what is the banker's claim? Certificates of deposit mature at 10 years. The average maturity term for a certificate of deposit is 10 years. The average maturity term for a certificate of deposit is different than 8 years. The average maturity term for a certificate of deposit is different than 10 years.

The average maturity term for a certificate of deposit is different than 10 years. Since a claim is a statement about the population parameter, the banker is making a statement about the population average maturity term for a certificate of deposit. Specifically, the banker is claiming that the average maturity term for a certificate of deposit is different than 10 years. By testing this claim, the banker can determine whether there is enough evidence to suggest the length of time quoted in the finance journal for a certificate of deposit to mature is correct or not.

A digital marketer is studying the use of digital video ads as a marketing tool. The digital marketer states the average amount of money a U.S. company spends on digital video ads is greater than 85 thousand dollars. If we would like to test the digital marketer's claim with a hypothesis test using a significance level of α=0.05 , which of the following choices are true? There is a 5% chance we will conclude μ=85, but is in fact μ>85. There is a 5% chance of rejecting the null hypothesis. There is a 5% chance we will conclude μ>85, but is in fact μ=85. There is a 5% chance that μ>85.

There is a 5% chance we will conclude μ>85, but is in fact μ=85. A significance level is the probability of rejecting the null hypothesis, H0, when it is in fact true. So, a significance level of α=0.05 means there is a 5% chance we will conclude from our data that μ>85 thousand dollars, but it is in fact μ=85 thousand dollars.

Which of the following results in a null hypothesis μ=31 and alternative hypothesis μ<31? A fitness center claims that the mean amount of time that a person spends at the gym per visit is less than 31 minutes. Some researchers do not think this is correct and want to show that the mean time is at least 31 minutes. A fitness center claims that the mean amount of time that a person spends at the gym per visit is 31 minutes. Some researchers do not think this is correct and want to show that the mean time is less than 31 minutes. A fitness center claims that the mean amount of time that a person spends at the gym per visit is 31 minutes. Some researchers do not think this is correct and want to show that the mean time is not equal to 31 minutes. A fitness center claims that the mean amount of time that a person spends at the gym per visit is more than 31 minutes. Some researchers do not think this is correct and want to show that the mean time is at most 31 minutes.

A fitness center claims that the mean amount of time that a person spends at the gym per visit is 31 minutes. Some researchers do not think this is correct and want to show that the mean time is less than 31 minutes. Remember that the null hypothesis is the statement that the researchers are trying to reject. The null hypothesis is μ=31, so this should be what the researchers are trying to reject. In other words, that should be the claim of the fitness center. So the fitness center claim (the null hypothesis) is that μ=31, and the researchers are trying to show (the alternative hypothesis) that μ<31, less than 31, which is the second answer choice.

Which of the following results in a null hypothesis μ=31 and alternative hypothesis μ<31? A hospital claims that the mean wait time for emergency room patients is at most 31 minutes. A group of researchers think this is inaccurate and wants to show that the mean wait time is more than 31 minutes. A hospital claims that the mean wait time for emergency room patients is more than 31 minutes. A group of researchers think this is inaccurate and wants to show that the mean wait time is less than 31 minutes. A hospital claims that the mean wait time for emergency room patients is 31 minutes. A group of researchers think this is inaccurate and wants to show that the mean wait time is less than 31 minutes. A hospital claims that the mean wait time for emergency room patients is 31 minutes. A group of researchers think this is inaccurate and wants to show that the mean wait time is not 31 minutes.

A hospital claims that the mean wait time for emergency room patients is 31 minutes. A group of researchers think this is inaccurate and wants to show that the mean wait time is less than 31 minutes. The null hypothesis, μ=31, is the claim that the researchers are trying to reject. In this case, the researchers are trying to reject the hospital's claim. In words, they claim that the average wait time is at least 31. The alternative hypothesis is the claim that the researchers are trying to demonstrate, μ<31. So the researchers want to show that the average wait time is less than 31 minutes. Thus, the third choice is correct.

If a hypothesis test were to be performed, in which of the following contexts would a t-test be appropriate? A sample of 21 dogs is taken from the dogs within a county with approximately normally distributed weights. The sample mean is 22 pounds and the sample standard deviation is 3 pounds. The population standard deviation is 2.5 pounds. A convenience sample of 29 two-inch candles is taken from the candles offered at a candle store and the burning times were collected. The sample is used to estimate the burning times for all the candles in the store. The sample mean is 4 hours and the sample standard deviation is 1.25 hours. A sample mean of 20 minutes and a sample standard deviation of 4 minutes are obtained using a simple random sample of 11 individuals from the residents of a city, which have approximately normally distributed commute times. A sample mean of 90 and sample standard deviation of 7 are obtained using a simple random sample of size 25 from a normally distributed population with standard deviation 2.

A sample mean of 20 minutes and a sample standard deviation of 4 minutes are obtained using a simple random sample of 11 individuals from the residents of a city, which have approximately normally distributed commute times. In order to perform a hypothesis test of a single population mean μ using a student's t-distribution, the data should be a simple random sample that comes from a population that is approximately normally distributed with an unknown population standard deviation. The only option meeting these criteria is the context in the third answer choice.

If a hypothesis test were to be performed, in which of the following contexts would a t-test be appropriate? A simple random sample of 29 books is taken from a public library and the lengths (pages) were collected. The sample mean is 166 pages and the population standard deviation is 27 pages. A sample mean of 30 pounds and a sample standard deviation of 5 pounds are obtained using a simple random sample of 27 dogs from the dogs within a county, which have approximately normally distributed weights. A sample mean of 31,994 dollars and a standard deviation of 4,658 dollars are obtained using a simple random sample of 22 individuals from the residents of a town having approximately normally distributed incomes. The population standard deviation is 3,959.3 dollars. A convenience sample of 16 vehicles is taken from a large group of vehicles with approximately normally distributed fuel consumption rates. The sample mean is 20 miles per gallon and the sample standard deviation is 2 miles per gallon.

A sample mean of 30 pounds and a sample standard deviation of 5 pounds are obtained using a simple random sample of 27 dogs from the dogs within a county, which have approximately normally distributed weights. In order to perform a hypothesis test of a single population mean μ using a Student's t-distribution, the data should be a simple random sample that comes from a population that is approximately normally distributed with an unknown population standard deviation. The only option meeting these criteria is the context in the second answer choice.

If a hypothesis test were to be performed, in which of the following contexts would a t-test be appropriate? A simple random sample of 19 students is taken from a school of 1,188 students with approximately normally distributed heights. The sample mean is 61 inches and the sample standard deviation is 2 inches. A sample mean of 19 miles per gallon and a standard deviation of 2 miles per gallon are obtained using a convenient sample of 10 vehicles from a large group of vehicles, which have approximately normally distributed fuel consumption rates. A sample mean of 84 and sample standard deviation of 2 are obtained using a simple random sample of size 14 from a large population. A sample mean of 112 minutes and a standard deviation of 9 minutes are obtained using a simple random sample of 10 movies from the collection of movies at a rental store having approximately normally distributed time lengths. The population standard deviation is 7.6 minutes.

A simple random sample of 19 students is taken from a school of 1,188 students with approximately normally distributed heights. The sample mean is 61 inches and the sample standard deviation is 2 inches. In order to perform a hypothesis test of a single population mean μ using a Student's t-distribution, the data should be a simple random sample that comes from a population that is approximately normally distributed with an unknown population standard deviation. The only option meeting these criteria is the context in the first answer choice.

Which of the following results in a null hypothesis p=0.61 and alternative hypothesis p>0.61? A study says that at least 61% of students study less than 5 hours per week. A researcher thinks this is incorrect, and wants to show that fewer than 61% of students study less than 5 hours per week. A study says that more than 61% of students study less than 5 hours per week. A researcher thinks this is incorrect, and wants to show that at least 61% of students study less than 5 hours per week. A study says that at most 61% of students study less than 5 hours per week. A researcher thinks this is incorrect, and wants to show that more than 61% of students study less than 5 hours per week. A study says that less than 61% of students study less than 5 hours per week. A researcher thinks this is incorrect, and wants to show that more than 61% of students study less than 5 hours per week.

A study says that at most 61% of students study less than 5 hours per week. A researcher thinks this is incorrect, and wants to show that more than 61% of students study less than 5 hours per week. Remember that the alternative hypothesis is the claim that the researcher is trying to show. The null hypothesis, p=0.61 corresponds to the claim in the study, and the alternative hypothesis, p>0.61 corresponds to what the researcher is trying to show (to reject the null hypothesis). So the third choice is the correct answer.

A former residential complex was found to be a Superfund site polluted with several hazardous materials. Tamela is an investigative journalist who would like to show whether the proportion of the residents of the complex who eventually died from cancer is greater than the regional average of 14%. She randomly selects 54 residents who lived in the complex and finds that 12 of those residents eventually died from cancer. Are all of the conditions for this hypothesis test met, and if so, what are the null and alternative hypotheses for this hypothesis test? Although the proportion follows a binomial model with two independent outcomes and the data are selected at random, the number of successes and the number of failures are not both greater than or equal to 5. Although the proportion follows a binomial model with two independent outcomes and the number of successes and the number of failures are both greater than or equal to 5, the data are not selected at random. Although the data are selected at random and the number of successes and the number of failures are both greater than or equal to 5, the proportion does not follow a binomial model. All of the conditions to conduct the hypothesis test have been met. The null and alternative hypotheses are {H0:p=0.14Ha:p<0.14. All of the conditions to conduct the hypothesis test have been met. The null and alternative hypotheses are {H0:p=0.14Ha:p>0.14. All of the conditions to conduct the hypothesis test have been met. The null and alternative hypotheses are {H0:p=0.14Ha:p≠0.14.

All of the conditions to conduct the hypothesis test have been met. The null and alternative hypotheses are {H0:p=0.14Ha:p>0.14. First verify whether all of the conditions have been met. Let p be the population proportion for the residents of the complex who eventually died from cancer. Since Tamela is completing a survey for which there are two independent outcomes, the proportion follows a binomial model. The question states that Tamela randomly selected the residents who formally resided in the complex. The expected number of successes, np=7.56, and the expected number of failures, nq=n(1−p)=46.44, are both greater than or equal to 5. Since all of the conditions for this hypothesis test have been satisfied, determine the null and alternative hypotheses. The null hypothesis would claim that p, the population proportion, is equal to 0.14. Since Tamela is determining whether the proportion of the residents of the complex who eventually died from cancer is greater than the regional average of 14%, the alternative hypothesis would be p is greater than 0.14. The null and alternative hypotheses are shown below. {H0:p=0.14Ha:p>0.14

An airline company claims in a recent advertisement that more than 94% of passenger luggage that is lost is recovered and reunited with the customer within 1 day. Hunter is a graduate student studying statistics. For a research project, Hunter wants to find out whether there is convincing evidence in support of the airline company's claim. He randomly selects 315 passengers of the airline whose luggage was lost by the airline and found that 276 of those passengers were reunited with their luggage within 1 day. Are all of the conditions for this hypothesis test met, and if so, what are the null and alternative hypotheses for this hypothesis test? Although the proportion follows a binomial model with two independent outcomes and the data are selected at random, the number of successes and the number of failures are not both greater than or equal to 5. Although the proportion follows a binomial model with two independent outcomes and the number of successes and the number of failures are both greater than or equal to 5, the data are not selected at random. Although the data are selected at random and the number of successes and the number of failures are both greater than or equal to 5, the proportion does not follow a binomial model. All of the conditions to conduct the hypothesis test have been met. The null and alternative hypotheses are {H0:p=0.94Ha:p≠0.94. All of the conditions to conduct the hypothesis test have been met. The null and alternative hypotheses are {H0:p=0.94Ha:p>0.94. All of the conditions to conduct the hypothesis test have been met. The null and alternative hypotheses are {H0:p=0.94Ha:p<0.94.

All of the conditions to conduct the hypothesis test have been met. The null and alternative hypotheses are {H0:p=0.94Ha:p>0.94. First verify whether all of the conditions have been met. Let p be the population proportion for the airline passengers whose luggage was lost by the airline and were reunited with their luggage within 1 day. Since Hunter is completing a survey where there are two independent outcomes, the proportion follows a binomial model. The question states that Hunter randomly selected the airline passengers whose luggage was lost by the airline. The expected number of successes, np=296.1, and the expected number of failures, nq=n(1−p)=18.9, are both greater than or equal to 5. Since all of the conditions for this hypothesis test have been satisfied, determine the null and alternative hypotheses. Since Hunter is determining whether the proportion for reuniting passengers with their luggage within 1 day is greater than 94%, the null hypothesis is that p is equal to 0.94 and the alternative hypothesis is that p is greater than 0.94. The null and alternative hypotheses are shown below. {H0:p=0.94Ha:p>0.94

Which of the following results in a null hypothesis p=0.47 and alternative hypothesis p>0.47? An online article claims that less than 47% of internet users participate in social media. A group of researchers think this is incorrect, and they want to show that at least 47% of internet users participate in social media. An online article claims that at least 47% of internet users participate in social media. A group of researchers think this is incorrect, and they want to show that less than 47% of internet users participate in social media. An online article claims that more than 47% of internet users participate in social media. A group of researchers think this is incorrect, and they want to show that at most 47% of internet users participate in social media. An online article claims that at most 47% of internet users participate in social media. A group of researchers think this is incorrect, and they want to show that more than 47% of internet users participate in social media.

An online article claims that at most 47% of internet users participate in social media. A group of researchers think this is incorrect, and they want to show that more than 47% of internet users participate in social media. Remember that the null hypothesis is the statement that the researchers are trying to reject, or show is wrong. The null hypothesis is p=0.47. The alternative hypothesis, p>0.47, should be what the researchers are trying to show. So the fourth answer choice is correct.

Which of the following results in a null hypothesis p=0.44 and alternative hypothesis p<0.44? An online article is trying to show that less than 44% of internet users participate in social media, contrary to an established figure saying that 44% of internet users participate in social media. An online article is trying to show that 44% of internet users participate in social media, contrary to an established figure saying that more than 44% of internet users participate in social media. An online article is trying to show that more than 44% of internet users participate in social media, contrary to an established figure saying 44% of internet users participate in social media. An online article is trying to show that at least 44% of internet users participate in social media, contrary to an established figure saying that less than 44% of internet users participate in social media.

An online article is trying to show that less than 44% of internet users participate in social media, contrary to an established figure saying that 44% of internet users participate in social media. Consider each of the options. The null hypothesis is p=0.44; this must be the established fact that the article is trying to reject. The first answer choice is the correct choice, because the article is trying to show that less than 44% of users participate in social media (p<0.44), which matches the alternative hypothesis given in the question.

Suppose a random sample of adult women has a sample mean height of x¯=64.3 inches, with a sample standard deviation of s=2.4 inches. Since height distribution are generally symmetric and bell-shaped, we can apply the Empirical Rule. Between what two heights are approximately 99.7% of the data? Round your answers to the nearest tenth.

Approximately 99.7%​ of the women's heights are between 57.1 and 71.5 inches. The Empirical Rule states that approximately 99.7% of the data is within three standard deviations of the mean. In terms of the sample mean x¯ and sample standard deviation s, that is between x¯−3s and x¯+3s. The value three standard deviations below the mean is x¯−3s=64.3−3(2.4)=57.1 The value three standard deviations above the mean is x¯+3s=64.3+3(2.4)=71.5 So by the Empirical Rule, we can say that Approximately 99.7% of the women's heights are between 57.1 and 71.5 inches.

Determine the Type I error if the null hypothesis, H0, is Carmin will receive a 95% on her statistics exam and the alternative hypothesis, Ha, is Carmin will not receive a 95% on her statistics exam. Carmin concludes that she will not receive a 95% on her statistics exam when, in fact, she will not. Carmin concludes that she will not receive a 95% on her statistics exam when, in fact, she will. Carmin does not conclude that she will not receive a 95% on her statistics exam when, in fact, she will not. Carmin does not conclude that she will not receive a 95% on her statistics exam when, in fact, she will.

Carmin concludes that she will not receive a 95% on her statistics exam when, in fact, she will. A Type I error is the decision to reject the null hypothesis when it is true. In this case, the Type I error is when Carmin concludes that she will not receive a 95% on her statistics exam when, in fact, she will.

A doctor is interested in the average number of adult patients who are overdue for a booster shot. Based on a recent study about daily visits to the doctor's office, an average number of 5 adult patients are overdue for a booster shot when they visit the doctor's office. The doctor believes this average is different than the average quoted in the study. After calculating the test statistic in a hypothesis test, what does the doctor compare the test statistic to in order to make a decision? Significance level Null hypothesis Critical value P-value

Critical value A critical value approach allows us to create a rejection region, which determines if our test statistic is enough to support a claim.

A money management firm, which specializes in debt reduction, is interested in the average amount of debt for a married couple. In previous years, married couples had 4500 dollars, on average, of debt. Due to increased costs of college and mortgage interest rates, the money management firm suspects this average has increased in recent years. If the money management firm carries out an hypothesis test, what forms the rejection region? Critical value Null hypothesis Test statistic P-value

Critical value The critical value is based on the significance level and test type, which is then used to create a rejection region. The test statistic is compared to the critical value in order to make a conclusion. If the test statistic falls in the rejection region, the money management firm can reject the null hypothesis.

Suppose the null hypothesis, H0, is Darrell has $500 in his bank account. And the alternative hypothesis, Ha, is Darrell does not have $500 in his bank account.What is the Type II error in this scenario? Darrell concludes that he does not have $500 in his bank account when, in fact, he does not have $500 in his bank account. Darrell concludes that he does not have $500 in his bank account when, in fact, he does have $500 in his bank account. Darrell cannot conclude that he does not have $500 in his bank account when, in fact, he does not have $500 in his bank account. Darrell cannot conclude that he does not have $500 in his bank account when, in fact, he does have $500 in his bank account.

Darrell cannot conclude that he does not have $500 in his bank account when, in fact, he does not have $500 in his bank account. A Type II error is the decision not to reject the null hypothesis when, in fact, it is false. In this case, the Type II error is when Darrell cannot conclude that he does not have $500 in his bank account when, in fact, he does not have $500 in his bank account.

The following data was calculated during a study on budget control. Now interpret the results for this hypothesis test: The test statistic is calculated as t0=−2.28 The significance level is α=0.01. The p-value is between 0.02 and 0.05. Reject H0. There is NOT strong evidence to conclude that the average monthly amount spent on travel expenses at a small company is different than 65 hundred dollars. Reject H0. There is strong evidence to conclude that the average monthly amount spent on travel expenses at a small company is different than 65 hundred dollars. Fail to reject H0. There is NOT strong evidence to conclude that the average monthly amount spent on travel expenses at a small company is different than 65 hundred dollars. Fail to reject H0. There is strong evidence to conclude that the average monthly amount spent on travel expenses at a small company is different than 65 hundred dollars.

Fail to reject H0. There is NOT strong evidence to conclude that the average monthly amount spent on travel expenses at a small company is different than 65 hundred dollars. In making the decision to reject or not reject H0, recall the criterion for comparing the p−value to α. If p≤α, reject H0. Otherwise, fail to reject H0. Since the p-value is between 0.02 and 0.05, this is greater than the significance level α=0.01. Hence, the decision is to fail to reject the null hypothesis. In conclusion, at the 1% significance level, the sample data does NOT provide strong evidence to conclude that the average monthly amount spent on travel expenses at a small company is different than 65 hundred dollars.

A cell phone manufacturer claims that the average battery life of its newest flagship smartphone is exactly 20 hours. Javier believes the population mean battery life is less than 20 hours. He tests this claim by selecting a random sample of 33 phones of this model. Javier found that the sample mean battery life is 19.5 hours with a sample standard deviation of 1.9 hours. Should Javier reject or fail to reject the null hypothesis given the sample data below? H0:μ=20 versus Ha:μ<20 α=0.05 (significance level) t0=−1.51 0.05<p-value<0.10 Reject the null hypothesis because −1.51<0.05. Fail to reject the null hypothesis because −1.51<0.05. Reject the null hypothesis because 0.05<p-value<0.10 is greater than the significance level α=0.05. Fail to reject the null hypothesis because 0.05<p-value<0.10 is greater than the significance level α=0.05.

Fail to reject the null hypothesis because 0.05<p-value<0.10 is greater than the significance level α=0.05. In making the decision to reject or not reject H0, recall the criterion for comparing the p−value to α. If p≤α, reject H0. Since 0.05<p-value<0.10 is greater than the significance level α=0.05, Javier fails to reject the null hypothesis.

Doctor Hector claims that the population mean birth weight of babies born at the hospital is exactly 8lb. Sheila, a nurse who assists Hector, wants to test this claim, so she takes a random sample of 55 babies born at the hospital. She determines that the sample mean weight is 7.8lb with a sample standard deviation of 0.9lb. Should Shelia reject or fail to reject the null hypothesis given the sample data below? H0:μ=8 versus Ha:μ≠8 α=0.05 (significance level) t0=−1.65 0.10<p-value<0.20 Reject the null hypothesis because −1.65<0.05. Fail to reject the null hypothesis because −1.65<0.05. Reject the null hypothesis because 0.10<p-value<0.20 is greater than the significance level α=0.05. Fail to reject the null hypothesis because 0.10<p-value<0.20 is greater than the significance level α=0.05.

Fail to reject the null hypothesis because 0.10<p-value<0.20 is greater than the significance level α=0.05. In making the decision to reject or not reject H0, recall the criterion for comparing the p−value to α. If p≤α, reject H0. Since 0.10<p-value<0.20 is greater than the significance level α=0.05, Shelia fails to reject the null hypothesis.

The USPS ships boxes that weigh exactly 2lb, according to their packaging. Elena, a USPS worker, tests the claim by hypothesizing that the population mean weight of the boxes is not equal to 2lb. She randomly selects 15 boxes and found that the sample mean weight is 1.9968lb with a standard deviation of 0.0286lb. Should Elena reject or fail to reject the null hypothesis given the sample data below? H0:μ=2 versus Ha:μ≠2 α=0.05 (significance level) t0=−0.43 p-value>0.20 Reject the null hypothesis because −0.43<0.05. Fail to reject the null hypothesis because −0.43<0.05. Reject the null hypothesis because p-value>0.20 is greater than the significance level α=0.05. Fail to reject the null hypothesis because p-value>0.20 is greater than the significance level α=0.05.

Fail to reject the null hypothesis because p-value>0.20 is greater than the significance level α=0.05. In making the decision to reject or not reject H0, recall the criterion for comparing the p−value to α. If p≤α, reject H0. Since p-value>0.20 is greater than the significance level α=0.05, Elena fails to reject the null hypothesis.

A linguistics expert is studying the proportion of work time people in his industry study a new language. He predicts it is more than 10%. To test this prediction, he surveys 500 random linguists and determines that 120 of them spend more than 10% of their work time studying a new language. The following is the setup for this hypothesis test: H0:p=0.10 Ha:p>0.10 The p-value for this hypothesis test is 0.09. At the 2.5% significance level, should she reject or fail to reject the null hypothesis? Reject the null hypothesis because 0.10>0.025. Fail to reject the null hypothesis because 0.10>0.025. Reject the null hypothesis because the p-value =0.09 is less than the significance level α=0.025. Fail to reject the null hypothesis because the p-value =0.09 is more than the significance level α=0.025.

Fail to reject the null hypothesis because the p-value =0.09 is more than the significance level α=0.025. The p-value is 0.009. Compare the p-value to the level of significance α to make a conclusion for the hypothesis test. In this example, the p-value of 0.09 is more than the level of significance which is 0.025. Since the p-value is more than the level of significance, fail to reject the null hypothesis.

Maddie is a vlogger who creates makeup tutorial videos. The site that hosts her videos recently demonetized her channel because her ad click-through rate fell below 4.3%. To test this claim, she randomly selected 500 subscribers and found that 16 of them had clicked on the last ad that they saw. Use a TI-83, TI-83 plus, or TI-84 calculator to test whether the website's claim is true and less than 4.3% of the ads shown during her videos are clicked on, and then draw a conclusion in the context of the problem. Use α=0.05. Reject the null hypothesis. There is sufficient evidence to conclude that the proportion is less than 4.3%. Reject the null hypothesis. There is insufficient evidence to conclude that the proportion is less than 4.3%. Fail to reject the null hypothesis. There is sufficient evidence to conclude that the proportion is less than 4.3%. Fail to reject the null hypothesis. There is insufficient evidence to conclude that the proportion is less than 4.3%.

Fail to reject the null hypothesis. There is insufficient evidence to conclude that the proportion is less than 4.3%. Step 1: Press the STAT button, highlight TESTS, scroll to 5: 1-PropZTest..., and hit ENTER. Step 2: Enter the hypothesized proportion p0=0.043, number of successes x=16, sample size n=500, and type of test (left-tailed). Step 3: Click CALCULATE. The calculator returns a p-value, rounding to three decimal places, of 0.113. Step 4: Since the p-value is greater than α=0.05, fail to reject the null hypothesis. There is insufficient evidence to conclude that the proportion is less than 4.3%.

A brokerage firm is curious about the proportion of clients who have high-risk stocks in their stock portfolio. Let the proportion of clients who have high-risk stocks be p. If the brokerage firm wants to know if the proportion of clients who have high-risk stocks is less than 15%, what are the null and alternative hypotheses? H0: p=0.15; Ha: p>0.15 H0: p<0.15; Ha: p=0.15 H0: p=0.15; Ha: p<0.15 H0: μ=0.15; Ha: μ<0.15

H0: p=0.15; Ha: p<0.15 The alternative hypothesis is what is to be tested. In this case, we are testing if the proportion of clients who have high-risk stocks is less than 15%. Ha never has a symbol with an equal in it. So, this is the alternative hypothesis: Ha:p<0.15 The null hypothesis is the statement believed to be true unless it can be shown to be incorrect beyond a reasonable doubt. H0 always has a symbol with an equal in it. So, this is the null hypothesis: H0:p=0.15.

A politician claims that at least 68% of voters support a decrease in taxes. A group of researchers are trying to show that this is not the case. Identify the researchers' null hypothesis, H0, and the alternative hypothesis, Ha, in terms of the parameter p. H0: p=0.68; Ha: p>0.68 H0: p=0.68; Ha: p≠0.68 H0: p=0.68; Ha: p≤0.68 H0: p=0.68; Ha: p<0.68

H0: p=0.68; Ha: p<0.68 Let the parameter p be used to represent the proportion. The null hypothesis is always stated with the equality symbol: equal (=). Therefore, the null hypothesis H0 is p=0.68. The alternative hypothesis is contradictory to the null hypothesis, so Ha is p<0.68. Also, remember that the alternative hypothesis is the statement that the research or study is trying to show. In this case, they are trying to show that the politician is wrong, so the alternative hypothesis is the opposite of what the politician said, which is at least 0.68. At least means 0.68 or more. Therefore, Ha is p<0.68.

A city wants to show that the mean number of public transportation users per day is more than 5,575. Identify the null hypothesis, H0, and the alternative hypothesis, Ha, in terms of the parameter μ. H0: μ=5,575; Ha: μ<5,575 H0: μ=5,575; Ha: μ≥5,575 H0: μ=5,575; Ha: μ>5,575 H0: μ=5,575; Ha: μ≠5,575

H0: μ=5,575; Ha: μ>5,575 Let the parameter μ be used to represent the mean. Remember that the null hypothesis is the statement already believed to be true, and the alternative hypothesis is the statement trying to be shown. In this case, the city is trying to show that there is more, so μ>5,575, so this is the alternative hypothesis. The null hypothesis is : μ=5,575.

A medical journal published new medical research about diastolic blood pressure. Let the mean diastolic blood pressure reading be μ mmHg. If a doctor wants to know if the new medical research discussed in the journal results in diastolic blood pressure readings that are, on average, less than 75 mmHg, what are the null and alternative hypotheses? H0: μ=75; Ha: μ<75 H0: μ<75; Ha: μ=75 H0: μ=75; Ha: μ>75 H0: μ=75; Ha: μ≠75

H0: μ=75; Ha: μ<75 The alternative hypothesis is what is to be tested. In this case, we are testing if diastolic blood pressure is, on average, less than 75 mmHg. Ha never has a symbol with an equal in it. So, this is the alternative hypothesis: Ha:μ<75 mmHg. The null hypothesis is the statement believed to be true unless it can be shown to be incorrect beyond a reasonable doubt. H0 always has a symbol with an equal in it. So, this is the null hypothesis: H0:μ=75 mmHg.

A tech company wishes to study the number of production facilities that rely on new technologies, like 3D printers. Let the mean number of 3D printers used in a facility be μ. If the tech company wants to know if the number of 3D printers is, on average, less than 9 3D printers per facility, what are the null and alternative hypotheses? H0: μ=9; Ha: μ≠9 H0: μ=9; Ha: μ>9 H0: μ<9; Ha: μ=9 H0: μ=9; Ha: μ<9

H0: μ=9; Ha: μ<9 The alternative hypothesis is what is to be tested. In this case, we are testing if the number of 3D printers used in a facility is less than 9, on average. Ha never has a symbol with an equal in it. So, this is the alternative hypothesis: Ha:μ<9 printers. The null hypothesis is the statement believed to be true unless it can be shown to be incorrect beyond a reasonable doubt. H0 always has a symbol with an equal in it. So, this is the null hypothesis: H0:μ=9 printers.

A mattress store advertises that their beds last at least 5 years, on average. A consumer group thinks that they do not last that long and wants to set up a hypothesis test. If μ denotes the average time, in years, that the mattresses last, what are the null and alternative hypotheses in this situation? H0:μ=5; Ha:μ<5 H0:μ=5; Ha:μ>5 H0:μ=5; Ha:μ≠5 H0:μ=5; Ha:μ≤5 H0:μ=5; Ha:μ≥5

H0:μ=5; Ha:μ<5 The store claims that the mattresses last at least 5 years. This is the assumption the mattress store has been making. The consumer group is trying to reject this claim. The alternative hypothesis, Ha, is the opposite of the assumption. So Ha:μ<5 while the null hypothesis always contains the equality symbol: H0:μ=5.

An assistant lecturer, at the sociology department of a university, claims that the average amount of time adults spend on social media daily is less than 3.6 hours. If the assistant sociology lecturer wants to conduct a hypothesis test, should they use a left-, right-, or two-tailed hypothesis test to analyze whether the average amount of time adults spent on social media daily is less than 3.6 hours? Left-tailed test Right-tailed test Two-tailed test

Left-tailed test To identify the null and alternative hypothesis, first determine the purpose of the hypothesis test. In this scenario, the assistant sociology lecturer is analyzing whether the average amount of time adults spend on social media daily is less than 3.6 hours. Since the primary concern of this test is dealing with whether a population parameter, p, is less than a specified value, p0, then the assistant sociology lecturer should conduct a left-tailed test.

A finance blog, which also sells money management tools, believes the use of personal finance software is decreasing. The finance blog would like to test the claim that the average number of personal finance software sales per month is under 2400 sales. Using the computed test statistic of −1.33 and the critical value of −1.96, is there enough evidence for the finance blog to reject the null hypothesis? Yes. There is sufficient evidence for the finance blog to reject the null hypothesis because the test statistic falls in the rejection region. No. There is NOT sufficient evidence for the finance blog to reject the null hypothesis because the test statistic does not fall in the rejection region. Yes. There is sufficient evidence for the finance blog to reject the null hypothesis because the critical value falls in the rejection region. No. There is NOT sufficient evidence for the finance blog to reject the null hypothesis because the critical value does not fall in the rejection region.

No. There is NOT sufficient evidence for the finance blog to reject the null hypothesis because the test statistic does not fall in the rejection region. This is a left-tailed test because the alternative hypothesis formed by the claim has a less than sign (under 2400 sales). The hypotheses are Ho:μ=2400 and Ha:μ<2400 . Since this is a left-tailed test, the rejection region is the shaded region on the left separated by one critical value, in this case −1.96. Since the test statistic is −1.33, it falls to the right of 1.96, which is not in the rejection region. So, there is NOT enough evidence to reject the null hypothesis.

Jonathan, a manager at an advertising firm, believes the proportion of businesses relying on print media for advertising is changing. He would like to test the claim that the proportion of businesses relying on print media for advertising is different than 18%. Using the computed test statistic of −1.00 and the critical values of −1.28 and 1.28, is there enough evidence for the manager to reject the null hypothesis? Yes. There is sufficient evidence for the manager to reject the null hypothesis because the test statistic falls in the rejection region. No. There is NOT sufficient evidence for the manager to reject the null hypothesis because the test statistic does not fall in the rejection region. Yes. There is sufficient evidence for the manager to reject the null hypothesis because the critical value falls in the rejection region. No. There is NOT sufficient evidence for the manager to reject the null hypothesis because the critical value does not fall in the rejection region.

No. There is NOT sufficient evidence for the manager to reject the null hypothesis because the test statistic does not fall in the rejection region. This is a two-tailed test because the alternative hypothesis formed by the claim has a not equals sign (different than 18%). The hypotheses are Ho:p=0.18 and Ha:p≠0.18 . Since this is a two-tailed test, the rejection region are the shaded regions on the left and right separated by two critical values, in this case −1.28 and 1.28. Since the test statistic is −1.00, it falls in the middle of −1.28 and 1.28, which is not in the rejection region. So, there is NOT enough evidence to reject the null hypothesis.

Ken, a market researcher, believes fees to use e-commerce content management systems are decreasing. He would like to test the claim that the average fee to use a e-commerce content management system is under 5 dollars. Using the computed test statistic of −1.77 and the critical value of −1.96, is there enough evidence for the market researcher to reject the null hypothesis? Use the graph below to interpret the critical value. First select the appropriate test type (left-, right-, or two- tailed), then plot the points along the x-axis (for the critical value and test statistic), and use them to choose the appropriate interpretation.

No. There is NOT sufficient evidence for the market researcher to reject the null hypothesis because the test statistic does not fall in the rejection region. This is a left-tailed test because the alternative hypothesis formed by the claim has a less than sign (under 5 dollars). The hypotheses are Ho:μ=5 and Ha:μ<5 . Since this is a left-tailed test, the rejection region is the shaded region on the left separated by one critical value, in this case −1.96. Since the test statistic is −1.77, it falls to the right of 1.96, which is not in the rejection region. So, there is NOT enough evidence to reject the null hypothesis.

A health insurance provider believes errors in hospital billing codes are increasing. The provider would like to test the proportion of incorrect billing codes from hospitals in their network. If the health insurance provider carries out a hypothesis test and writes out their alternative hypothesis, what is the complement to the alternative hypothesis? Critical value Null hypothesis Significance level Test statistic

Null hypothesis The null hypothesis is an equality statement. Writing the null and alternative hypotheses are a first step in carrying out a hypothesis test.

The following data was calculated during a study on mobile apps. Now interpret the results for this hypothesis test: The test statistic is calculated as t0=−3.56 The significance level is α=0.10. The p-value is less than 0.01. Fail to reject H0. There is NOT strong evidence to conclude that the average number of mobile apps on a person's cell phone is different than 9 mobile apps. Fail to reject H0. There is strong evidence to conclude that the average number of mobile apps on a person's cell phone is different than 9 mobile apps. Reject H0. There is NOT strong evidence to conclude that the average number of mobile apps on a person's cell phone is different than 9 mobile apps. Reject H0. There is strong evidence to conclude that the average number of mobile apps on a person's cell phone is different than 9 mobile apps.

Reject H0. There is strong evidence to conclude that the average number of mobile apps on a person's cell phone is different than 9 mobile apps. In making the decision to reject or not reject H0, recall the criterion for comparing the p−value to α. If p≤α, reject H0. Otherwise, fail to reject H0. Since the p-value is less than 0.01, this is less than the significance level α=0.10. Hence, the decision is to reject the null hypothesis. In conclusion, at the 10% significance level, the sample data does provide strong evidence to conclude that the average number of mobile apps on a person's cell phone is different than 9 mobile apps.

A dietitian is interested in the average amount of time teenagers exercise per week. Based on previous studies, teenagers exercise an average of 6.5 hours per week. The dietitian suspects this average has increased. During the process of hypothesis testing, the dietitian calculates a test statistic of 3.30. If this test statistic is higher than the critical value, what can the dietitian do next? Fail to reject the null hypothesis Reject the null hypothesis Fail to reject the test statistic Reject the test statistic

Reject the null hypothesis Since the test statistic is higher than the critical value, there is enough evidence to reject the null hypothesis. The dietitian has sufficient evidence to conclude teenagers exercise more than an average of 6.5 hours per week.

Rebecca is a real estate agent who would like to find evidence supporting the claim that the population mean market value of houses in the neighborhood where she works is greater than $250,000. To test the claim, she randomly selects 35 houses in the neighborhood and finds that the sample mean market value is $259,860 with a sample standard deviation of $24,922. Should Rebecca reject or fail to reject the null hypothesis given the sample data below? H0:μ=250,000 versus Ha:μ>250,000 α=0.05 (significance level) t≈2.34 , which has 34 degrees of freedom 0.01<p-value<0.025 Reject the null hypothesis because 2.34<0.05. Fail to reject the null hypothesis because 2.34>0.05. Fail to reject the null hypothesis because 0.01<p-value<0.025is less than the significance level α=0.05. Reject the null hypothesis because 0.01<p-value<0.025 is less than the significance level α=0.05.

Reject the null hypothesis because 0.01<p-value<0.025 is less than the significance level α=0.05. In making the decision to reject or not reject H0, recall the criterion for comparing the p−value to α. If p≤α, reject H0. Since 0.01<p-value<0.025 is less than the significance level α=0.05, Rebecca should reject the null hypothesis.

Mrs. Obama implemented a program to get "children playing outside" and predicts it increased the proportion of time children spent outside on the weekends to more than 15%. To test this prediction, she surveyed 300 random family participants and found that children of 205 families spent more than 15% of their weekend time outside. The following is the setup for this hypothesis test: H0:p=0.15 Ha:p>0.15 The p-value for this hypothesis test is 0.002. At the 2.5% significance level, should she reject or fail to reject the null hypothesis? Reject the null hypothesis because 0.15>0.02. Fail to reject the null hypothesis because 0.15>0.02. Reject the null hypothesis because the p-value =0.002 is less than the significance level α=0.025. Fail to reject the null hypothesis because the p-value =0.002 is less than the significance level α=0.025.

Reject the null hypothesis because the p-value =0.002 is less than the significance level α=0.025. The p-value is 0.002. Compare the p-value to the level of significance α to make a conclusion for the hypothesis test. In this example, the p-value of 0.002 is less than the level of significance which is 0.025. Since the p-value is less than the level of significance, reject the null hypothesis.

Joanna, a machine learning engineer is studying what programming languages are the most effective to help systems learn from data. Based on previous research, more than 50% of machine learning involves the use of Python, a programming language. She suspects this percent has increased in recent years. In the process of hypothesis testing, what set value will allow Joanna to make a conclusion about whether her sample proportion is "different enough" than the claimed population proportion? Null hypothesis P-value Test statistic Significance level

Significance level Before calculating a test statistic, the significance level is usually determined. Deciding on a significance level allows us to make a conclusion about whether our sample parameter is "different enough" than the claimed population parameter.

After running a mile a day over a period of two weeks, the average amount of weight loss is 2.5 pounds. A dietitian, who publishes health articles in a newspaper, states their new diet program helps with additional weight loss when combining their special diet with running a mile a day over a period of two weeks. Interested in studying the dietitian's article further, you ask friends who have tried the dietitian's new program and you determine their weight loss to be 3.0 pounds in a two week period, on average. As you set up a hypothesis test to determine if the dietitian's article is correct, what is the dietitian's claim? Adults should run every day to lose weight. The average amount of weight loss is less than 2.5 pounds. The average amount of weight loss is greater than 3.0 pounds. The average amount of weight loss is greater than 2.5 pounds.

The average amount of weight loss is greater than 2.5 pounds. Since a claim is a statement about the population parameter, you are making a statement about the population average amount of weight loss. Specifically, the dietitian is claiming that the average amount of weight loss is greater than 2.5 pounds. By testing this claim, you can determine whether there is enough evidence to suggest the dietitian's article is correct or not.

A social media platform states that a social media post from a marketing agency has 7 hashtags, on average. A digital marketing specialist studying social media advertising believes the average number of hashtags used in a post from a marketing agency is different than the number stated by the social media platform. After completing a study, the digital marketing specialist found that the average number of hashtags used by a marketing agency in a social media post is 7.9 hashtags on average. As the digital marketing specialist sets up a hypothesis test to determine if their belief is correct, what is their claim? The average number of hashtags used in a social media post from a marketing agency is different than 7 hashtags. The average number of hashtags used in a social media post from a marketing agency is different than 7.9 hashtags. Marketing agencies use too many hashtags in a social media post. The average number of hashtags used in a social media post from a marketing agency is 7 hashtags.

The average number of hashtags used in a social media post from a marketing agency is different than 7 hashtags. Since a claim is a statement about the population parameter, the digital marketing specialist is making a statement about the population average number of hashtags used in a social media post from a marketing agency. Specifically, the digital marketing specialist is claiming that the average number of hashtags used in a social media post from a marketing agency is different than 7 hashtags. By testing this claim, the digital marketing specialist can determine whether there is enough evidence to suggest the statement made by the social media platform is correct or not.

Suppose the null hypothesis, H0, is a surgical procedure is successful at least 80% of the time. And the alternative hypothesis, Ha, states the doctors' claim, which is a surgical procedure is successful less than 80% of the time.Which of the following gives β, the probability of a Type II error? The doctors conclude that the surgical procedure is successful less than 80% of the time when, in fact, the surgical procedure is successful at least 80% of the time. The doctors conclude that the surgical procedure is successful less than 80% of the time when, in fact, the surgical procedure is successful less than 80% of the time. The doctors do not conclude that the surgical procedure is successful less than 80% of the time when, in fact, the surgical procedure is successful less than 80% of the time. The doctors do not conclude that the surgical procedure is successful less than 80% of the time when, in fact, the surgical procedure is successful at least 80% of the time.

The doctors do not conclude that the surgical procedure is successful less than 80% of the time when, in fact, the surgical procedure is successful less than 80% of the time. A Type II error is the decision not to reject the null hypothesis when, in fact, it is false. In this case, the Type II error is when the doctors do not conclude that the surgical procedure is successful less than 80% of the time when, in fact, the surgical procedure is successful less than 80% of the time.

An article reports that young adults from a certain city spend, on average, more than $150 per week on food. To test this claim, Eric, a college student who resides in the city, selects a representative sample of 55 young adults. The following is the data from this study: The alternative hypothesis Ha:μ>150. The sample mean amount spent by the 55 young adults is $156.45. The sample standard deviation is $26.62. The test statistic is calculated as 1.80. Using the information above and the portion of the t− table below, choose the correct p− value and interpretation for this hypothesis test. Values for right-tail areas under the t-distribution curve The probability of observing a value of t0=1.80 or less if the null hypothesis is true is between 0.025 and 0.05. The probability of observing a value of t0=1.80 or more if the null hypothesis is true is between 0.005 and 0.01. The probability of observing a value of t0=1.80 or more if the null hypothesis is true is between 0.025 and 0.05. The probability of observing a value of t0=1.80 or more if the null hypothesis is true is between 0.05 and 0.10.

The probability of observing a value of t0=1.80 or more if the null hypothesis is true is between 0.025 and 0.05. Notice that the test statistic has 55−1=54 degrees of freedom and that this is a right-tailed test because the alternative hypothesis is Ha:μ>150. Find the p-value for right-tailed test of a t-distribution with 54 degrees of freedom, where t=1.80. Using the table of areas in the right tail for the t-distribution, in the row for 54 degrees of freedom, the t-test statistic, 1.80, is greater than 1.674 and less than 2.005. Since this is a right-tailed test, the p-value is less than 0.05 and greater than 0.025. Thus, 0.025<p-value<0.05.

Preston, an official for a state's department of safety, claims that passenger vehicles traveling on a certain expressway, on average, travel faster than the posted speed limit of 65 miles per hour. Preston collects a random sample of 59 passenger vehicles traveling on the expressway. The following is the data from this study: The alternative hypothesis Ha:μ>65. The sample mean speed of the 59 passenger vehicles is 67.1 miles per hour. The sample standard deviation is 7.5 miles per hour. The test statistic is calculated as 2.15. Using the information above and the portion of the t− table below, choose the correct p− value and interpretation for this hypothesis test.

The probability of observing a value of t0=2.15 or more if the null hypothesis is true is between 0.01 and 0.025. Notice that the test statistic has 59−1=58 degrees of freedom and that this is a right-tailed test because the alternative hypothesis is Ha:μ>65. Find the p-value for right-tailed test of a t-distribution with 58 degrees of freedom, where t=2.15. Using the table of areas in the right tail for the t-distribution, in the row for 58 degrees of freedom, the t-test statistic, 2.15, is greater than 2.002 and less than 2.392. Since this is a right-tailed test, the p-value is less than 0.025 and greater than 0.01. Thus, 0.01<p-value<0.025.

Kurtis is a statistician who claims that the average salary of an employee in the city of Yarmouth is no more than $55,000 per year. Gina, his colleague, believes this to be incorrect, so she randomly selects 61 employees who work in Yarmouth and records their annual salaries. The following is the data from this study: The alternative hypothesis Ha:μ>55,000. The sample mean salary of the 61 employees is $56,500 per year. The sample standard deviation is $3,750. The test statistic is calculated as 3.12. Using the information above and the portion of the t− table below, choose the correct p− value and interpretation for this hypothesis test. The probability of observing a value of t0=3.12 or more if the null hypothesis is true is between 0.025 and 0.05. The probability of observing a value of t0=3.12 or less if the null hypothesis is true is less than 0.005. The probability of observing a value of t0=3.12 or more if the null hypothesis is true is greater than 0.10. The probability of observing a value of t0=3.12 or more if the null hypothesis is true is less than 0.005.

The probability of observing a value of t0=3.12 or more if the null hypothesis is true is less than 0.005. Notice that the test statistic has 61−1=60 degrees of freedom and that this is a right-tailed test because the alternative hypothesis is Ha:μ>55,000. Find the p-value for right-tailed test of a t-distribution with 60 degrees of freedom, where t=3.12. Using the table of areas in the right tail for the t-distribution, in the row for 60 degrees of freedom, the t-test statistic, 3.12, is greater than 2.660. Since this is a right-tailed test, the p-value is less than 0.005.

The following data was calculated during a study on budget control. Now find and interpret the p− value for this hypothesis test.: A budget controller would like to test the claim that the average monthly amount spent on travel expenses at a small company is different than 65 hundred dollars. Sample size =18 months Sample mean =61.5 hundred dollars Sample standard deviation =6.5 hundred dollars The test statistic is calculated as t0=−2.28 Use the table below to help estimate the p− value. Based on the p-value, choose the correct interpretation. The probability of observing a value of t0=±2.28 or more extreme if the null hypothesis is true is between 2% and 5%. The probability of observing a value of t0=−2.28 or less if the null hypothesis is true is between 2% and 5%. The probability of observing a value of t0=±2.28 or more extreme if the null hypothesis is true is between 95% and 98%. The probability of observing a value of t0=−2.28 or greater if the null hypothesis is true is between 95% and 98%.

The probability of observing a value of t0=±2.28 or more extreme if the null hypothesis is true is between 2% and 5%. The budget controller would like to test the claim that the average monthly amount spent on travel expenses at a small company is different than 65 hundred dollars. "Different than" means this is a two-tailed test. Since this is a two-tailed test, the p−value will have a probability that is to the left of t0=−2.28 and to the right of t0=2.28. Using the table of areas in the right tail for the t-distribution, in the row for n−1=17 degrees of freedom, the absolute value of the t-test statistic, 2.28, is between 2.110 and 2.567 in the row for 17 degrees of freedom. So, the area in one tail of the distribution is between 0.01 and 0.025. Therefore, the p−value for this two-tailed test is twice the area in one tail, which will be between 0.02 and 0.05. This p−value means the probability of observing a value of t0=±2.28 or more extreme if the null hypothesis is true is between 2% and 5%.

Dr. Hector claims that the mean birth weight of babies born at the hospital where he works is 8 pounds. Sheila, a nurse who assists Dr. Hector, wants to test this claim, so she takes a random sample of 55 babies born at the hospital. The following is the data from this study: The alternative hypothesis Ha:μ≠8. The sample mean weight of the 55 babies is 7.8 pounds. The sample standard deviation is 0.9 pounds. The test statistic is calculated as −1.65. Using the information above and the portion of the t− table below, choose the correct p− value and interpretation for this hypothesis test. The probability of observing a value of t0=1.65 or more if the null hypothesis is true is between 0.05 and 0.20. The probability of observing a value of t0=−1.65 or less or observing a value of t0=1.65 or more if the null hypothesis is true is between 0.10 and 0.20. The probability of observing a value of t0=−1.65 or less or observing a value of t0=1.65 or more if the null hypothesis is true is between 0.01 and 0.02. The probability of observing a value of t0=−1.65 or less if the null hypothesis is true is between 0.005 and 0.01.

The probability of observing a value of t0=−1.65 or less or observing a value of t0=1.65 or more if the null hypothesis is true is between 0.10 and 0.20. Notice that the test statistic has 55−1=54 degrees of freedom and that this is a two-tailed test because the alternative hypothesis is Ha:μ≠8. Find the p-value for a two-tailed test of a t-distribution with 54 degrees of freedom, where t≈−1.65. Using the table of areas in the right tail for the t-distribution, in the row for 54 degrees of freedom, the absolute value of the t-test statistic, 1.65, is greater than 1.297 and less than 1.674. So, since this is a two-tailed test, the p−value is less than 2∗0.10 and greater than 2∗0.05. Thus, 0.10<p-value<0.20.

Suppose the null hypothesis, H0, is: no more than 70% of customers at a sporting goods store do not shop at any other sporting goods stores.And the alternative hypothesis, Ha, is: the sporting goods store claims more than 70% of its customers do not shop at any other sporting goods stores.What is β, the probability of a Type II error in this scenario? The probability that the sporting goods store can conclude that more than 70% of its customers do not shop at any other sporting goods stores when, in fact, more than 70% of its customers do not shop at any other sporting goods stores. The probability that the sporting goods store can conclude that more than 70% of its customers do not shop at any other sporting goods stores when, in fact, no more than 70% of its customers do not shop at any other sporting goods stores. The probability that the sporting goods store cannot conclude that more than 70% of its customers do not shop at any other sporting goods stores when, in fact, no more than 70% of its customers do not shop at any other sporting goods stores. The probability that the sporting goods store cannot conclude that more than 70% of its customers do not shop at any other sporting goods stores when, in fact, more than 70% of its customers do not shop at any other sporting goods stores.

The probability that the sporting goods store cannot conclude that more than 70% of its customers do not shop at any other sporting goods stores when, in fact, more than 70% of its customers do not shop at any other sporting goods stores. A Type II error is the decision not to reject the null hypothesis when, in fact, it is false. In this case, the Type II error is when the store cannot conclude that more than 70% of its customers do not shop at any other sporting goods stores when, in fact, more than 70% do not shop at any other sporting goods stores.

Using the information obtained from the previous study about banking fees, the following data was obtained for the hypothesis test: H0: μ=9, Ha: μ>9 Assume the significance level is α=0.10. The z− test statistic is calculated as 1.77. What is the p−value and conclusion for this hypothesis test? Move the blue dot to choose the appropriate test (left-, right, or two-tailed). Use the graph below to move the slider to the test statistic in order to find the p-value. Then, make a conclusion based on those results. The p− value is 0.0384. Fail to reject the null hypothesis H0. The banker does NOT have sufficient evidence that the average percent of banking fees is greater than 9 dollars. The p− value is 0.9616. Fail to reject the null hypothesis H0. The banker does NOT have sufficient evidence that the average percent of banking fees is greater than 9 dollars. The p− value is 0.0384. Reject the null hypothesis H0. The banker has sufficient evidence that the average percent of banking fees is greater than 9 dollars. The p− value is 0.9616. Reject the null hypothesis H0. The banker has sufficient evidence that the average percent of banking fees is greater than 9 dollars

The p− value is 0.0384. Reject the null hypothesis H0. The banker has sufficient evidence that the average percent of banking fees is greater than 9 dollars. The banker is testing the claim that the average percent of banking fees is greater than 9 dollars, which is our alternative hypothesis, Ha. The null hypothesis H0 would be the equality statement that the average percent of banking fees is 9 dollars.H0: μ=9Ha: μ>9Since the alternative hypothesis Ha has a greater than symbol, this is a right-tailed test. Moving the slider to the test statistic of z=1.77 results in a p− value of 0.0384. Since the p−value is less than α=0.10 , reject the null hypothesis H0. There is enough evidence at the α=0.10 significance level to support the alternative hypothesis. Therefore, based on a significance level at 0.10, the banker has sufficient evidence that the average percent of banking fees is greater than 9 dollars, which is our alternative hypothesis.

Using the information obtained from the previous study about programming languages, the following data was obtained for the hypothesis test: H0: μ=52, Ha: μ≠52 Assume the significance level is α=0.05. The z− test statistic is calculated as 2.07. What is the p−value and conclusion for this hypothesis test? Move the blue dot to choose the appropriate test (left-, right, or two-tailed). Use the graph below to move the slider to the test statistic in order to find the p-value. Then, make a conclusion based on those results. The p− value is 0.0384. Fail to reject the null hypothesis H0. The computer scientist does NOT have sufficient evidence that the average percent of companies that use Python is different than 52 percent. The p− value is 0.0384. Reject the null hypothesis H0. The computer scientist has sufficient evidence that the average percent of companies that use Python is different than 52 percent. The p− value is 0.9616. Fail to reject the null hypothesis H0. The computer scientist does NOT have sufficient evidence that the average percent of companies that use Python is different than 52 percent. The p− value is 0.9616. Reject the null hypothesis H0. The computer scientist has sufficient evidence that the average percent of companies that use Python is different than 52 percent.

The p− value is 0.0384. Reject the null hypothesis H0. The computer scientist has sufficient evidence that the average percent of companies that use Python is different than 52 percent. The computer scientist is testing the claim that the average percent of companies that use Python is different than 52 percent, which is our alternative hypothesis, Ha. The null hypothesis H0 would be the equality statement that the average percent of companies that use Python is 52 percent.H0: μ=52Ha: μ≠52Since the alternative hypothesis Ha has a not equals symbol, this is a two-tailed test. Moving the slider to the test statistic of z=2.07 results in a p− value of 0.0384. Since the p−value is less than α=0.05 , reject the null hypothesis H0. There is enough evidence at the α=0.05 significance level to support the alternative hypothesis. Therefore, based on a significance level at 0.05, the computer scientist has sufficient evidence that the average percent of companies that use Python is different than 52 percent.

Using the information obtained from the previous study about email marketing, the following data was obtained for the hypothesis test: H0: p=0.52, Ha: p>0.52 Assume the significance level is α=0.01. The z− test statistic is calculated as 0.97. What is the p−value and conclusion for this hypothesis test? Move the blue dot to choose the appropriate test (left-, right, or two-tailed). Use the graph below to move the slider to the test statistic in order to find the p-value. Then, make a conclusion based on those results. The p− value is 0.8340. Fail to reject the null hypothesis H0. The business owner does NOT have sufficient evidence that the percent of businesses that use email marketing to help grow sales is more than 52%. The p− value is 0.8340. Reject the null hypothesis H0. The business owner has sufficient evidence that the percent of businesses that use email marketing to help grow sales is more than 52%. The p− value is 0.1660. Fail to reject the null hypothesis H0. The business owner does NOT have sufficient evidence that the percent of businesses that use email marketing to help grow sales is more than 52%. The p− value is 0.1660. Reject the null hypothesis H0. The business owner has sufficient evidence that the percent of businesses that use email marketing to help grow sales is more than 52%.

The p− value is 0.1660. Fail to reject the null hypothesis H0. The business owner does NOT have sufficient evidence that the percent of businesses that use email marketing to help grow sales is more than 52%. The business owner is testing the claim that the percent of businesses that use email marketing to help grow sales is more than 52% which is our alternative hypothesis, Ha. The null hypothesis H0 would be the equality statement that the percent of businesses that use email marketing to help grow sales is 52%.H0: p=0.52 Ha: p>0.52Since the alternative hypothesis Ha has a greater than symbol, this is a right-tailed test. Moving the slider to the test statistic of z=0.97 results in a p− value of 0.1660. Since the p−value is greater than α=0.01 , fail to reject the null hypothesis H0. There is NOT enough evidence at the α=0.01 significance level to support the alternative hypothesis. Therefore, based on a significance level at 0.01, the business owner does NOT have sufficient evidence that the percent of businesses that use email marketing to help grow sales is more than 52%.

Using the information obtained from the previous study about social media posts, the following data was obtained for the hypothesis test: H0: μ=3, Ha: μ≠3 Assume the significance level is α=0.01. The z− test statistic is calculated as −1.17. What is the p−value and conclusion for this hypothesis test? Move the blue dot to choose the appropriate test (left-, right, or two-tailed). Use the graph below to move the slider to the test statistic in order to find the p-value. Then, make a conclusion based on those results The p− value is 0.7580. Reject the null hypothesis H0. The social media content associate has sufficient evidence that the average number of social media posts per day that helps to build an audience is different than 3 daily posts. The p− value is 0.2420. Reject the null hypothesis H0. The social media content associate has sufficient evidence that the average number of social media posts per day that helps to build an audience is different than 3 daily posts. The p− value is 0.7580. Fail to reject the null hypothesis H0. The social media content associate does NOT have sufficient evidence that the average number of social media posts per day that helps to build an audience is different than 3 daily posts. The p− value is 0.2420. Fail to reject the null hypothesis H0. The social media content associate does NOT have sufficient evidence that the average number of social media posts per day that helps to build an audience is different than 3 daily posts.

The p− value is 0.2420. Fail to reject the null hypothesis H0. The social media content associate does NOT have sufficient evidence that the average number of social media posts per day that helps to build an audience is different than 3 daily posts. The social media content associate is testing the claim that the average number of social media posts per day that helps to build an audience is different than 3 daily posts, which is our alternative hypothesis, Ha. The null hypothesis H0 would be the equality statement that the average number of social media posts per day that helps to build an audience is 3 daily posts.H0: μ=3Ha: μ≠3 Since the alternative hypothesis Ha has a not equals symbol, this is a two-tailed test. Moving the slider to the test statistic of z=−1.17 results in a p− value of 0.2420. Since the p−value is greater than α=0.01 , fail to reject the null hypothesis H0. There is not enough evidence at the α=0.01 significance level to support the alternative hypothesis. Therefore, based on a significance level at 0.01, the social media content associate does NOT have sufficient evidence that the average number of social media posts per day that helps to build an audience is different than 3 daily posts.

Determine the Type I error if the null hypothesis, H0, is: 65% of college students will graduate with debt.And, the alternative hypothesis, Ha, is: that researchers claim more than 65% of college students will graduate with debt. The researchers conclude that more than 65% of college students will graduate with debt when, in fact, 65% will graduate with debt. The researchers cannot conclude that more than 65% of college students will graduate with debt when, in fact, 65% will graduate with debt. The researchers conclude that 65% of college students will graduate with debt when, in fact, more than 65% will graduate with debt. The researchers cannot conclude that 65% of college students will graduate with debt when, in fact, more than 65% will graduate with debt.

The researchers conclude that more than 65% of college students will graduate with debt when, in fact, 65% will graduate with debt. A Type I error is the decision to reject the null hypothesis when it is true. In this case, the Type I error is when the researchers conclude that more than 65% of college students will graduate with debt when, in fact, 65% will graduate with debt.

A consumer protection company is testing a seat belt to see how much force it can hold. The null hypothesis, H0, is that the seat belt can hold at least 5000 pounds of force. The alternative hypothesis, Ha, is that the seat belt can hold less than 5000 pounds of force. What is a Type II error in this scenario? The researchers states that there is insufficient evidence to conclude that the seat belt holds less than 5000 pounds of force, but the seat belt actually holds less than 5000 pounds. The researchers states that there is insufficient evidence to conclude that the seat belt holds more than 5000 pounds of force, but the seat belt actually holds more than 5000 pounds. The researchers states that there is sufficient evidence to conclude that the seat belt holds less than 5000 pounds of force, and the seat belt actually holds less than 5000 pounds. The researchers states that there is sufficient evidence to conclude that the seat belt holds more than 5000 pounds of force, and the seat belt actually holds more than 5000 pounds.

The researchers states that there is insufficient evidence to conclude that the seat belt holds less than 5000 pounds of force, but the seat belt actually holds less than 5000 pounds. Remember that a Type I error is rejecting the null hypothesis when the null hypothesis is true. A Type II error is not rejecting the null hypothesis when it is false. We are asked for the Type II error in this scenario. Failing to reject the null hypothesis means failing to reject the statement that the seat belt can hold at least 5000 pounds. Therefore, a Type II error is: The researchers states that there is insufficient evidence to conclude that the seat belt holds less than 5000 pounds of force, but the seat belt actually holds less than 5000 pounds.

A consumer protection company is testing a towel rack to see how much force it can hold. The null hypothesis, H0, is that the rack can hold 100 pounds of force. The alternative hypothesis, Ha, is that the rack can hold less than 100 pounds of force. What is a Type I error in this scenario? The researchers states that there is sufficient evidence to conclude that the rack holds greater than 100 pounds of force, but the rack actually holds 100 pounds. The researchers states that there is sufficient evidence to conclude that the rack holds less than 100 pounds of force, but the rack actually holds 100 pounds. The researchers states that there is insufficient evidence to conclude that the rack holds less than 100 pounds of force, but the rack actually holds 100 pounds. The researchers states that there is insufficient evidence to conclude that the rack holds more than 100 pounds of force, but the rack actually holds 100 pounds.

The researchers states that there is sufficient evidence to conclude that the rack holds less than 100 pounds of force, but the rack actually holds 100 pounds. Remember that a Type I error is rejecting the null hypothesis when the null hypothesis is true. A Type II error is not rejecting the null hypothesis when it is false. We are asked for the Type I error in this scenario. Rejecting the null hypothesis means rejecting the statement that the rack can hold 100 pounds. Therefore, a Type I error is: The researchers states that there is sufficient evidence to conclude that the rack holds less than 100 pounds of force, but the rack actually holds 100 pounds.

Using the information above, choose the correct conclusion for this hypothesis test. There is NOT sufficient evidence to conclude that the population mean birth weight of babies born at the hospital is not equal to 7.9lb. There is sufficient evidence to conclude that the true population mean birth weight of babies born at the hospital is not equal to 7.9lb. There is sufficient evidence to conclude that the true population mean birth weight of babies born at the hospital is not equal to 8lb. There is NOT sufficient evidence to conclude that the population meanbirth weight of babies born at the hospital is not equal to 8lb.

There is NOT sufficient evidence to conclude that the population mean birth weight of babies born at the hospital is not equal to 8lb. Compare the p-value to α=0.05. Since the p-value is greater than α, Shelia fails to reject the null hypothesis H0. Therefore, there is NOT enough evidence at the α=0.05 level of significance to support the claim that the population mean birth weight of babies born at the hospital is not equal to 8lb.

Using the information above, choose the correct conclusion for this hypothesis test. There is NOT sufficient evidence to conclude that the population mean weight of the boxes is not equal to 1.9lb. There is sufficient evidence to conclude that the true population mean weight of the boxes is not equal to 1.9lb. There is sufficient evidence to conclude that the true population mean weight of the boxes is not equal to 2lb. There is NOT sufficient evidence to conclude that the population mean weight of the boxes is not equal to 2lb.

There is NOT sufficient evidence to conclude that the population mean weight of the boxes is not equal to 2lb. Compare the p-value to α=0.05. Since the p-value is greater than α, Elena fails to reject the null hypothesis H0. Therefore, there is NOT enough evidence at the α=0.05 level of significance to support the claim that the population mean weight of the boxes is not equal to 2lb.

A doctor is researching the average resting heart rate for a healthy female adult. The doctor states the average resting heart rate for a healthy female adult is less than 85 beats per minute. If we would like to test the doctor's claim with a hypothesis test using a significance level of α=0.01 , which of the following choices are true? There is a 1% chance we will conclude μ=85, but is in fact μ<85. There is a 1% chance we will conclude μ<85, but is in fact μ=85. There is a 1% chance of rejecting the null hypothesis. There is a 1% chance that μ<85.

There is a 1% chance we will conclude μ<85, but is in fact μ=85. A significance level is the probability of rejecting the null hypothesis, H0, when it is in fact true. So, a significance level of α=0.01 means there is a 1% chance we will conclude from our data that μ<85 beats per minute, but it is in fact μ=85 beats per minute

A marketing analyst is researching how to increase organic growth to company websites. During their research, the marketing analyst states the average number of words on a website's first page is longer than 1500 words. If we would like to test the marketing analyst's claim with a hypothesis test using a significance level of α=0.025 , which of the following choices are true? There is a 2.5% chance we will conclude μ>1500, but is in fact μ=1500. There is a 2.5% chance of rejecting the null hypothesis. There is a 2.5% chance we will conclude μ=1500, but is in fact μ>1500. There is a 2.5% chance that μ>1500.

There is a 2.5% chance we will conclude μ>1500, but is in fact μ=1500. A significance level is the probability of rejecting the null hypothesis, H0, when it is in fact true. So, a significance level of α=0.025 means there is a 2.5% chance we will conclude from our data that μ>1500 words, but it is in fact μ=1500 words.

An e-commerce marketplace associate is researching ways to increase sales for their clients' websites. During their research, the associate states the average percent of customers who do not make a purchase on a non-secure website is greater than 75 percent. If we would like to test the associate's claim with a hypothesis test using a significance level of α=0.025 , which of the following choices are true? There is a 2.5% chance that μ>75. There is a 2.5% chance we will conclude μ=75, but is in fact μ>75. There is a 2.5% chance of rejecting the null hypothesis. There is a 2.5% chance we will conclude μ>75, but is in fact μ=75.

There is a 2.5% chance we will conclude μ>75, but is in fact μ=75. A significance level is the probability of rejecting the null hypothesis, H0, when it is in fact true. So, a significance level of α=0.025 means there is a 2.5% chance we will conclude from our data that μ>75 percent, but it is in fact μ=75 percent.

Determine the Type II error if the null hypothesis, H0, is: the percentage of college students that will graduate with debt is no more than 65%. And, the alternative hypothesis, Ha, is: the percentage of college students that will graduate with debt is more than 65%. There is sufficient evidence to conclude that less than 65% of college students will graduate with debt when, in fact, less than 65% of college students will graduate with debt. There is sufficient evidence to conclude that more than 65% of college students will graduate with debt when, in fact, more than 65% of college students will graduate with debt. There is insufficient evidence to conclude that less than 65% of college students will graduate with debt when, in fact, less than 65% of college students will graduate with debt. There is insufficient evidence to conclude that more than 65% of college students will graduate with debt when, in fact, more than 65% of college students will graduate with debt.

There is insufficient evidence to conclude that more than 65% of college students will graduate with debt when, in fact, more than 65% of college students will graduate with debt. A Type II error is the decision not to reject the null hypothesis when, in fact, it is false. In this case, the Type II error is that there is insufficient evidence to conclude that more than 65% of college students will graduate with debt when, in fact, more than 65% of college students will graduate with debt.

Karen, a medical researcher, is studying ways to lower blood pressure in patients. She would like to test the claim that reducing sodium in a patient's diet can lower blood pressure by more than 9 mmHg, on average. If the z− test statistic was calculated as z=1.05, does the medical researcher have enough evidence to reject the null hypothesis? Assume α=0.025. Move the blue dot to choose the appropriate test (left-, right, or two-tailed). Then, use the graph below to show the test statistic, p-value, and the rejection region to make a conclusion about the hypothesis test. There is enough evidence to suggest that reducing sodium in a patient's diet can lower blood pressure by less than 9 mmHg, on average. There is not enough evidence to suggest that reducing sodium in a patient's diet can lower blood pressure by less than 9 mmHg, on average. There is enough evidence to suggest that reducing sodium in a patient's diet can lower blood pressure by more than 9 mmHg, on average. There is not enough evidence to suggest that reducing sodium in a patient's diet can lower blood pressure by more than 9 mmHg, on average.

There is not enough evidence to suggest that reducing sodium in a patient's diet can lower blood pressure by more than 9 mmHg, on average. Since the medical researcher would like to test the claim that reducing sodium in a patient's diet can lower blood pressure by more than 9 mmHg, on average, this is a right-tailed test because of the inequality "greater than". The z− test statistic is 1.05, which is not far away from the center of the distribution. This leads us to believe there is not enough evidence to reject the null hypothesis. To state the results more accurately, we know, from the graph, that the p−value for this z− test statistic is 0.1469. Since the p− value of 0.1469 is greater than α=0.025, we can not reject the null hypothesis. There is NOT enough evidence to suggest that reducing sodium in a patient's diet can lower blood pressure by more than 9 mmHg, on average.

The check-in nurse is concerned about the number of patients coming into the doctor's office with pneumonia. The check-in nurse would like to test the claim that the proportion of patients coming into the doctor's office, who are later diagnosed with pneumonia, is different than 0.10. If the z− test statistic was calculated as z=−1.07, does the check-in nurse have enough evidence to reject the null hypothesis? Assume α=0.05. There is enough evidence to suggest that the proportion of patients coming into the doctor's office, who are later diagnosed with pneumonia, is not 0.10. There is not enough evidence to suggest that the proportion of patients coming into the doctor's office, who are later diagnosed with pneumonia, is not 0.10. There is enough evidence to suggest that the proportion of patients coming into the doctor's office, who are later diagnosed with pneumonia, is less than 0.10. There is not enough evidence to suggest that the proportion of patients coming into the doctor's office, who are later diagnosed with pneumonia, is less than 0.10.

There is not enough evidence to suggest that the proportion of patients coming into the doctor's office, who are later diagnosed with pneumonia, is not 0.10. Since the check-in nurse would like to test the claim that the proportion of patients coming into the doctor's office, who are later diagnosed with pneumonia, is different than 0.10, this is a two-tailed test because of the "not equals."The z− test statistic is −1.07, which is not far away from the center of the distribution. This leads us to believe there is not enough evidence to reject the null hypothesis. To state the results more accurately, we know, from the graph, that the p−value for this z− test statistic is 0.2846. Since the p− value of 0.2846 is greater than α=0.05, we can not reject the null hypothesis. There is NOT enough evidence to suggest the proportion of patients coming into the doctor's office, who are later diagnosed with pneumonia, is different than 0.10.

A debt collector manager would like to test the claim that the average amount of credit card debt for an individual is more than 3.2 thousand dollars. If the z− test statistic was calculated as z=0.85, does the debt collector manager have enough evidence to reject the null hypothesis? Assume α=0.025. There is enough evidence to suggest the average amount of credit card debt for an individual is less than 3.2 thousand dollars. There is not enough evidence to suggest the average amount of credit card debt for an individual is less than 3.2 thousand dollars. There is not enough evidence to suggest the average amount of credit card debt for an individual is greater than 3.2 thousand dollars. There is enough evidence to suggest the average amount of credit card debt for an individual is greater than 3.2 thousand dollars.

There is not enough evidence to suggest the average amount of credit card debt for an individual is greater than 3.2 thousand dollars. Since the debt collector manager would like to test the claim that the average amount of credit card debt for an individual is more than 3.2 thousand dollars, this is a right-tailed test because of the inequality "greater than". The z− test statistic is 0.85, which is not far away from the center of the distribution. This leads us to believe there is not enough evidence to reject the null hypothesis. To state the results more accurately, we know, from the graph, that the p−value for this z− test statistic is 0.1977. Since the p− value of 0.1977 is greater than α=0.025, we can not reject the null hypothesis. There is NOT enough evidence to suggest the average amount of credit card debt for an individual is more than 3.2 thousand dollars.

A marketing manager would like to test the claim that the percent increase of company sales after an advertising campaign is different than 17 percent. If the z− test statistic was calculated as z=2.43, does the marketing manager have enough evidence to reject the null hypothesis? Assume α=0.01. Move the blue dot to choose the appropriate test (left-, right, or two-tailed). Then, use the graph below to show the test statistic, p-value, and the rejection region to make a conclusion about the hypothesis test. There is enough evidence to suggest the average percent increase of company sales after an advertising campaign is greater than 17 percent. There is not enough evidence to suggest the average percent increase of company sales after an advertising campaign is greater than 17 percent. There is enough evidence to suggest the average percent increase of company sales after an advertising campaign is not 17 percent. There is not enough evidence to suggest the average percent increase of company sales after an advertising campaign is not 17 percent.

There is not enough evidence to suggest the average percent increase of company sales after an advertising campaign is not 17 percent. Since the marketing manager would like to test the claim that the average percent increase of company sales after an advertising campaign is different than 17 percent, this is a two-tailed test because of the "not equals."The z− test statistic is 2.43, which is far away from the center of the distribution. This leads us to believe there is enough evidence to reject the null hypothesis. To state the results more accurately, we know, from the graph, that the p−value for this z− test statistic is 0.015. Since the p− value of 0.015 is greater than α=0.01, we can fail to reject the null hypothesis. There is NOT enough evidence to suggest the average percent increase of company sales after an advertising campaign is different than 17 percent.

Suppose the null hypothesis, H0, is: Darrell has enough money in his bank account to purchase a new television.And, the alternative hypothesis, Ha, is: Darrell does not have enough money in his bank account to purchase a new television. What is the Type I error in this scenario? There is sufficient evidence to conclude that Darrell does not have enough money in his bank account to purchase the television when, in fact, he does. There is sufficient evidence to conclude that Darrell does not have enough money in his bank account to purchase the television when, in fact, he does not. There is insufficient evidence to conclude that Darrell does not have enough money in his bank account to purchase the television when, in fact, he does. There is insufficient evidence to conclude that Darrell does not have enough money in his bank account to purchase the television when, in fact, he does not.

There is sufficient evidence to conclude that Darrell does not have enough money in his bank account to purchase the television when, in fact, he does. A Type I error is the decision to reject the null hypothesis when it is true. In this case, the Type I error is when there is sufficient evidence to conclude that Darrell does not have enough money in his bank account to purchase the television when, in fact, he does.

Using the information above, choose the correct conclusion for this hypothesis test. H0:p=0.15 ; Ha:p>0.15 The p-value for this hypothesis test is 0.002. The level of significance is α=0.025 There is sufficient evidence to conclude that the proportion of time children spent outside on the weekends to more than 15%. There is NOT sufficient evidence to conclude that the proportion of time children spent outside on the weekends to more than 15%. There is sufficient evidence to conclude that proportion of time children spent outside on the weekends to more than 25%. There is NOT sufficient evidence to conclude that the proportion of time children spent outside on the weekends to more than 25%.

There is sufficient evidence to conclude that the proportion of time children spent outside on the weekends to more than 15%. Compare the p-value to α=0.025. Since the p-value is less than α, reject the null hypothesis H0. Therefore, there is enough evidence at the α=0.025 level of significance to suggest that the proportion of time children spent outside on the weekends to more than 15%.

A sociology professor, who teaches at the same sociology department of the university, thinks adults spend a different amount of time on social media daily.The sociology professor would like to carry out a hypothesis test and test the claim that the average amount of time adults spend on social media daily is different from 3.6 hours. Why is their hypothesis test two-tailed? This is a two-tailed test because no direction is specified. This is a two-tailed test because a direction is specified. The population parameter is greater than the specified value. This is a two-tailed test because a direction is specified. The population parameter is less than the specified value. More information is needed.

This is a two-tailed test because no direction is specified. The sociology professor claims that the average amount of time adults spend on social media daily is different from 3.6 hours. Based on this claim, a direction is not specified, which means this is a two-tailed test. If this were a one-tailed test, then the claim may read "the average amount of time adults spend on social media daily is greater than 3.6 hours."

An office manager, who works at a different department at the finance company, believes employees spend a different amount of time on the phone daily. This office manager would like to carry out a hypothesis test and test the claim that the amount of time employees spent on the phone per day is different from 5 hours, on average. Why is their hypothesis test two-tailed? This is a two-tailed test because no direction is specified. This is a two-tailed test because a direction is specified. The population parameter is greater than the specified value. This is a two-tailed test because a direction is specified. The population parameter is less than the specified value. More information is needed.

This is a two-tailed test because no direction is specified. The office manager claims that the amount of time employees spent on the phone per day is different from 5 hours, on average. Based on this claim, a direction is not specified, which means this is a two-tailed test. If this were a one-tailed test, then the claim may read "the amount of time employees spent on the phone per day is greater than 5 hours, on average".

A trader claims that the proportion of stocks that offer dividends is different from 0.14. If the trader wants to conduct a hypothesis test, should they use a left-, right-, or two-tailed hypothesis test to analyze whether the proportion of stocks that offer dividends is different from 0.14? Left-tailed test Right-tailed test Two-tailed test

Two-tailed test To identify the null and alternative hypothesis, first determine the purpose of the hypothesis test. In this scenario, the trader is analyzing whether the proportion of stocks that offer dividends is different from 0.14. Since the primary concern of this test is dealing with whether a population parameter, p, is different from a specified value, p0 and no direction is specified, then the trader should conduct a two-tailed test.

A government official wanted to know if more than 90% of the residents in a region had an income that was less than 100 thousand dollars. The official randomly sampled 500 residents and found that 461 of them had an income that was less than 100 thousand dollars. The official conducts a one-proportion hypothesis test at the 5% significance level, to test whether the true proportion of residents in the region that have an income that is less than 100 thousand dollars is greater than 90%. (a) H0:p=0.9; Ha:p>0.9, which is a right-tailed test. (b) z0=1.64, p-value is = 0.051 (c) Which of the following are appropriate conclusions for this hypothesis test? Select all that apply. We reject H0. We fail to reject H0. At the 5% significance level, the data provide sufficient evidence to conclude the true proportion of residents in the region that have an income that is less than 100 thousand dollars is greater than 90%. At the 5% significance level, the data do not provide sufficient evidence to conclude the true proportion of residents in the region that have an income that is less than 100 thousand dollars is greater than 90%.

We fail to reject H0. At the 5% significance level, the data do not provide sufficient evidence to conclude the true proportion of residents in the region that have an income that is less than 100 thousand dollars is greater than 90%. In this case, p=0.051 and α=0.05, so p>α. Therefore, we fail to reject the null hypothesis. This means that at the 5% significance level, the test results are not statistically significant, and do not provide evidence against the null hypothesis. The official cannot conclude that at the 5% significance level that the true proportion of residents in the region that have an income that is less than 100 thousand dollars is greater than 90%.

A financial institution believes the average interest rate for a high interest savings account is different than the previous year. The financial institution would like to test the claim that the average interest rate for a high interest savings account is different than 1.75%. Using the computed test statistic of 2.11 and the critical values of −1.96 and 1.96, is there enough evidence for the financial institution to reject the null hypothesis? Yes. There is sufficient evidence for the financial institution to reject the null hypothesis because the test statistic falls in the rejection region. No. There is NOT sufficient evidence for the financial institution to reject the null hypothesis because the test statistic does not fall in the rejection region. Yes. There is sufficient evidence for the financial institution to reject the null hypothesis because the critical value falls in the rejection region. No. There is NOT sufficient evidence for the financial institution to reject the null hypothesis because the critical value does not fall in the rejection region.

Yes. There is sufficient evidence for the financial institution to reject the null hypothesis because the test statistic falls in the rejection region. This is a two-tailed test because the alternative hypothesis formed by the claim has a not equals sign (different than 1.75%). The hypotheses are Ho:μ=1.75 and Ha:μ≠1.75 . Since this is a two-tailed test, the rejection region are the shaded regions on the left and right separated by two critical values, in this case −1.96 and 1.96. Since the test statistic is 2.11, it falls to the right of 1.96, which is in the rejection region. So, there is enough evidence to reject the null hypothesis.

An online marketplace believes the markup for wholesale items is increasing this fiscal year. The online marketplace would like to test the claim that the average percent of wholesale markup is over 25%. Using the computed test statistic of 3.20 and the critical value of 1.65, is there enough evidence for the online marketplace to reject the null hypothesis? Use the graph below to interpret the critical value. First select the appropriate test type (left-, right-, or two- tailed), then plot the points along the x-axis (for the critical value and test statistic), and use them to choose the appropriate interpretation. Yes. There is sufficient evidence for the online marketplace to reject the null hypothesis because the test statistic falls in the rejection region. No. There is NOT sufficient evidence for the online marketplace to reject the null hypothesis because the test statistic does not fall in the rejection region. Yes. There is sufficient evidence for the online marketplace to reject the null hypothesis because the critical value falls in the rejection region. No. There is NOT sufficient evidence for the online marketplace to reject the null hypothesis because the critical value does not fall in the rejection region.

Yes. There is sufficient evidence for the online marketplace to reject the null hypothesis because the test statistic falls in the rejection region. This is a right-tailed test because the alternative hypothesis formed by the claim has a greater than sign (more than 25%). The hypotheses are Ho:μ=25 and Ha:μ>25 . Since this is a right-tailed test, the rejection region is the shaded region on the right separated by one critical value, in this case 1.65. Since the the test statistic is 3.20, it falls to the right of 1.65, which is in the rejection region. So, there is enough evidence to reject the null hypothesis.

Suppose the null hypothesis, H0, is: the mean age of the horses on a ranch is 6 years. What is the Type I error in this scenario? You cannot conclude that the mean age of the horses on a ranch is 6 years when, in fact, it is. You cannot conclude that the mean age of the horses on a ranch is 6 years when, in fact, it is not. You cannot conclude that the mean age of the horses is not 6 years when, in fact, it is. You cannot conclude that the mean age of the horses is not 6 years when, in fact, it is not.

You cannot conclude that the mean age of the horses on a ranch is 6 years when, in fact, it is. A Type I error is the decision to reject the null hypothesis when it is true. In this case, the Type I error is rejecting the null hypothesis that the mean is 6 years when, in fact, it is.

Suppose the null hypothesis, H0, is a weightlifting bar can withstand weights of 800 pounds and less.And the alternative hypothesis, Ha, is a weightlifting bar can withstand weights of greater than 800 pounds.What is α, the probability of a Type I error in this scenario? You cannot conclude the weightlifting bar can withstand weights of greater than 800 pounds when, in fact, the weightlifting bar can withstand weights of greater than 800 pounds. You cannot conclude the weightlifting bar can withstand weights of greater than 800 pounds when, in fact, the weightlifting bar can withstand weights of less than or equal to 800 pounds. You conclude the weightlifting bar can withstand weights of greater than 800 pounds when, in fact, the weightlifting bar can withstand weights of less than or equal to 800 pounds. You conclude the weightlifting bar can withstand weights of greater than 800 pounds when, in fact, the weightlifting bar can withstand weights of greater than 800 pounds

You conclude the weightlifting bar can withstand weights of greater than 800 pounds when, in fact, the weightlifting bar can withstand weights of less than or equal to 800 pounds. A Type I error is the decision to reject the null hypothesis when it is true. In this case, the Type I error is concluding the weightlifting bar can withstand weights of greater than 800 pounds when, in fact, the weightlifting bar can withstand weights of less than or equal to 800 pounds.

Determine the Type II error if the null hypothesis, H0, is: the mean price of a loaf of bread is $1.67. And the alternative hypothesis, Ha, states the claim, which is the mean price of a loaf of bread is not $1.67. You conclude that the mean price of a loaf of bread is not $1.67 when, in fact, the mean price of a loaf of bread is $1.67. You conclude that the mean price of a loaf of bread is not $1.67 when, in fact, the mean price of a loaf of bread is not $1.67. You do not conclude that the mean price of a loaf of bread is not $1.67 when, in fact, the mean price of a loaf of bread is $1.67. You do not conclude that the mean price of a loaf of bread is not $1.67 when, in fact, the mean price of a loaf of bread is not $1.67.

You do not conclude that the mean price of a loaf of bread is not $1.67 when, in fact, the mean price of a loaf of bread is not $1.67. A Type II error is the decision not to reject the null hypothesis when, in fact, it is false. In this case, the Type II error is that you cannot conclude that the mean price of the bread is $1.67 when, in fact, it is not.

Determine the Type II error if the null hypothesis, H0, is: a wooden ladder can withstand weights of 250 pounds and less. You do not reject the null hypothesis that the ladder can withstand weight of 250 pounds and less when, in fact, it cannot. You conclude that the ladder cannot withstand weight of 250 pounds and less when, in fact, it really can. You do not reject the null hypothesis that the ladder can withstand weight of 250 pounds and less when, in fact, it can. You reject the null hypothesis that the ladder can withstand weight of 250 pounds and less when, in fact, it can.

You do not reject the null hypothesis that the ladder can withstand weight of 250 pounds and less when, in fact, it cannot. A Type II error is the decision not to reject the null hypothesis when, in fact, it is false. In this case, the Type II error is stating that there is insufficient evidence to conclude that the ladder can withstand weight of 250 pounds, in fact, it cannot.

Suppose the null hypothesis, H0, is: a weightlifting bar can withstand weights of 800 pounds and less. And the alternative hypothesis, Ha, is: a weightlifting bar can withstand weights of more than 800 pounds. What is the Type I error in this scenario? You say there is insufficient evidence to conclude that the weightlifting bar can withstand weights of more than 800 pounds when, in fact, it can not. You say there is insufficient evidence to conclude that the weightlifting bar can withstand weights of more than 800 pounds when, in fact, it can. You say there is sufficient evidence to conclude that the weightlifting bar can withstand weights of more than 800 pounds when, in fact, it can not. You say there is sufficient evidence to conclude that the weightlifting bar can withstand weights of more than 800 pounds when, in fact, it can.

You say there is sufficient evidence to conclude that the weightlifting bar can withstand weights of more than 800 pounds when, in fact, it can not. A Type I error is the decision to reject the null hypothesis when it is true. In this case, the Type I error is saying there is sufficient evidence to conclude that the weightlifting bar can withstand weights of more than 800 pounds when, in fact, it can not.

Determine the Type I error if the null hypothesis, H0, is: the mean price of a loaf of bread is $1.67. You state that there is insufficent evidence to conclude that the mean price of a loaf of bread is $1.67 when, in fact, it is. You state that there is sufficient evidence to conclude that the mean price of bread is not $1.67 , when, in fact, it is. You state that there is insufficient evidence to conclude that the mean price of a loaf of bread is not $1.67 when, in fact, it is not. You cannot conclude that the mean price of a loaf of bread is $1.67 when, in fact, it is not

You state that there is sufficient evidence to conclude that the mean price of bread is not $1.67 , when, in fact, it is. A Type I error is the decision to reject the null hypothesis when it is true. In this case, the Type I error is that you conclude that the mean price of bread is not $1.67 , when, in fact, it is.

Determine the Type I error if the null hypothesis, H0, is: a wooden ladder can withstand weights of 250 pounds and less. You think the ladder can withstand weight of 250 pounds and less when, in fact, it cannot. You think the ladder cannot withstand weight of 250 pounds and less when, in fact, it really can. You think the ladder can withstand weight of 250 pounds and less when, in fact, it can. You think the ladder cannot withstand weight of 250 pounds and less when, in fact, it cannot.

You think the ladder cannot withstand weight of 250 pounds and less when, in fact, it really can. A Type I error is the decision to reject the null hypothesis when it is true. In this case, the Type I error is thinking the wooden ladder cannot withstand the weights of 250 pounds or less when it really can.

Suppose the null hypothesis, H0, is: the mean age of the horses on a ranch is 6 years. What is the Type II error in this scenario? You think the mean age of the horses on a ranch is 6 years when, in fact, it is. You think the mean age of the horses on a ranch is 6 years when, in fact, it is not. You think the mean age of the horses is not 6 years when, in fact, it is. You think the mean age of the horses is not 6 years when, in fact, it is not.

You think the mean age of the horses on a ranch is 6 years when, in fact, it is not. A Type II error is the decision not to reject the null hypothesis when, in fact, it is false. In this case, the Type II error is thinking that the mean age of the horses is 6 years when, in fact, it is not.

A social media content associate, who works for a growing start up business, is looking for ways to use social media platforms to build an audience. It is known that the average number of social media posts per day that helps to build an audience is 3 daily posts. After posting on social media for a few months, the social media content associate believes the average number of social media posts per day that helps to build an audience is different. What are the hypotheses? Fill in the blanks with the correct symbol (=,≠,<, or >) to represent the correct hypothesis.

null hypothesis : mean = 3 alternative hypothesis : mean ≠ 3 The social media content associate would be testing the claim that the average number of social media posts per day that helps to build an audience is different than 3 daily posts, which is our alternative hypothesis, Ha. The null hypothesis H0 would be the equality statement that the average number of social media posts per day that helps to build an audience is 3 daily posts.

A computer scientist is studying popular programming languages used in 2019. It is known that the average percent of companies that use Python is 52 percent. The computer scientist believes the average percent of companies that use Python is different. What are the hypotheses? Fill in the blanks with the correct symbol (=,≠,<, or >) to represent the correct hypothesis.

null hypothesis : mean = 52 alternative hypothesis : mean ≠ 52

A budget controller would like to test the claim that the average monthly amount spent on travel expenses at a small company is different than 65 hundred dollars. To test this claim, at the 1% significance level, the budget controller collects the following data on a sample of 18 months and records the amount spent on travel expenses. The following is the data from this study: Sample size =18 monthsSample mean =61.5 hundred dollarsSample standard deviation =6.5 hundred dollars Identify the null and alternative hypothesis for this study by filling in the blanks with the correct symbol (=,≠,<, or > to represent the correct hypothesis.)

null hypothesis : mean = 65 alternative hypothesis : mean ≠ 65 This is a two-tailed test, because Ha would be if the average monthly amount spent on travel expenses at a small company is different than 65 hundred dollars. H0 : μ=65; Ha : μ≠65

A banker, who works at a credit union, is interested in studying the average percent of banking fees customers per on a monthly basis. It is known that the average monthly banking fee is 9 dollars. The banker believes the average monthly banking fee has increased. What are the hypotheses? Fill in the blanks with the correct symbol (=,≠,<, or >) to represent the correct hypothesis.

null hypothesis : mean = 9 alternative hypothesis : mean > 9 The banker would be testing the claim that the average percent of banking fees is greater than 9 dollars, which is our alternative hypothesis, Ha. The null hypothesis H0 would be the equality statement that the average percent of banking fees is 9 dollars.H0: μ=9Ha: μ>9

A business owner is looking for ways to increase sales over the next year. It is known that the percent of businesses that use email marketing to help grow sales is 52%. The business owner believes the percent of businesses that use email marketing to help grow sales has increased. What are the hypotheses? Fill in the blanks with the correct symbol (=,≠,<, or >) to represent the correct hypothesis.

null hypothesis : p = 0.52 alternative hypothesis : p > 0.52

A civil engineer is interested in the number of planning permits submitted by cities annually. It is known that the average number of planning permits submitted by a city every year is 670 permits. After studying this further, the civil engineer believes the average number of planning permits submitted by a city every year is different. What are the hypotheses? Fill in the blanks with the correct symbol (=,≠,<, or >) to represent the correct hypothesis.

null hypothesis : μ​ = 670 alternative hypothesis : μ ≠ 670

Austin works in the marketing department for a large cable company. Recently, he heard that 56% of cable TV subscribers are considering dropping their cable TV subscription. He tests the claim by randomly selecting a sample of cable TV subscribers and asking whether they are considering dropping their cable TV subscription. Austin finds that 122 of the 224 he surveyed are considering dropping their cable TV subscription. What are the null and alternative hypotheses for this hypothesis test? {H0:p≠0.56Ha:p=0.56 {H0:p=0.56Ha:p>0.56 {H0:p=0.56Ha:p≠0.56 {H0:p=0.56Ha:p<0.56

{H0:p=0.56Ha:p≠0.56 First verify whether all of the conditions have been met. Let p be the population proportion for cable TV subscribers who are considering dropping their cable TV subscription. Since there are two independent outcomes for each trial, the proportion follows a binomial model. The question states that the sample was collected randomly. The expected number of successes, np=125.44, and the expected number of failures, nq=n(1−p)=98.56, are both greater than or equal to 5. Since Austin is looking for evidence that supports the claim that 56% of cable TV subscribers are considering dropping their cable TV subscription, the null hypothesis is that p is equal to 0.56 and the alternative hypothesis is that p is not equal to 0.56. The null and alternative hypotheses are shown below. {H0:p=0.56Ha:p≠0.56

Kylie works for a large nursery and is investigating whether to use a new brand of seeds. The new brand of seeds advertises that 93% of the seeds germinate, which is higher than the germination rate of the seeds she is currently using. She will change over to this new brand unless the actual germination rate is less than what is advertised. Kylie conducts an experiment by randomly selecting 76 seeds of the new brand and plants them. She finds that 70 of those seeds germinated. What are the null and alternative hypotheses for this hypothesis test? {H0:p=0.93Ha:p>0.93 {H0:p=0.93Ha:p<0.93 {H0:p≠0.93Ha:p=0.93 {H0:p=0.93Ha:p≠0.93

{H0:p=0.93Ha:p<0.93 First verify whether all of the conditions have been met. Let p be the population proportion for the germination rate of the new seeds. Since there are two independent outcomes for each trial, the proportion follows a binomial model. The question states that the sample was collected randomly. The expected number of successes, np=70.68, and the expected number of failures, nq=n(1−p)=5.32, are both greater than or equal to 5. Since Kylie is testing whether the germination rate is less than 93%, the null hypothesis is that p is equal to 0.93 and the alternative hypothesis is that p is less than 0.93. The null and alternative hypotheses are shown below. {H0:p=0.93Ha:p<0.93