Exam 4

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The​ _________ hypothesis is a statement that the value of a population parameter is equal to some claimed value.

Rejection of the claim about aspirin is more serious because the wrong aspirin dosage could cause more serious adverse reactions than a wrong vitamin C dosage. It would be wise to use a smaller significance level for testing the claim about the aspirin.

A bottle contains a label stating that it contains pills with 500 mg of vitamin​ C, and another bottle contains a label stating that it contains pills with 325 mg of aspirin. When testing claims about the mean contents of the​ pills, which would have more serious​ implications: rejection of the vitamin C claim or rejection of the aspirin​ claim? Considering only a type I error and using the same sample​ size, is it wise to use the same significance level for hypothesis tests about the mean amount of vitamin C and the mean amount of​ aspirin?

Left-tailed test z= -1.18 p= 0.1195 H0 : p = 0.08 Fail to reject the null hypothesis because the​P-value is greater than the significance ​level, α. There is not sufficient evidence to support the claim that less than 8​% of treated subjects experienced headaches.

A certain drug is used to treat asthma. In a clinical trial of the​ drug, 16 of 265 treated subjects experienced headaches​ (based on data from the​ manufacturer). The accompanying calculator display shows results from a test of the claim that less than 8​% of treated subjects experienced headaches. Use the normal distribution as an approximation to the binomial distribution and assume a 0.01 significance level to complete parts​ (a) through​ (e) below.

Stat Proportion stat one sample with summary Does the method appear to be​ effective? Yes, the proportion of girls is significantly from 0.5.

A clinical trial tests a method designed to increase the probability of conceiving a girl. In the study 335 babies were​ born, and 268 of them were girls. Use the sample data to construct a 99​% confidence interval estimate of the percentage of girls born. Based on the​ result, does the method appear to be​ effective?

z-scored with an area of α to its right.

A critical​ value, zα​, denotes the​ _______.

Does it appear that there is too much mercury in tuna​ sushi? Yes, because it is possible that the mean is greater than 1 ppm.​ Also, at least one of the sample values exceeds 1​ ppm, so at least some of the fish have too much mercury.

A food safety guideline is that the mercury in fish should be below 1 part per million​ (ppm). Listed below are the amounts of mercury​ (ppm) found in tuna sushi sampled at different stores in a major city. Construct a 90​% confidence interval estimate of the mean amount of mercury in the population. Does it appear that there is too much mercury in tuna​ sushi?

a. )421 + 156 = 577 =n Stat Proportion stat one sample with summary 0.240 < p < 0.301 b. )Given that the percentage of offspring yellow peas is not​ 25%, do the results contradict​ expectations? No, the confidence interval includes​ 0.25, so the true percentage could easily equal​ 25%

A genetic experiment with peas resulted in one sample of offspring that consisted of 421 green peas and 156 yellow peas. a. Construct a 90​% confidence interval to estimate of the percentage of yellow peas. b. It was expected that​ 25% of the offspring peas would be yellow. Given that the percentage of offspring yellow peas is not​ 25%, do the results contradict​ expectations?

The value of p is the sample proportion the sample proportion. The value of q is found from evaluating 1−p. The value of n is the sample size.the sample size. The value of E is the margin of error. The value of p is the population proportion. If the confidence level is 90​%, what is the value of α​? α= 0.1

A magazine provided results from a poll of 2000 adults who were asked to identify their favorite pie. Among the 2000 ​respondents, 13​% chose chocolate​ pie, and the margin of error was given as ±5 percentage points. What values do p​, q​, ​n, E, and p​ represent? If the confidence level is 90​%, what is the value of α​?

A 99​% confidence interval must be wider than an 80​% confidence interval in order to be more confident that it captures the true value of the population proportion.

A magazine provided results from a poll of 1000 adults who were asked to identify their favorite pie. Among the 1000 ​respondents, 11​% chose chocolate​ pie, and the margin of error was given as ±5 percentage points. Given specific sample​ data, which confidence interval is​ wider: the 99​% confidence interval or the 80​% confidence​ interval? Why is it​ wider?

The statement indicates that the interval 14​% ±4​% is likely to contain the true population percentage of people that prefer chocolate pie.

A magazine provided results from a poll of 500 adults who were asked to identify their favorite pie. Among the 500 respondents, 14​% chose chocolate​ pie, and the margin of error was given as ±4 percentage points. Describe what is meant by the statement that​ "the margin of error was given as ±4 percentage​ points."

The confidence level

A newspaper provided a​ "snapshot" illustrating poll results from 1910 professionals who interview job applicants. The illustration showed that​ 26% of them said the biggest interview turnoff is that the applicant did not make an effort to learn about the job or the company. The margin of error was given as ±3 percentage points. What important feature of the poll was​ omitted?

H0 : p = .92 Fail to reject the null hypothesis because the​P-value is greater than the significance​level, α. There is not sufficient evidence to warrant rejection of the claim that 92​% of adults own a cell phone.

A poll of 2,035 randomly selected adults showed that 93​% of them own cell phones. The technology display below results from a test of the claim that 92​% of adults own cell phones. Use the normal distribution as an approximation to the binomial​ distribution, and assume a 0.01 significance level to complete parts​ (a) through​ (e).

b. Do cell phone users appear to have a rate of cancer of the brain or nervous system that is different from the rate of such cancer among those not using cell​ phones? Why or why​ not? No​, because 0.0345​% is included in the confidence interval. If yes, it is not included

A study of 420,087 cell phone users found that 133 of them developed cancer of the brain or nervous system. Prior to this study of cell phone​ use, the rate of such cancer was found to be 0.0345​% for those not using cell phones. Complete parts​ (a) and​ (b).

Would it be reasonable to sample this number of​ students? Yes. This number of IQ test scores is a fairly small number.

An IQ test is designed so that the mean is 100 and the standard deviation is 14 for the population of normal adults. Find the sample size necessary to estimate the mean IQ score of statistics students such that it can be said with 95​% confidence that the sample mean is within 6 IQ points of the true mean. Assume that σ=14 and determine the required sample size using technology. Then determine if this is a reasonable sample size for a real world calculation.

Select the correct choice below​ and, if​ necessary, fill in the answer box to complete your choice. Neither the normal distribution nor the t distribution applies.

Assume that we want to construct a confidence interval. Do one of the​ following, as​ appropriate: (a) find the critical value tα/2​, ​(b) find the critical value zα/2​, or​ (c) state that neither the normal distribution nor the t distribution applies. The confidence level is 95​%, σ is not​ known, and the normal quantile plot of the 17 salaries​ (in thousands of​ dollars) of basketball players on a team is as shown.

point estimate

A​ _______ is a single value used to approximate a population parameter.

hypothesis test

A​ _____________ is a procedure for testing a claim about a property of a population.

The value of the test statistic is 6.15

Claim: Most adults would erase all of their personal information online if they could. A software firm survey of 420 randomly selected adults showed that 65​% of them would erase all of their personal information online if they could. Find the value of the test statistic.

a. Express the original claim in symbolic form. μ = 68.6 bpm b. Identify the null and alternative hypotheses. ​PICTURE

Claim: The mean pulse rate​ (in beats per​ minute) of adult males is equal to 68.6 bpm. For a random sample of 130 adult​ males, the mean pulse rate is 67.9 bpm and the standard deviation is 10.7 bpm. Complete parts​ (a) and​ (b) below.

p = (UCL + LCL) /2 E= (UCL - LCL) / 2 p = (.666 + .222) /2 = .444 E = (.666 - .222) /2 = .222 p±E= 0.444 ± 0.222

Express the confidence interval 0.222<p<0.666 in the form p±E.

p±E = 0.444 ± 0.222

Express the confidence interval 0.222<p<0.666 in the form p±E.

1-.89= .11 / 2 = 0.055 StatCrunch Stat Calculators Normal Standard P(X> ____) = 0.055 zα/2 = 1.60

Find the critical value zα/2 that corresponds to the given confidence level. 89​%

What is the confidence interval for the population mean μ​? Stat T stats one sample with summary 27.1 hg < u < 28.7 Are the results between the two confidence intervals very​ different? ​No, because the confidence interval limits are similar.

Here are summary statistics for randomly selected weights of newborn​ girls: n=248​, x=27.9 hg, s=7.8 hg. Construct a confidence interval estimate of the mean. Use a 90​% confidence level. Are these results very different from the confidence interval 26.7 hg<μ<29.7 hg with only 16 sample​ values, x=28.2 ​hg, and s=3.5 ​hg?

The sample observations are not a random​ sample, so a test about a population proportion using the normal approximating method cannot be used. This statement means that if the​ P-value is very​ low, the null hypothesis should be rejected. This statement seems to suggest that with a high​ P-value, the null hypothesis has been proven or is​ supported, but this conclusion cannot be made. Choosing this specific of a significance level could give the impression that the significance level was chosen specifically to reach a desired conclusion.

In a certain​ survey, 522 people chose to respond to this​ question: "Should passwords be replaced with biometric security​ (fingerprints, etc)?" Among the​ respondents, 54​% said​ "yes." We want to test the claim that more than half of the population believes that passwords should be replaced with biometric security. Complete parts​ (a) through​ (d) below.

a. Assume that nothing is known about the percentage to be estimated. n = (((za/2)^2).25)/(0.04^2) b. look at camera c. Does the added knowledge in part​ (b) have much of an effect on the sample​ size? ​No, using the additional survey information from part​ (b) only slightly reduces the sample size.

In a study of government financial aid for college​ students, it becomes necessary to estimate the percentage of​ full-time college students who earn a​ bachelor's degree in four years or less. Find the sample size needed to estimate that percentage. Use a 0.04 margin of error and use a confidence level of 90​%. Complete parts​ (a) through​ (c) below.

Stat t stat one sample with data. What does the confidence interval tell about the population of all adult​ females? Select the correct choice below​ and, if​ necessary, fill in the answer​ box(es) to complete your choice. The results tell nothing about the population of all adult​ females, because participants in speed dating are not a representative sample of the population of all adult females.

In a study of speed​ dating, male subjects were asked to rate the attractiveness of their female​ dates, and a sample of the results is listed below ​(1=not ​attractive; 10=extremely ​attractive). Construct a confidence interval using a 90​% confidence level. What do the results tell about the mean attractiveness ratings of the population of all adult​ females?

247+61= 308 Stat Proportion stat one sample with summary 0.161 < p < 0.235 b. Choose the correct answer below. Since the two confidence intervals​ overlap, neither restaurant appears to have a significantly different percentage of orders that are not accurate.

In a study of the accuracy of fast food​ drive-through orders, Restaurant A had 247 accurate orders and 61 that were not accurate. a. Construct a 90​% confidence interval estimate of the percentage of orders that are not accurate. b. Compare the results from part​ (a) to this 90​% confidence interval for the percentage of orders that are not accurate at Restaurant​ B: 0.179<p<0.256. What do you​ conclude?

3238*.803 = 2600.114 = 2600 c. What do the results tell us about the proportion of college students who use at least one prescription​ medication? The results tell us nothing about the proportion of college students who use at least one prescription medication

In a survey of 3238 adults aged 57 through 85​ years, it was found that 80.3​% of them used at least one prescription medication. Complete parts​ (a) through​ (c) below.

degrees of freedom

The number of​ _______ for a collection of sample data is the number of sample values that can vary after certain restrictions have been imposed on all data values

What does the result tell us about the population of all​ celebrities? Select the correct choice below​ and, if​ necessary, fill in the answer​ box(es) to complete your choice. Because the ten wealthiest celebrities are not a representative​ sample, this​ doesn't provide any information about the population of all celebrities. Do the data appear to be from a normally distributed population as​ required? ​No, because the points lie reasonable close to a straight​ line, but there is a systematic pattern that is not a straight line pattern

Listed below are the amounts of net worth​ (in millions of​ dollars) of the ten wealthiest celebrities in a country. Construct a 99​% confidence interval. What does the result tell us about the population of all​ celebrities? Do the data appear to be from a normally distributed population as​ required?

The sample is not unusual if the claim is true. The sample is unusual if the claim is false.​ Therefore, there is sufficient evidence to support the claim.

Make a decision about the given claim. Use only the rare event​ rule, and make subjective estimates to determine whether events are likely. For​ example, if the claim is that a coin favors heads and sample results consist of 11 heads in 20​ flips, conclude that there is not sufficient evidence to support the claim that the coin favors heads​ (because it is easy to get 11 heads in 20 flips by chance with a fair​ coin). ​Claim: The mean IQ score of students in a large math class is greater than 106. A simple random sample of the students has a mean IQ score of 128.8.

a. ) What is the number of degrees of freedom that should be used for finding the critical value tα/2​? df= 50-1= 49 b. ) Find the critical value tα/2 corresponding to a​ 95% confidence level. T calculator 2.01 c.) Give a brief general description of the number of degrees of freedom. The number of degrees of freedom for a collection of sample data is the number of sample values that can vary after certain restrictions have been imposed on all data values.

Refer to the accompanying data display that results from a sample of airport data speeds in Mbps. Complete parts​ (a) through​ (c) below.

a.) Express the confidence interval in the format that uses the​ "less than" symbol. Given that the original listed data use one decimal​ place, round the confidence interval limits accordingly. 13.05 Mbps < u < 22.15 Mbps b. Identify the best point estimate of μ is 17.60 Mbps. The margin of error is E = 4.55 Mbps. c. In constructing the confidence interval estimate of μ​, why is it not necessary to confirm that the sample data appear to be from a population with a normal​ distribution? Because the sample size of 50 is greater than​ 30, the distribution of sample means can be treated as a normal distribution.

Refer to the accompanying data display that results from a sample of airport data speeds in Mbps. Complete parts​ (a) through​ (c) below. TInterval ​(13.046,22.15) x=17.598 Sx=16.01712719 n=50

We have​ 95% confidence that the limits of 13.05 Mbps and 22.15 Mbps contain the true value of the mean of the population of all data speeds at the airports.

Refer to the accompanying data display that results from a sample of airport data speeds in Mbps. The results in the screen display are based on a​ 95% confidence level. Write a statement that correctly interprets the confidence interval.

The​ P-value of 0.001 is preferred because it corresponds to the sample evidence that most strongly supports the alternative hypothesis that the method is effective.

The Ericsson method is one of several methods claimed to increase the likelihood of a baby girl. In a clinical​ trial, results could be analyzed with a formal hypothesis test with the alternative hypothesis of p​0.5, which corresponds to the claim that the method increases the likelihood of having a​ girl, so that the proportion of girls is greater than 0.5. If you have an interest in establishing the success of the​ method, which of the following​ P-values would you​ prefer: 0.999,​ 0.5, 0.95,​ 0.05, 0.01,​ 0.001? Why?

sample mean

The​ _______ is the best point estimate of the population mean.

test statistic

The​ ___________ is a value used in making a decision about the null hypothesis and is found by converting the sample statistic to a score with the assumption that the null hypothesis is true.

The number of degrees of freedom df = 34

Thirty-five different video games showing alcohol use were observed. The duration times of alcohol use ​(in seconds) were recorded. When using this sample for a t test of the claim that the population mean is greater than 73 ​sec, what does df​denote, and what is its​value?

The confidence level should be 98% The given confidence interval contains the value of 90 ​sec, so there is not sufficient evidence to support the claim that the mean is greater than 90 sec.

Twelve different video games showing alcohol use were observed. The duration times of alcohol use were​ recorded, with the times​ (seconds) listed below. Assume that these sample data are used with a 0.01 significance level in a test of the claim that the population mean is greater than 90 sec. If we want to construct a confidence interval to be used for testing that​ claim, what confidence level should be used for a confidence​ interval? If the confidence interval is found to be −44.6 sec<μ<258.1 ​sec, what should we conclude about the​ claim?

The sample observations must be a simple random sample. Either the population is normally​ distributed, or n>​30, or both. No. The sample size is not greater than​ 30, the sample does not appear to be from a normally distributed​ population, and there is not enough information given to determine whether the sample is a simple random sample

Twelve different video games showing drug use were observed. The duration times of drug use were​ recorded, with the times​ (seconds) listed below. What requirements must be satisfied to test the claim that the sample is from a population with a mean greater than 90 ​sec? Are the requirements all​ satisfied?

a.) Find the best point estimate of the population proportion p. x/ n = p 579 / 1079 = .537 z(a/2) 1-.9= .1 / 2 = .05 Stat Calculator Normal P(X > ___) = 0.05 1.645 ​b.) Identify the value of the margin of error E. E = z(a/2)[(pq/n) E = 1.645[ ((.537(1-.537))/ 1079) ​c) Construct the confidence interval. Stat Proportion stats one sample with summary L. Limit < p < U. Limit 0.512 < p < 0.562 ​d) Write a statement that correctly interprets the confidence interval. Choose the correct answer below. One has 90​% confidence that the interval from the lower bound to the upper bound actually does contain the true value of the population proportion.

Use the sample data and confidence level given below to complete parts​ (a) through​ (d). A research institute poll asked respondents if they felt vulnerable to identity theft. In the​ poll, n=1079 and x=579 who said​ "yes." Use a 90% confidence level

The​ P-value separates the critical region from the values that do not lead to rejection of the null hypothesis.

Which of the following is NOT true about​ P-values in hypothesis​ testing?

The confidence interval methods for the mean are robust against departures from​ normality, meaning they work well with distributions that​ aren't normal, provided that departures from normality are not too extreme.

What does it mean to say that the confidence interval methods for the mean are robust against departures from​ normality?

The sample should be a voluntary response or convenience sample.

When analyzing​ polls, which of the following is NOT a​ consideration?

The population​ mean, μ= (upper confidence limit) + (lower confidence limit)

Which of the following calculations is NOT derived from the confidence​ interval?

​Percentage, Probability, and Proportion

Which of the following groups has terms that can be used interchangeably with the​ others?

If the​ P-value is less than​ 0.05, the decision is to reject the null hypothesis.​ Otherwise, we fail to reject the null hypothesis.

Which of the following is NOT a criterion for making a decision in a hypothesis​ test?

The standard deviation of the Student t distribution is s=1.

Which of the following is NOT a property of the Student t​ distribution?

The confidence level is​ 95%

Which of the following is NOT a requirement for constructing a confidence interval for estimating a population mean with σ ​known?

The trials are done without replacement.

Which of the following is NOT a requirement for constructing a confidence interval for estimating the population​ proportion?

The sample​ mean, x is greater than 30.

Which of the following is NOT a requirement for testing a claim about a population mean with σ ​known?

The lowercase​ symbol, p, represents the probability of getting a test statistic at least as extreme as the one representing sample data and is needed to test the claim.

Which of the following is NOT a requirement of testing a claim about a population proportion using a formal method of hypothesis​ testing?

A type I error is making the mistake of rejecting the null hypothesis when it is actually false.

Which of the following is NOT a true statement about error in hypothesis​ testing?

161.7±27.8

Which of the following is NOT an equivalent expression for the confidence interval given by 161.7<μ<​189.5?

A critical value is the area in the​ right-tail region of the standard normal curve.

Which of the following is NOT an observation about critical​ values?

Standard Deviation

Which of the following is NOT needed to determine the minimum sample size required to estimate a population​ proportion?

The size of the​ population, N

Which of the following is NOT required to determine minimum sample size to estimate a population​ mean?

A conclusion based on a confidence interval estimate will be the same as a conclusion based on a hypothesis test

Which of the following is NOT true when testing a claim about a​ proportion?

The Student t distribution has a mean of t=0 and a standard deviation of s=1

Which of the following is not a characteristic of the t​ test?

The population​ mean, μ​, is equal to 1.

Which of the following is not a requirement for testing a claim about a population with σ not​ known?

The​ P-value method and the classical method are not equivalent to the confidence interval method in that they may yield different results.

Which of the following is not true when using the confidence interval method for testing a claim about μ when σ is​ unknown?

We are​ 99% confident that the interval from 4.1 to 5.6 actually does contain the true value of μ.

Which of the following would be a correct interpretation of a​ 99% confidence interval such as 4.1<μ<​5.6?

a. Assume that nothing is known about the percentage of passengers who prefer aisle seats. n = look at picture

You are the operations manager for an airline and you are considering a higher fare level for passengers in aisle seats. How many randomly selected air passengers must you​ survey? Assume that you want to be 95​% confident that the sample percentage is within 2.5 percentage points of the true population percentage. Complete parts​ (a) and​ (b) below.


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