math chapt 4
Define simple random sampling.
A sample of size n from a population of size N is obtained through simple random sampling if every possible sample of size n has an equally likely chance of occurring. The sample is then called a simple random sample.
Suppose a simple random sample of size n=150 is obtained from a population whose size is N=25,000 and whose population proportion with a specified characteristic is p=0.4. Complete parts (a) through (c) below. Describe the sampling distribution of p. Choose the phrase that best describes the shape of the sampling distribution below. What is the probability of obtaining x=66 or more individuals with the characteristic? That is, what is P(p≥0.44)? What is the probability of obtaining x=51 or fewer individuals with the characteristic? That is, what is P(p≤0.34)?
Approximately normal because n≤0.05N and np(1-p)>10 stat calc normal mean = 0.4 std= 0.04 0.1587 0.0668
In a survey, 45% of the respondents stated that they talk to their pets on the telephone. A veterinarian believed this result to be too high, so he randomly selected 230 pet owners and discovered that 97 of them spoke to their pet on the telephone. Does the veterinarian have a right to be skeptical? Use the α=0.1level of significance.
Because np01−p0=56.9>10, the sample size is less than 5% of the population size, and the sample is given to be random, the requirements for testing the hypothesis are satisfied. What are the null and alternative hypotheses? H0:p=0.45 versus H1: p<0.45 Find the test statistic, z0. z0=negative 0.86 Find the P-value. P-value=0.195 Does the veterinarian have a right to be skeptical? A. The veterinarian does not have a right to be skeptical. There is not sufficient evidence to conclude that the true proportion of pet owners who talk to their pets on the telephone is less than 45%
To cut the standard error of the mean in half, the sample size must be doubled.
False. The sample size must be increased by a factor of four to cut the standard error in half.
Two researchers, Jaime and Mariya, are each constructing confidence intervals for the proportion of a population who is left-handed. They find the point estimate is 0.26. Each independently constructed a confidence interval based on the point estimate, but Jaime's interval has a lower bound of 0.199 and an upper bound of 0.321, while Mariya's interval has a lower bound of 0.227 and an upper bound of 0.346. Which interval is wrong? Why?
Mariya's interval is wrong because it is not centered on the point estimate.
Explain the difference between statistical significance and practical significance.
Statistical significance means that the sample statistic is not likely to come from the population whose parameter is stated in the null hypothesis. Practical significance refers to whether the difference between the sample statistic and the parameter stated in the null hypothesis is large enough to be considered important in an application.
The sample proportion, denoted p, is given by the formula p=x/n where x is the number of individuals with a specified characteristic in a sample of n individuals
The sample proportion, denoted p, is given by the formula p=x/n where x is the number of individuals with a specified characteristic in a sample of n individuals
By how many times does the sample size have to be increased to decrease the margin of error by a factor of 1/3
The sample size must be increased by a factor of 9 to decrease the margin of error by a factor of 1/3 ( 3x3) = 9 What is the general relationship, if any, between the sample size and the margin of error? Increasing the sample size by a factor M results in the margin of error decreasing by a factor of 1/ square root M
σx is 10 If the sample size is n=9, what is likely true about the shape of the population? If the sample size is n=9, what is the standard deviation of the population from which the sample was drawn? The standard deviation of the population from which the sample was drawn is
The shape of the population is approximately normal. 30. 10 x square 9
Suppose a polling agency reported that 46.9% of registered voters were in favor of raising income taxes to pay down the national debt. The agency states that results are based on telephone interviews with a random sample of 1048 registered voters. Suppose the agency states the margin of error for 95% confidence is 3.0%. Determine and interpret the confidence interval for the proportion of registered voters who are in favor of raising income taxes to pay down the national debt.
We are 95% confident that the proportion of registered voters in favor of raising income taxes to pay down the national debt is between 0.439 and 0.499 0.469-0.03 = 0.439 0.469 + 0.03= 0.499
A 95% confidence interval may be interpreted by saying there is a 95% probability that the interval includes the unknown parameter. A 95% confidence interval does not mean that there is a 95% probability that the interval contains the parameter. The 95% in a 95% confidence interval represents the proportion of all samples that will result in intervals that include the population proportion. OK
false
In a survey of 2085 adults in a certain country conducted during a period of economic uncertainty, 52% thought that wages paid to workers in industry were too low. The margin of error was 3 percentage points with 95% confidence. For parts (a) through (d) below, which represent a reasonable interpretation of the survey results? For those that are not reasonable, explain the flaw.
We are 95% confident 52% of adults in the country during the period of economic uncertainty felt wages paid to workers in industry were too low. Is the interpretation reasonable? The interpretation is flawed. The interpretation provides no interval about the population proportion. We are 92% to 98% confident 52% of adults in the country during the period of economic uncertainty felt wages paid to workers in industry were too low. Is the interpretation reasonable? The interpretation is flawed. The interpretation indicates that the level of confidence is varying. We are 95% confident that the interval from 0.49 to 0.55 contains the true proportion of adults in the country during the period of economic uncertainty who believed wages paid to workers in industry were too low. Is the interpretation reasonable? The interpretation is reasonable. In 95% of samples of adults in the country during the period of economic uncertainty, the proportion who believed wages paid to workers in industry were too low is between 0.49 and 0.55. Is the interpretation reasonable? The interpretation is flawed. The interpretation suggests that this interval sets the standard for all the other intervals, which is not true.
Several years ago, 47% of parents who had children in grades K-12 were satisfied with the quality of education the students receive. A recent poll asked 1,055 parents who have children in grades K-12 if they were satisfied with the quality of education the students receive. Of the 1,055 surveyed, 483 indicated that they were satisfied. Construct a 95% confidence interval to assess whether this represents evidence that parents' attitudes toward the quality of education have changed. What are the null and alternative hypotheses?
What are the null and alternative hypotheses? H0:p=0.47 versus H1: p≠0.47 Use technology to find the 95% confidence interval. The lower bound is 0.43 The upper bound is 0.49 (stat porp stat with one with summary click Confidence interval for p) What is the correct conclusion? Since the interval contains the proportion stated in the null hypothesis, there is insufficient evidence that parents' attitudes toward the quality of education have changed.
Put the following in order for the most area in the tails of the distribution. (a) Standard Normal Distribution (b) Student's t-Distribution with 5 degrees of freedom. (c) Student's t-Distribution with 10 degrees of freedom.
b c a
(a) When constructing 95% confidence intervals for the mean when the parent population is right skewed and the sample size is small, the proportion of intervals that include the population mean is (above, below, equal to) 0.95. (b) When constructing 95% confidence intervals for the mean when the parent population is right skewed and the sample size is small, the proportion of intervals that include the population mean approaches _____ as the sample size, n, increases.
below 0.95
The level of confidence represents the expected proportion of intervals that will contain the parameter if a large number of different samples of size n is obtained. It is denoted left parenthesis 1 minus alpha right parenthesis times 100 %.
level of confidence, (1-a)x 100
Find the sample variance and standard deviation. 21,13,4,9,10
s2- 39.3 s- 6.3
A simple random sample of size n=46 is obtained from a population with μ=67 and σ=15. (a) What must be true regarding the distribution of the population in order to use the normal model to compute probabilities involving the sample mean? Assuming that this condition is true, describe the sampling distribution of x. (b) Assuming the normal model can be used, determine P(x<71.4). (c) Assuming the normal model can be used, determine P(x≥68.4).
since the sample size is large enough the population size does not have to be normal Approximately normal with mean 67, std 15/ square 46 stat calc normal mean 67. std= 15/square root 46 = 2.2116 (b) P(x<71.4)=0.9767 (c) P(x≥68.4)=0.2634
The Student's t-distribution is symmetric about 0.
symmetric about 0.
Test the hypothesis using the P-value approach. Be sure to verify the requirements of the test. H0: p=0.87 versus H1: p≠0.87 n=500, x=430, α=0.1 Is np01−p0≥10?
yes=56.55 (500 (0.87)(1-0.87) Now find p hat . 430/500 =0.86
Sample: 24,14,4,13,20
μ- population mean xˉ - sample mean - 15
Katrina wants to estimate the proportion of adults who read at least 10 books last year. To do so, she obtains a simple random sample of 100 adults and constructs a 95% confidence interval. Matthew also wants to estimate the proportion of adults who read at least 10 books last year. He obtains a simple random sample of 400 adults and constructs a 99% confidence interval. Assuming both Katrina and Matthew obtained the same point estimate, whose estimate will have the smaller margin of error? Justify your answer.
Matthew's estimate will have the smaller margin of error because the larger sample size more than compensates for the higher level of confidence. Your answer is correct. C.
Suppose a random sample of n=390 teenagers 13 to 17 years of age was asked if they use social media. Of those surveyed, 277 stated that they do use social media. Find the sample proportion of teenagers 13 to 17 years of age who use social media.
0.710 ( 277/390)
Determine the point estimate of the population mean and margin of error for the confidence interval. Lower bound is 18, upper bound is 28. The point estimate of the population mean is 23 The margin of error for the confidence interval is 5
23 18+28/2 5 28-23
A researcher studying public opinion of proposed Social Security changes obtains a simple random sample of 25 adult Americans and asks them whether or not they support the proposed changes. To say that the distribution of the sample proportion of adults who respond yes, is approximately normal, how many more adult Americans does the researcher need to sample in the following cases? (a) 20% of all adult Americans support the changes
37.5 ronded to 38 10/ 0.20 (1-0.20) = 62.5 62.5-25 =
In a trial of 175 patients who received 10-mg doses of a drug daily, 42 reported headache as a side effect. Obtain a point estimate for the population proportion of patients who received 10-mg doses of a drug daily and reported headache as a side effect.
42/175 = 0.24
The horizontal axis in the sampling distribution of p represents all possible sample proportions from a simple random sample of size n. (a) What percent of sample proportions results in a 75% confidence interval that includes the population proportion? (b) What percent of sample proportions results in a 75% confidence interval that does not include the population proportion?
75% 25%
If one hundred 90% confidence intervals are constructed for a population parameter, we would expect.... 90 of the intervals to capture the unknown parameter. A (1−α)•100% confidence interval indicates that (1−α)•100% of all simple random samples of size n from the population whose parameter is unknown will result in an interval that contains the parameter. In this case, the proportion of intervals that should capture the unknown parameter in a 90% confidence interval is 0.9, so we would expect 0.9•100=90 of the intervals to capture the unknown parameter.
90
Some have argued that throwing darts at the stock pages to decide which companies to invest in could be a successful stock-picking strategy. Suppose a researcher decides to test this theory and randomly chooses 50 companies to invest in. After 1 year, 26 of the companies were considered winners; that is, they outperformed other companies in the same investment class. To assess whether the dart-picking strategy resulted in a majority of winners, the researcher tested H0:p=0.5 versus H1: p>0.5 and obtained a P-value of 0.3886. Explain what this P-value means and write a conclusion for the researcher. (Assume α is 0.1 or less.)
About 39 in 100 samples will give a sample proportion as high or higher than the one obtained if the population proportion really is 0.5. Because the P-value is large, do not reject the null hypothesis. There is not sufficient evidence to conclude that the dart-picking strategy resulted in a majority of winners.
Describe the sampling distribution of p. Assume the size of the population is 10,000. n=400, p=0.4 Choose the phrase that best describes the shape of the sampling distribution of p below. Determine the mean of the sampling distribution of p. Determine the standard deviation of the sampling distribution of p.
Approximately normal because n≤0.05N and np(1-p)>10 o.4 0.024. square root 0.4(1-0.4)/ 400
A doctor wants to estimate the mean HDL cholesterol of all 20- to 29-year-old females. How many subjects are needed to estimate the mean HDL cholesterol within 2 points with 99% confidence assuming s=17.5 based on earlier studies? Suppose the doctor would be content with 90% confidence. How does the decrease in confidence affect the sample size required?
A 99% confidence level requires 508 subjects. A 90% confidence level requires 208subjects. e=2 Confidence = 90 width 4 (2.x2) std 17.5 stat z stat one sample with sample size
Explain what a P-value is. What is the criterion for rejecting the null hypothesis using the P-value approach?
A P-value is the probability of observing a sample statistic as extreme or more extreme than the one observed under the assumption that the statement in the null hypothesis is true. If P-value<α, reject the null hypothesis.
In a clinical trial, 18 out of 867 patients taking a prescription drug daily complained of flulike symptoms. Suppose that it is known that 1.6% of patients taking competing drugs complain of flulike symptoms. Is there sufficient evidence to conclude that more than 1.6% of this drug's users experience flulike symptoms as a side effect at the α=0.05 level of significance?
Because np01−p0=13.71 >10, the sample size is less than 5% of the population size, and the sample can be reasonably assumed to be random, the requirements for testing the hypothesis are satisfied. (Round to one decimal place as needed.) What are the null and alternative hypotheses? H0:p = 0.016 versus H1: p>0.0160 Find the test statistic, z0. z0=1.12 Find the P-value. P-value=0.132 Choose the correct conclusion below. A. Since P-value>α, do not reject the null hypothesis and conclude that there is not sufficient evidence that more than 1.6% of the users experience flulike symptoms.
Twenty years ago, 56% of parents of children in high school felt it was a serious problem that high school students were not being taught enough math and science. A recent survey found that 257 of 700 parents of children in high school felt it was a serious problem that high school students were not being taught enough math and science. Do parents feel differently today than they did twenty years ago? Use the α=0.05 level of significance.
Because np01−p0=172.5>10, the sample size is less than 5% of the population size, and the sample can be reasonably assumed to be random, the requirements for testing the hypothesis are satisfied. What are the null and alternative hypotheses? H0:p=0.56 versus H1: p≠0.56 Find the test statistic. z0=negative -10.28 Find the P-value. P-value=0.001 Determine the conclusion for this hypothesis test. Choose the correct answer below. Since P-value<α, reject the null hypothesis and conclude that there is sufficient evidence that parents feel differently today.
The mean waiting time at the drive-through of a fast-food restaurant from the time an order is placed to the time the order is received is 86.7 seconds. A manager devises a new drive-through system that he believes will decrease wait time. As a test, he initiates the new system at his restaurant and measures the wait time for 10 randomly selected orders. The wait times are provided in the table to the right. Complete parts (a) and (b) below.
Because the sample size is small, the manager must verify that the wait time is normally distributed and the sample does not contain any outliers. The normal probability plot is shown below and the sample correlation coefficient is known to be r=0.991. Are the conditions for testing the hypothesis satisfied? Yes, the conditions are satisfied. The normal probability plot is linear enough, since the correlation coefficient is greater than the critical value. In addition, a boxplot does not show any outliers. Is the new system effective? Conduct a hypothesis test using the P-value approach and a level of significance of α=0.05. First determine the appropriate hypotheses. H0:muμ= 86.7 H1:uμ< 86.7 Find the test statistic. ((stat t stat one smaple with data)) t0=negative 1.66 Find the P-value. The P-value is 0.0650.065. Use the α=0.05 level of significance. What can be concluded from the hypothesis test? The P-value is greater than the level of significance so there is not sufficient evidence to conclude the new system is effective.
What does "95% confidence" mean in a 95% confidence interval?
If 100 different confidence intervals are constructed, each based on a different sample of size n from the same population, then we expect 95 of the intervals to include the parameter and 5 to not include the parameter.
In a survey conducted by the Gallup Organization, 1100 adult Americans were asked how many hours they worked in the previous week. Based on the results, a 95% confidence interval for the mean number of hours worked had a lower bound of 42.7 and an upper bound of 44.5. Provide two recommendations for decreasing the margin of error of the interval.
Increase the sample size. Decrease the confidence level.
Test the hypothesis using the P-value approach. Be sure to verify the requirements of the test. H0: p=0.3 versus H1: p>0.3 n=100; x=40, α=0.05
Is np01−p0≥10? yes P-value=0.0145 ( stat proportion stat one sample with summ # of succ 40 out of 100. Ho: 0.03 Ha >0.03 Reject the null hypothesis, because the P-value is less than α.
A simple random sample of size n=57 is obtained from a population that is skewed left with μ=33 and σ=2. Does the population need to be normally distributed for the sampling distribution of x to be approximately normally distributed? Why? What is the sampling distribution of x?
No. The central limit theorem states that regardless of the shape of the underlying population, the sampling distribution of x becomes approximately normal as the sample size, n, increases.
A trade magazine routinely checks the drive-through service times of fast-food restaurants. A 90% confidence interval that results from examining 756 customers in one fast-food chain's drive-through has a lower bound of 174.4 seconds and an upper bound of 178.0 seconds. What does this mean?
One can be 90 confident that the mean drive-through service time of this fast-food chain is between 174.4 seconds and 178.0 seconds.
A credit score is used by credit agencies (such as mortgage companies and banks) to assess the creditworthiness of individuals. Values range from 300 to 850, with a credit score over 700 considered to be a quality credit risk. According to a survey, the mean credit score is 708.3. A credit analyst wondered whether high-income individuals (incomes in excess of $100,000 per year) had higher credit scores. He obtained a random sample of 32 high-income individuals and found the sample mean credit score to be 726.6 with a standard deviation of 83.3. Conduct the appropriate test to determine if high-income individuals have higher credit scores at the α=0.05 level of significance.
State the null and alternative hypotheses. H0: μequals=708.3 H1: μgreater than>708.3 Identify the t-statistic. t0=1.24 Identify the P-value. P-value=0.112 Make a conclusion regarding the hypothesis. Fail to reject the null hypothesis. There is not sufficient evidence to claim that the mean credit score of high-income individuals is greater than 708.3
In a study, researchers wanted to measure the effect of alcohol on the hippocampal region, the portion of the brain responsible for long-term memory storage, in adolescents. The researchers randomly selected 24 adolescents with alcohol use disorders to determine whether the hippocampal volumes in the alcoholic adolescents were less than the normal volume of 9.02 cm3. An analysis of the sample data revealed that the hippocampal volume is approximately normal with no outliers and x=8.19 cm3 and s=0.8 cm3. Conduct the appropriate test at the α=0.01 level of significance.
State the null and alternative hypotheses. H0: μ=9.029.02 H1: μless than<9.029.02 Identify the t-statistic. t0=negative -5.08 (Round to two decimal places as needed.) Identify the P-value. P-value=0.0000.000 (Round to three decimal places as needed.) Make a conclusion regarding the hypothesis. Reject the null hypothesis. There is sufficient evidence to claim that the mean hippocampal volume is less than 9.029.02 cm3.
A group conducted a poll of 2064 likely voters just prior to an election. The results of the survey indicated that candidate A would receive 47% of the popular vote and candidate B would receive 45% of the popular vote. The margin of error was reported to be 4%. The group reported that the race was too close to call. Use the concept of a confidence interval to explain what this means. What does it mean to say the race was too close to call?
The margin of error suggests candidate A may receive between 43% and 51% of the popular vote and candidate B may receive between 41% and 49% of the popular vote. Because the poll estimates overlap when accounting for margin of error, the poll cannot predict the winner.
The data from a simple random sample with 25 observations was used to construct the plots given below. The normal probability plot that was constructed has a correlation coefficient of 0.936. Judge whether a t-interval could be constructed using the data in the sample.
The normal probability plot does not suggest. the data could come from a normal population because 0.936less than<0.959 and the boxplot shows outliers, so a t-interval could not. be constructed.
Determine the point estimate of the population proportion, the margin of error for the following confidence interval, and the number of individuals in the sample with the specified characteristic, x, for the sample size provided. Lower bound=0.136, upper bound=0.364, n=1500
The point estimate of the population proportion is 0.25. ( 0.364+0.136 / 2 ) The margin of error is 0.114. ( 0.364- 0.136 / 2 ) The number of individuals in the sample with the specified characteristic is 375. (1500x 0.25) (point estimate number)
An agricultural researcher is interested in estimating the mean length of the growing season in a region. Treating the last 10 years as a simple random sample, he obtains the following data, which represent the number of days of the growing season. What is the point estimate of the population mean number of days of the growing season? 157 158 150142169186193182 163 156 The point estimate is 165.6
stat summary stat columms mean = 165.6
A simple random sample of size n is drawn from a population that is normally distributed. The sample mean, x, is found to be 113, and the sample standard deviation, s, is found to be 10. (a) Construct a 96% confidence interval about μ if the sample size, n, is 22. (b) Construct a 96% confidence interval about μ if the sample size, n, is 12. (c) Construct a 90% confidence interval about μ if the sample size, n, is 22. (d) Should the confidence intervals in parts (a)-(c) have been computed if the population had not been normally distributed?
stat t stats one sample with summary
In a national survey college students were asked, "How often do you wear a seat belt when riding in a car driven by someone else?" The response frequencies appear in the table to the right. (a) Construct a probability model for seat-belt use by a passenger. (b) Would you consider it unusual to find a college student who never wears a seat belt when riding in a car driven by someone else?
to find probaility add up all the numbers and divide Would you consider it unusual to find a college student who never wears a seat belt when riding in a car driven by someone else? A. Yes, because P(never)<0.05.
Determine μx and σx from the given parameters of the population and sample size. μ=71, σ=16, n=64
μ=71 σ=2 16/ square root 64
The average daily volume of a computer stock in 2011 was μ=35.1 million shares, according to a reliable source. A stock analyst believes that the stock volume in 2014 is different from the 2011 level. Based on a random sample of 40 trading days in 2014, he finds the sample mean to be 25.7 million shares, with a standard deviation of s=11.9 million shares. Test the hypotheses by constructing a 95% confidence interval. Complete parts (a) through (c) below. (a) State the hypotheses for the test. H0: muμ equals= 35.1 million shares H1: muμ not equals≠ 35.1 million shares (b) Construct a 95% confidence interval about the sample mean of stocks traded in 2014. The lower bound is 21.89421.894 million shares. The upper bound is 29.50629.506 million shares. (Round to three decimal places as needed.) (c) Will the researcher reject the null hypothesis? A. Reject the null hypothesis because μ=35.1 million shares does not fall in the confidence interval. Your answer is correct.
(a) State thehypotheses for the test.H0:muμequals=35.1million sharesH1:muμnot equals≠35.1million shares(b) Construct a 95% confidence interval about the sample mean of stocks traded in 2014.The lower bound is 21.89421.894 million shares.The upper bound is 29.50629.506 million shares.(Round to three decimal places as needed.)(c) Will the researcher reject the null hypothesis?A.Rejectthe null hypothesis because μ=35.1 million shares does not fall in the confidence interval.Your answer is correct.
To test H0: μ=107 versus H1: μ≠107 a simple random sample of size n=35 is obtained. Complete parts a through e below. LOADING... Click here to view the t-Distribution Area in Right Tail. (a) Does the population have to be normally distributed to test this No, because n≥30. Yes, because the sample is random. (b) If x=103.9 and s=5.9, compute the test statistic. The test statistic is t0=negative 3.11 (
(d) Approximate the P-value. Choose the correct answer below. 0.002<P-value<0.005 Your answer is correct. Interpret the P-value. Choose the correct answer below. A If 1000 random samples of size n=35 are obtained, about 4 samples are expected to result in a mean as extreme or more extreme than the one observed if μ=107. (e) If the researcher decides to test this hypothesis at the α=0.01 level of significance, will the researcher reject the null hypothesis? Yes
The head of institutional research at a university believed that the mean age of full-time students was declining. In 1995, the mean age of a full-time student was known to be 27.4 years. After looking at the enrollment records of all 4934 full-time students in the current semester, he found that the mean age was 27.1 years, with a standard deviation of 7.3 years. He conducted a hypothesis of H0: μ=27.4 years versus H1: μ<27.4 years and obtained a P-value of 0.0020. He concluded that the mean age of full-time students did decline. Is there anything wrong with his research?
Yes, the head of institutional research has access to the entire population, inference is unnecessary. He can say with 100% confidence that the mean age has decreased.
One of the more popular statistics reported in the media is the president's job approval rating. The approval rating is reported as the proportion of the population who approve of the job that the sitting president is doing and is typically based on a random sample of registered voters where the sample size is the same week to week. Complete parts (a) and (b) below.
This proportion tends to fluctuate from week to week. Name some reasons for the fluctuation in the statistic. Select all valid reasons below. The proportion could be changing due to sampling error; different people are in the sample. The proportion could be changing because people's attitudes are changing. b) A poll showed the approval rating to be 0.43(43%). A second poll based on 2000 randomly selected voters showed that 892 approved of the job the president was doing. Do the results of the second poll indicate that the proportion of voters who approve of the job the president is doing is significantly higher than the original level? Explain. Assume the α=0.1 level of significance. H1:p>0.43 .H0:p=0.43 Find the test statistic for this hypothesis test. z=1.45 Determine the P-value for this hypothesis test. P-value=0.074 State the conclusion for this hypothesis test. A. Reject H0. There is sufficient evidence at the α=0.1 level of significance to conclude that the proportion of voters who approve of the job the president is doing is significantly higher than the original level.
A college entrance exam company determined that a score of 21 on the mathematics portion of the exam suggests that a student is ready for college-level mathematics. To achieve this goal, the company recommends that students take a core curriculum of math courses in high school. Suppose a random sample of 150 students who completed this core set of courses results in a mean math score of 21.7 on the college entrance exam with a standard deviation of 3.4. Do these results suggest that students who complete the core curriculum are ready for college-level mathematics? That is, are they scoring above 21 on the math portion of the exam? Complete parts a) through d) below.
The appropriate null and alternative hypotheses are H0:μ=21 versus H1:μ> 21 Verify that the requirements to perform the test using the t-distribution are satisfied. Check all that apply. The students' test scores were independent of one another. The sample size is larger than 30. The students were randomly sampled. Use the P-value approach at the α=0.05 level of significance to test the hypotheses in part (a). Identify the test statistic. t0 (stat t stat one with sum = 2.52 Identify the P-value. P-value=0.006 d) Write a conclusion based on the results. Choose the correct answer below. Rejectt he null hypothesis and claim that there is sufficient evidence to conclude that the population mean is greater than 21.
Suppose the mean IQ score of people in a certain country is 101. Suppose the director of a college obtains a simple random sample of 43 students from that country and finds the mean IQ is 105.6 with a standard deviation of 13.1. Complete parts (a) through (d) below. (a) Consider the hypotheses H0: μ=101 versus H1: μ>101. Explain what the director is testing. Perform the test at the α=0.05 level of significance. Write a conclusion for the test.
The director is testing if the sample provided sufficient evidence that the population mean IQ score is actually greater than 101. Find the test statistic for this hypothesis test. 2.31 ( stat t stat one sam sum Find the P-value for this hypothesis test. 0.013. D. Reject H0.There is sufficient evidence to conclude that the mean IQ score of people in the country is greater than 102 at the α=0.05 level of significance. Consider the hypotheses H0: μ=103 versus H1: μ>103. Explain what the director is testing. Perform the test at the α=0.05 level of significance. Write a conclusion for the test. Explain what the director is testing. Choose the correct answer below. The director is testing if the sample provided sufficient evidence that the population mean IQ score is actually greater than 103. Find the test statistic for this hypothesis test. 1.301.30 (Round to two decimal places as needed.) Find the P-value for this hypothesis test. 0.1000.100 (Round to three decimal places as needed.) . Do not reject H0. There is not sufficient evidence to conclude that the mean IQ score of people in the country is greater than 103 at the α=0.05 level of significance. Based on the results of parts (a)-(c), write a few sentences that explain the difference between "accepting" the statement in the null hypothesis versus "not rejecting" the statement in the null hypothesis. One should not reject rather than accept the null hypothesis. If one accepts the null hypothesis, this indicates that the population mean is a specific value, such as 101, 102, or 103, and so the same data have been used to conclude that the population mean is three different values. If one does not reject the null hypothesis, this indicates that the population mean could be 101, 102, or 103 or even some other value; we are simply not ruling them out as the value of the population mean. Therefore, accepting the null hypothesis can lead to contradictory conclusions, whereas not rejecting does not.
According to a study conducted by a statistical organization, the proportion of people who are satisfied with the way things are going in their lives is 0.80. Suppose that a random sample of 100 people is obtained. Complete parts (a) through (e) below. Suppose the random sample of 100 people is asked, "Are you satisfied with the way things are going in your life?" Is the response to this question qualitative or quantitative? Explain. Describe the sampling distribution of p, the proportion of people who are satisfied with the way things are going in their life. Be sure to verify the model requirements. Since the sample size is no more than 5% of the population size and np(1−p)=16≥10, the distribution of p is approximately normal with μp=0.800 and σp=0.04 In the sample obtained in part (a), what is the probability that the proportion who are satisfied with the way things are going in their life exceeds 0.83? The probability that the proportion who are satisfied with the way things are going in their life exceeds 0.83 is Using the distribution from part (c), would it be unusual for a survey of 100 people to reveal that 74 or fewer people in the sample are satisfied with their lives? The probability that 74 or fewer people in the sample are satisfied is 0.0668, which is not unusual because this probability is not less than 5%.
The response is qualitative because the responses can be classified based on the characteristic of being satisfied or not. The sample proportion p is a random variable because the value of p varies from sample to sample. The variability is due to the fact that different people feel differently regarding their satisfaction. no more, 16 (. 100(0.80)(1-0.80), approximately normal, 0.800, 0.04 0.2266. stat calc normal mean 0.800 std 0.04 >0.83 0.0668 stat calc mean 0.800 std 0.04 <74, is not , is not 5 %
