1. Customers can replace 1, 2, 3, or 4 tires at a time for their vehicle at Tire Nation, a nationwide tire shop chain. For customers who have had their tires replaced at Tire Nation, the mean number of tires replaced is =μ3.67, with a standard deviation of =σ0.86. Suppose that we will take a random sample of =n6 tire replacement orders. (Each order gives the number of tires being replaced.) Let x represent the sample mean of the 6 tire replacement orders. Consider the sampling distribution of the sample mean x.Complete the following. Do not round any intermediate computations. Write your answers with two decimal places, rounding if needed.(a)Find μx (the mean of the sampling distribution of the sample mean).2. Alcohol withdrawal occurs when a person who uses alcohol excessively suddenly stops the alcohol use. Studies have shown that the onset of withdrawal is experienced a mean of 40 hours after the last drink, with a standard deviation of 17 hours. A sample of 40 people who use alcohol excessively is to be taken. What is the probability that the sample mean time between the last drink and the onset of withdrawal will be 38.2 hours or more?Carry your intermediate computations to at least four decimal places. Round your answer to at least three decimal places.

• Enter the lower and upper limits on the graph to show your confidence interval.
• For the point (), enter the manufacturer’s claim of 7.22 hours.

6. The standard deviation of test scores on a certain achievement test is 11.5. A random sample of 90 scores on this test had a mean of 75.3. Based on this sample, find a 90% confidence interval for the true mean of all scores. Then give its lower limit and upper limit.Carry your intermediate computations to at least three decimal places. Round your answers to one decimal place. (If necessary, consult a list of formulas.)Lower limit:Upper limit:   7. A psychologist is studying the self image of smokers, which she measures by the self-image (SI) score from a personality inventory. She would like to estimate the mean SI score, μ, for the population of all smokers. She plans to take a random sample of SI scores for smokers and estimate μ via this sample. Assuming that the standard deviation of SI scores for the population of all smokers is 90, what is the minimum sample size needed for the psychologist to be 99% confident that her estimate is within 10 of μ?Carry your intermediate computations to at least three decimal places. Write your answer as a whole number (and make sure that it is the minimum whole number that satisfies the requirements).(If necessary, consult a list of formulas.)
8. Do pregnant women give birth the week of their due date? A study claims that 12% of the population of all pregnant women actually gave birth the week of their due date. You are a researcher who wants to test this claim, so you will select a random sample of 60 women who have recently given birth.Follow the steps below to construct a 95% confidence interval for the population proportion of all pregnant women who gave birth the week of their due date. Then state whether the confidence interval you construct contradicts the study’s claim. (If necessary, consult a list of formulas.)(a)Click on “Take Sample” to see the results from the random sample.  Take SampleNumberProportionGave birth the week of due dateDid not give birth the week of due dateEnter the values of the sample size, the point estimate of the population proportion, and the critical value you need for your 95% confidence interval. (Choose the correct critical value from the table of critical values provided.) When you are done, select “Compute”.Critical values=z0.0052.576=z0.0102.326=z0.0251.960=z0.0501.645=z0.1001.282Sample size: Point estimate: Critical value: ComputeStandard error: Margin of error: 95% confidence interval:     (b)Based on your sample, graph the 95% confidence interval for the population proportion of all pregnant women who gave birth the week of their due date.

• Enter the values for the lower and upper limits on the graph to show your confidence interval.
• For the point (), enter the claim 0.12 from the study.

95% confidence interval:01    (c)Does the 95% confidence interval you constructed contradict the claim from the study? Choose the best answer from the choices below. No, the confidence interval does not contradict the claim. The proportion 0.12 from the study is inside the 95% confidence interval. No, the confidence interval does not contradict the claim. The proportion 0.12 from the study is outside the 95% confidence interval. Yes, the confidence interval contradicts the claim. The proportion 0.12 from the study is inside the 95% confidence interval. Yes, the confidence interval contradicts the claim. The proportion 0.12 from the study is outside the 95% confidence interval.
9. A researcher wishes to estimate the proportion of X-ray machines that malfunction. A random sample of 200 machines is taken, and 70 of the machines in the sample malfunction. Based upon this, compute a 90% confidence interval for the proportion of all X-ray machines that malfunction. Then find the lower limit and upper limit of the 90% confidence interval.  Carry your intermediate computations to at least three decimal places. Round your answers to two decimal places. (If necessary, consult a list of formulas.)Lower limit:Upper limit:  10. A coin-operated drink machine was designed to discharge a mean of 6 ounces of coffee per cup. Suppose that we want to carry out a hypothesis test to see if the true mean discharge differs from 6 . State the null hypothesis H0 and the alternative hypothesis H1 that we would use for this test.H0 :  H1 :  11. A decade-old study found that the proportion of high school seniors who felt that “getting rich” was an important personal goal was 70% . Suppose that we have reason to believe that this proportion has changed, and we wish to carry out a hypothesis test to see if our belief can be supported. State the null hypothesis H0 and the alternative hypothesis H1 that we would use for this test.H0 :  H1 :  12. The records of a casualty insurance company show that, in the past, its clients have had a mean of 1.7 auto accidents per day with a variance of 0.0016 . The actuaries of the company claim that the variance of the number of accidents per day is no longer equal to 0.0016 . Suppose that we want to carry out a hypothesis test to see if there is support for the actuaries’ claim. State the null hypothesis H0 and the alternative hypothesis H1 that we would use for this test.H0 :  H1 :  13. Some college graduates employed full-time work more than 40 hours per week, and some work fewer than 40 hours per week. We suspect that the mean number of hours worked per week by college graduates, μ, is different from 40 hours and wish to do a statistical test. We select a random sample of college graduates employed full-time and find that the mean of the sample is 38 hours and that the standard deviation is 6 hours.Based on this information, answer the questions below.  What are the null hypothesis (H0) and the alternative hypothesis (H1) that should be used for the test?  H0: μ is ?less thanless than or equal togreater thangreater than or equal tonot equal toequal to ?40638 H1: μ is ?less thanless than or equal togreater thangreater than or equal tonot equal toequal to ?40638  In the context of this test, what is a Type I error?  A Type I error is ?rejectingfailing to reject the hypothesis that μ is ?less thanless than or equal togreater thangreater than or equal tonot equal toequal to ?40638 when, in fact, μ is ?less thanless than or equal togreater thangreater than or equal tonot equal toequal to ?40638.  Suppose that we decide to reject the null hypothesis. What sort of error might we be making? ?Type IType II
14. Suppose there is a claim that a certain population has a mean, μ, that is different than 9. You want to test this claim. To do so, you collect a large random sample from the population and perform a hypothesis test at the 0.10 level of significance. To start this test, you write the null hypothesis, H0, and the alternative hypothesis, H1, as follows.H0: =μ9H1: ≠μ9Suppose you also know the following information.The value of the test statistic based on the sample is −1.873 (rounded to 3 decimal places). The p-value is 0.061 (rounded to 3 decimal places).(a)Complete the steps below for this hypothesis test.Normal DistributionStep 1: Select one-tailed or two-tailed.One-tailedTwo-tailedStep 2: Enter the test statistic. (Round to 3 decimal places.)Step 3: Shade the area represented by the p-value. Step 4: Enter the p-value. (Round to 3 decimal places.)0.10.20.30.41−12−23−3   (b)Based on your answer to part (a), which statement below is true?Since the p-value is less than (or equal to) the level of significance, the null hypothesis is rejected.Since the p-value is less than (or equal to) the level of significance, the null hypothesis is not rejected.Since the p-value is greater than the level of significance, the null hypothesis is rejected.Since the p-value is greater than the level of significance, the null hypothesis is not rejected.
15. A presidential candidate’s aide estimates that, among all college students, the proportion p who intend to vote in the upcoming election is at least 70%. If 146 out of a random sample of 235 college students expressed an intent to vote, can we reject the aide’s estimate at the 0.1 level of significance?  Perform a one-tailed test. Then complete the parts below.  Carry your intermediate computations to three or more decimal places and round your answers as specified in the table. (If necessary, consult a list of formulas.)(a)State the null hypothesis H0 and the alternative hypothesis H1.H0:H1:(b)Determine the type of test statistic to use.▼(Choose one)(c)Find the value of the test statistic. (Round to three or more decimal places.)(d)Find the critical value. (Round to three or more decimal places.)(e)Can we reject the aide’s estimate that the proportion of college students who intend to vote is at least 70%?Yes No