7. Sampling Distributions & Confidence Intervals: Mean

In Exercises 29–32, determine the minimum sample size n needed to estimate for the values of c, σ, and E.

c = 0.90, σ = 6.8, E = 1.

Long Life? In a survey of 35 adult Americans, it was found that the mean age (in years) that people would like to live to is 87.9 with a standard deviation of 15.5. An analysis of the raw data indicates the distribution is skewed left.

b. Construct and interpret a 95% confidence interval for the mean.

Gas prices are getting more and more expensive. The average gas price, from a random sample of 100 gas stations, was \$3.50. It is assumed that gas prices have a standard deviation of \$0.04. Construct an 80% confidence interval for the true mean gas price in the United States.

In Exercises 35–40, use the standard normal distribution or the t-distribution to construct a 95% confidence interval for the population mean. Justify your decision. If neither distribution can be used, explain why. Interpret the results.

In a random sample of 18 months from January 2011 through December 2020, the mean interest rate for 30-year fixed rate home mortgages was 3.95% and the standard deviation was 0.49%. Assume the interest rates are normally distributed. (Source: Freddie Mac)

Mensa Membership in Mensa requires a score in the top 2% on a standard intelligence test. The Wechsler IQ test is designed for a mean of 100 and a standard deviation of 15, and scores are normally distributed.

c. If 4 subjects take the Wechsler IQ test and they have a mean of 131 but the individual scores are lost, can we conclude that all 4 of them have scores of at least 131?

Freshman 15 Here is a sample of amounts of weight change (kg) of college students in their freshman year (from Data Set 13 “Freshman 15” in Appendix B): 11, 3, 0, , where represents a loss of 2 kg and positive values represent weight gained. Here are ten bootstrap samples:

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a. Using only the ten given bootstrap samples, construct an 80% confidence interval estimate of the mean weight change for the population.

Ages of Moviegoers Find the sample size needed to estimate the mean age of movie patrons, given that we want 98% confidence that the sample mean is within 1.5 years of the population mean. Assume that sigma=19.6 years, based on a previous report from the Motion Picture Association of America. Could the sample be obtained from one movie at one theater?

A nutritionist wants to estimate the average grams of protein in a brand of protein bars. She takes a random sample of 40 protein bars with $\bar{x}=210$ g & knows from prior data that $\sigma =12$. Make a 95% conf. int. for $\mu$.

In Exercises 27–30, find the critical values and for the level of confidence c and sample size n.

c = 0.99, n = 10

Mean IQ of Data Scientists See the preceding exercise, in which we can assume that sigma=15 for the IQ scores. Data scientists are a group with IQ scores that vary less than the IQ scores of the general population. Find the sample size needed to estimate the mean IQ of data scientists, given that we want 98% confidence that the sample mean is within 2 IQ points of the population mean. Does the sample size appear to be practical?

You want to take a trip to Paris. You randomly select 225 flights to Europe and find a mean and sample standard deviation of \$1500 and \$900, respectively. Construct and interpret a 95% confidence interval for the true mean price for a trip to Paris.

[DATA] Family Size A random sample of 60 married couples who have been married 7 years was asked the number of children they have. The results of the survey are as follows:

Note: x̄ = 2.27, s = 1.22.

c. Compute a 99% confidence interval for the mean number of children of all couples who have been married 7 years. Interpret this interval.

Use the standard normal distribution or the t-distribution to construct the indicated confidence interval for the population mean of each data set. Justify your decision. If neither distribution can be used, explain why. Interpret the results.

a. In a random sample of 40 patients, the mean waiting time at a dentist’s office was 20 minutes and the standard deviation was 7.5 minutes. Construct a 95% confidence interval for the population mean.

Mean Assume that we want to use the sample data given in Exercise 1 with the bootstrap method to estimate the population mean. The mean of the values in Exercise 1 is 54.3 seconds, and the mean of all of the tobacco times in Data Set 20 “Alcohol and Tobacco in Movies” from Appendix B is 57.4 seconds. If we use 1000 bootstrap samples and find the corresponding 1000 means, do we expect that those 1000 means will target 54.3 seconds or 57.4 seconds? What does that result suggest about the bootstrap method in this case?

In Exercises 53 and 54, find the mean and standard deviation of the indicated sampling distribution of sample means. Then sketch a graph of the sampling distribution.

The test scores for the Law School Admission Test (LSAT) in a recent year are normally distributed, with a mean of 151.88 and a standard deviation of 9.95. Random samples of size 40 are drawn from this population, and the mean of each sample is determined.