Section 7.2 Homework

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

161.7 ± 27.8

Assume that all​ grade-point averages are to be standardized on a scale between 0 and 4. How many​ grade-point averages must be obtained so that the sample mean is within 0.012 of the population​ mean? Assume that a 95​% confidence level is desired. If using the range rule of​ thumb, σ can be estimated as range/4 = 4-0/4=1 Does the sample size seem​ practical?

26678 No, because the required sample size is a fairly large number.

Listed below are student evaluation ratings of​ courses, where a rating of 5 is for​ "excellent." The ratings were obtained at one university in a state. Construct a confidence interval using a 90​% confidence level. What does the confidence interval tell about the population of all college students in the​ state? Ratings: 3.5, 2.9, 4.3, 4.5, 2.9, 4.3, 3.4, 4.3, 4.3, 4.2, 4.2, 3.5, 3.5, 3.8, 3.8 What is the confidence interval for the population mean μ​? What does the confidence interval tell about the population of all college students in the​ state? Select the correct choice below​ and, if​ necessary, fill in the answer​ box(es) to complete your choice.

3.59 <μ​< 4.12 The results tell nothing about the population of all college students in the​ state, since the sample is from only one university.

Here are summary statistics for randomly selected weights of newborn​ girls: n= 195​, x̄= 31.4 hg, s= 7.7 hg. Construct a confidence interval estimate of the mean. Use a 99​% confidence level. Are these results very different from the confidence interval 28.7 hg <μ< 33.1 hg with only 19 sample​ values, x̄ =30.9 hg, and s= 3.4 hg?

30.0 hg <μ< 32.8 hg ​ ​No, because the confidence interval limits are similar.

An IQ test is designed so that the mean is 100 and the standard deviation is 22 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 99​% confidence that the sample mean is within 6 IQ points of the true mean. Assume that σ= 22 and determine the required sample size using technology. Then determine if this is a reasonable sample size for a real world calculation.

90 Yes, this number of IQ test scores is a fairly small number.

Assume that we want to construct a confidence interval. Do one of the​ following, as​ appropriate: (a) find the critical value 2tα/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.

Neither the normal distribution nor the t distribution applies.

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

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.

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

The confidence level is​ 95%.

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

The size of the​ population, N

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

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

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

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

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. TInterval ​(13.046,22.15) x̄=17.598 Sx=16.01712719 n=50 Choose the correct answer below.

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.

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 98​% confidence interval estimate of the mean amount of mercury in the population. Does it appear that there is too much mercury in tuna​ sushi? Mercury (ppm) 0.62, 0.78, 0.10, 0.89, 1.24, 0.53, 0.95

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.

Refer to the accompanying data display that results from a sample of airport data speeds in Mbps. TInterval ​(13.046,22.15) x̄=17.598 Sx=16.01712719 n=50 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. b. Identify the best point estimate of μ and the margin of error. 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?

a. 13.05 Mbps<μ< 22.15Mbps b. The point estimate of μ is 17.60 Mbps. The margin of error is E= 4.55 Mbps. c. 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 set and construct a 90​% confidence interval estimate of the mean pulse rate of adult​ females; then do the same for adult males. Compare the results. a. Construct a 90​% confidence interval of the mean pulse rate for adult females. b. Construct a 90​% confidence interval of the mean pulse rate for adult males.

a. 72.6 bpm <μ< 80.0 bpm b. 63.8 bpm <μ< 70.6 bpm c. The confidence intervals do not​ overlap, so it appears that adult females have a significantly higher mean pulse rate than adult males.

Refer to the accompanying data display that results from a sample of airport data speeds in Mbps. TInterval ​(13.046,22.15) x̄= 17.598 Sx= 16.01712719 n= 50 a. What is the number of degrees of freedom that should be used for finding the critical value tα/2​? b. Find the critical value tα/2 corresponding to a​ 95% confidence level. c. Give a brief general description of the number of degrees of freedom.

a. df= 49 b. tα/2​= 2.01 c. 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.

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.

degree of freedom

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

sample mean

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. Here are summary statistics for randomly selected weights of newborn​ girls: n=263​, x̄=30.2 ​hg, s=7.9 hg. The confidence level is 95​%.

tα/2​= 1.97

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 99​%, sigmaσ is not​ known, and the histogram of 57 player salaries​ (in thousands of​ dollars) of football players on a team is as shown.

tα/2​= 2.660

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 90​%, σ= 34033403 thousand​ dollars, and the histogram of 60 player salaries​ (in thousands of​ dollars) of football players on a team is as shown.

zα/2= 1.65

In a test of the effectiveness of garlic for lowering​ cholesterol, 45 subjects were treated with garlic in a processed tablet form. Cholesterol levels were measured before and after the treatment. The changes (before−​after) in their levels of LDL cholesterol​ (in mg/dL) have a mean of 3.4 and a standard deviation of 19.1. Construct a 99​% confidence interval estimate of the mean net change in LDL cholesterol after the garlic treatment. What does the confidence interval suggest about the effectiveness of garlic in reducing LDL​ cholesterol?

−4.26 ​mg/dL <μ< 11.06 ​mg/dL The confidence interval limits contain 0, suggesting that the garlic treatment did not affect the LDL cholesterol levels.