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# Confidence Interval

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8.20
425 510 629 236 654 200
276 501 811 332 424 674
676 694 710 662 633

a. Find a 90% confidence interval for the mean number of tongue flicks per 20 minutes for all juvenile common lizards. Assume a population standard deviation of 190.0 (note: The sum of the data is 9047.)

8.24 Refer to 8.20

a. Find a 99% confidence interval for m.
b. Why is the confidence interval you found in part (a) longer than the one in Exercise 8.20.
c Which confidence interval yields a more precise estimate of m? Explain your answer.

3.36 A confidence interval for a population mean has a margin of error of 3.4.

a. Determine the length of the confidence interval.
b. If the sample mean is 52.8, obtain the confidence interval.

8.40 A sample of 27 female graduate physical therapy students had a mean of 22.46 percent body fat.

a. Assuming that percent body fat of female graduate physical-therapy students is normally distributed with standard deviation 4.10 percent body fat, determine a 95% confidence interval for the mean percent body fat of all female graduate physical-therapy students.
b. Obtain the margin of error, E, for the confidence interval you found in part (a)
c. Explain the meaning of E in this context in terms of the accuracy of the estimate.
d. Determine the sample size required to have a margin of error of 1.55 percent body fat with a 99% confidence level.

8.46 Professor Stanley obtains estimates for the mean age, m, of all U.S. millionaires. Suppose that one year's study involved a simple random sample of 36 millionaires who mean age was 58.36 years with a sample standard deviation of 13.36 years.

a. If, for next year's study, a confidence interval for m is to have a margin of error of 2 years and a confidence level of 95%, determine the required sample size.
b. Why did you use the sample standard deviation, s = 13.36, in place of o in your solution to part (a)? Why is it permissible to do so?

8.48 Suppose that a simple random sample is taken from a normal population having a standard deviation of 10 for the purpose of obtaining a 95% confidence interval for the mean of the population.
a. If the sample size is 4, obtain the margin of error.
b. Repeat part (a) for a sample size of 16.

8.60. For a t-curve with df = 8, find each t-value and illustrate your results graphically.

a. The t-value having area 0.05 to its right.
b. t0.10
c. The t-value having area 0.01 to its left (Hint: a t-curve is symmetric about 0)
d. The two t-values that divide the area under the curve into a middle 0.95 area and two outside 0.025 areas.

8.66 Gosset introduced what later became know as Student's t-distribution. Gosset used the following data set, which gives the additional sleep in hours obtained by a sample of 10 patients using laevohysocyamine hydrobromide.

1.9 0.8 1.1 0.1 -0.1
4.4 5.5 1.6 4.6 3.4

a. Find a 95% confidence interval for the additional sleep that would be obtained on average for all people using laevohysocyamine hydrobromide. (Note: x = 2.33 hr; s = 2.002 hr.)

8.70 A variable of a population has mean m and standard deviation o. For a sample of size n, under what conditions are the observed values of the studentized and standardized versions of x equal? Explain your answers.

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#### Solution Summary

8.20
425 510 629 236 654 200
276 501 811 332 424 674
676 694 710 662 633

a. Find a 90% confidence interval for the mean number of tongue flicks per 20 minutes for all juvenile common lizards. Assume a population standard deviation of 190.0 (note: The sum of the data is 9047.)

8.24 Refer to 8.20

a. Find a 99% confidence interval for m.
b. Why is the confidence interval you found in part (a) longer than the one in Exercise 8.20.
c Which confidence interval yields a more precise estimate of m? Explain your answer.

3.36 A confidence interval for a population mean has a margin of error of 3.4.

a. Determine the length of the confidence interval.
b. If the sample mean is 52.8, obtain the confidence interval.

8.40 A sample of 27 female graduate physical therapy students had a mean of 22.46 percent body fat.

a. Assuming that percent body fat of female graduate physical-therapy students is normally distributed with standard deviation 4.10 percent body fat, determine a 95% confidence interval for the mean percent body fat of all female graduate physical-therapy students.
b. Obtain the margin of error, E, for the confidence interval you found in part (a)
c. Explain the meaning of E in this context in terms of the accuracy of the estimate.
d. Determine the sample size required to have a margin of error of 1.55 percent body fat with a 99% confidence level.

8.46 Professor Stanley obtains estimates for the mean age, m, of all U.S. millionaires. Suppose that one year's study involved a simple random sample of 36 millionaires who mean age was 58.36 years with a sample standard deviation of 13.36 years.

a. If, for next year's study, a confidence interval for m is to have a margin of error of 2 years and a confidence level of 95%, determine the required sample size.
b. Why did you use the sample standard deviation, s = 13.36, in place of o in your solution to part (a)? Why is it permissible to do so?

8.48 Suppose that a simple random sample is taken from a normal population having a standard deviation of 10 for the purpose of obtaining a 95% confidence interval for the mean of the population.
a. If the sample size is 4, obtain the margin of error.
b. Repeat part (a) for a sample size of 16.

8.60. For a t-curve with df = 8, find each t-value and illustrate your results graphically.

a. The t-value having area 0.05 to its right.
b. t0.10
c. The t-value having area 0.01 to its left (Hint: a t-curve is symmetric about 0)
d. The two t-values that divide the area under the curve into a middle 0.95 area and two outside 0.025 areas.

8.66 Gosset introduced what later became know as Student's t-distribution. Gosset used the following data set, which gives the additional sleep in hours obtained by a sample of 10 patients using laevohysocyamine hydrobromide.

1.9 0.8 1.1 0.1 -0.1
4.4 5.5 1.6 4.6 3.4

a. Find a 95% confidence interval for the additional sleep that would be obtained on average for all people using laevohysocyamine hydrobromide. (Note: x = 2.33 hr; s = 2.002 hr.)