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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.)
b. interpret your answer from part (a).

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.)
b. Was the drug effective in increasing sleep? Explain your answer.

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.)
b. interpret your answer from part (a).

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.)
b. Was the drug effective in increasing sleep? Explain your answer.

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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  • BSc , Wuhan Univ. China
  • MA, Shandong Univ.
Recent Feedback
  • "Your solution, looks excellent. I recognize things from previous chapters. I have seen the standard deviation formula you used to get 5.154. I do understand the Central Limit Theorem needs the sample size (n) to be greater than 30, we have 100. I do understand the sample mean(s) of the population will follow a normal distribution, and that CLT states the sample mean of population is the population (mean), we have 143.74. But when and WHY do we use the standard deviation formula where you got 5.154. WHEN & Why use standard deviation of the sample mean. I don't understand, why don't we simply use the "100" I understand that standard deviation is the square root of variance. I do understand that the variance is the square of the differences of each sample data value minus the mean. But somehow, why not use 100, why use standard deviation of sample mean? Please help explain."
  • "excellent work"
  • "Thank you so much for all of your help!!! I will be posting another assignment. Please let me know (once posted), if the credits I'm offering is enough or you ! Thanks again!"
  • "Thank you"
  • "Thank you very much for your valuable time and assistance!"
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