Statistics Question

CHAPTER 9

ESTIMATION AND CONFIDENCE INTERVALS

1. See definition on page 362 of the text.

3. See definition on page 363 of the text.

5. (a) Standard error of the mean = [pic]; (1 - () = (1 – 0.1) = 0.9. Since the population distribution is approximately normal, the distribution of [pic]is approximately normal. Hence, margin of error = [pic]; 90% confidence interval = [pic]

(b) Standard error of the mean = [pic]; (1 - () = (1 – 0.025) = 0.975. Since sample size is large enough (n = 50 > 30), the distribution of [pic]is approximately normal. Hence, margin of error = [pic]; 97.5% confidence interval = [pic]

(c) Standard error of the mean = [pic]; (1 - () = (1 – 0.01) = 0.99. Since sample size is large enough (n = 60), the distribution of [pic]is approximately normal. Hence, margin of error = [pic]; 99% confidence interval = [pic]

7. Since the population is approximately normally distributed, the distribution of [pic]is approximately normal. Hence, a 90 percent confidence interval estimate for ( is

=[pic]

= [pic]

9. (a) It is the number in the table of Student’s t Distribution in row corresponding to “18” and column corresponding to “[pic]” . It equals 34.8053.

(b) Since the required left tail probability is 0.025, the right tail probability is 1 – 0.025 = 0.955. It is the number in the table of Student’s t Distribution in row corresponding to “30” and column corresponding to “[pic]”. So, L = 16.7908.

(c) Total area to the right of [pic] is 0.99. We want area between [pic] and U to be 0.98. Hence, area to the right of U should be (0.99 – 0.98) = 0.01. Thus, [pic]. It is the number in the table of Student’s t Distribution in row “5” and column “[pic]” and equals 15.0863.

(d) We want area between L and [pic] to be 0.925. Area to the left of [pic] is 0.05. Hence, total area to the right of L should be 0.925 + 0.05 = 0.975. Thus, [pic]. It is the number in...