Question

A simple random sample of size n is drawn from a population that is normally distributed. The sample mean, x, is found to be 113, and the sample standard deviation, s, is found to be 10 (a) Construct a 95% confidence interval about if the sample size, n, is 25. (b) Construct a 95% confidence interval about if the sample size, n, is 13 (c) Construct a 90% confidence interval about if the sample size, n, is 25. (d) Could we have computed the confidence intervals in parts (a)-(c) if the population had not been normally distributed? Click the icon to view the table of areas under the t-distribution. (a) Construct a 95% confidence interval about ? if the sample size, n, is 25. Lower bound Upper bound: (Use ascending order. Round to one decimal place as needed.) (b) Construct a 95% confidence interval about if the sample size, n, is 13 Lower bound Upper bound: (Use ascending order. Round to one decimal place as needed.) How does decreasing the sample size affect the margin of error, E? 0 A. As the sample size decreases, the margin of error increases. O B. As the sample size decreases, the margin of error decreases O C. As the sample size decreases, the margin of error stays the same (c) Construct a 90% confidence interval about if the sample size, n, is 25. Lower bound::Upper bound: (Use ascending order. Round to one decimal place as needed.) Compare the results to those obtained in part (a). How does decreasing the level of confidence affect the size of the margin of error, E? O A. As the percent confidence decreases, the size of the interval decreases O B. As the percent confidence decreases, the size of the interval stays the same

Answer 1

mean = 113 , s= 10

a)
n = 25 , t value at 95% = 2.0639

CI = mean +/- t *(s/sqrt(n))
= 113 +/- 2.0639 *(10/sqrt(25))
= (108.9,117.1)

Lower Bound = 108.9 , Upper Bound = 117.1

b)
n = 13 . t value at 95% = 2.1788
CI = mean +/- t *(s/sqrt(n))
= 113 +/- 2.1788 *(10/sqrt(25))
= (106.9,119)

Lower Bound = 106.9 , Upper Bound = 119

As the sample size decreases the margin of error increases

c)
n = 25 , t value at 90% = 1.7109

CI = mean +/- t *(s/sqrt(n))
= 113 +/- 1.7109 *(10/sqrt(25))
= (109.6,116.4)

Lower Bound = 109.6 , Upper Bound = 116.4

as the percent confidence decreases,the size of interval decreases.

d)
N0, the population needs to be normally distributed