Question

A simple random sample of size n is drawn from a population that is normally distributed. The sample​ mean, x overbar x​, is found to be 109 ​, and the sample standard​ deviation, s, is found to be 10 10. ​(a) Construct a 95​% confidence interval about mu μ if the sample​ size, n, is 15. ​(b) Construct a 95​% confidence interval about mu μ if the sample​ size, n, is 29. ​(c) Construct a a 96​% confidence interval about mu μ if the sample​ size, n, is 15. ​(d) Could we have computed the confidence intervals in parts​ (a)-(c) if the population had not been normally​ distributed? LOADING... Click the icon to view the table of areas under the​ t-distribution. ​(a) Construct a a 95​% confidence interval about mu μ if the sample​ size, n, is 15. Lower​ bound: nothing ​; Upper​ bound: nothing ​(Use ascending order. Round to one decimal place as​ needed.) ​(b) Construct a a 95 95​% confidence interval about mu μ if the sample​ size, n, is 29 29. Lower​ bound: nothing ​; Upper​ bound: nothing ​(Use ascending order. Round to one decimal place as​ needed.) How does increasing increasing the sample size affect the margin of​ error, E? A. As the sample size increases increases​, the margin of error increases increases. B. As the sample size increases increases​, the margin of error decreases decreases. C. As the sample size increases increases​, the margin of error stays the same. ​(c) Construct a a 96 96​% confidence interval about mu μ if the sample​ size, n, is 15 15. Lower​ bound: nothing ​; Upper​ bound: nothing ​(Use ascending order. Round to one decimal place as​ needed.) Compare the results to those obtained in part​ (a). How does increasing increasing the level of confidence affect the size of the margin of​ error, E? A. As the level of confidence increases increases​, the size of the interval increases increases. B. As the level of confidence increases increases​, the size of the interval stays the same. C. As the level of confidence increases increases​, the size of the interval decreases decreases. ​(d) Could we have computed the confidence intervals in parts​ (a)-(c) if the population had not been normally​ distributed? A. ​Yes, the population does not need to be normally distributed. B. ​No, the population needs to be normally distributed. C. ​Yes, the population needs to be normally distributed. D. ​No, the population does not need to be normally distributed.

Answer 1

a)

sample mean 'x̄=	109.000
sample size   n=	15.00
sample std deviation s=	10.000
std error 'sx=s/√n=	2.582

for 95% CI; and 14 df, value of t=		2.145
margin of error E=t*std error    =		5.538
lower bound=sample mean-E =		103.462
Upper bound=sample mean+E =		114.538
from above 95% confidence interval for population mean =(103.5 ,114.5)

b)

for 95% CI; and 28 df, value of t=		2.048
margin of error E=t*std error    =		3.803
lower bound=sample mean-E =		105.197
Upper bound=sample mean+E =		112.803
from above 95% confidence interval for population mean =(105.2,112.8)

c)

for 96% CI; and 14 df, value of t=		2.264
margin of error E=t*std error    =		5.846
lower bound=sample mean-E =		103.154
Upper bound=sample mean+E =		114.846
from above 96% confidence interval for population mean =(103.2 ,114.8)

d)

  B. As the sample size increases, the margin of error decreases

A. As the level of confidence increases increases​, the size of the interval increases s.

B. ​No, the population needs to be normally distributed (since sample size <30)