Question

A simple random sample of size n is drawn from a population that is normally distributed. The sample​ mean, x overbar​, is found to be 105​, and the sample standard​ deviation, s, is found to be 10. ​(a) Construct a 90​% confidence interval about mu if the sample​ size, n, is 24. ​(b) Construct a 90​% confidence interval about mu if the sample​ size, n, is 20. ​(c) Construct an 80​% confidence interval about mu if the sample​ size, n, is 24. ​(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 90​% confidence interval about mu if the sample​ size, n, is 24. Lower​ bound: nothing​; Upper​ bound: nothing ​(Use ascending order. Round to one decimal place as​ needed.)

Answer 1

a)

sample mean 'x̄=	105.000
sample size   n=	24.00
sample std deviation s=	10.000
std error 'sx=s/√n=	2.0412

for 90% CI; and 23 df, value of t=		1.714
margin of error E=t*std error    =		3.50
lower bound=sample mean-E =		101.50
Upper bound=sample mean+E =		108.50
from above 90% confidence interval for population mean =(101.5, 108.5 )

b)

90% confidence interval for population mean =(101.1 , 108.9 )

c)

80% confidence interval for population mean =(102.3 , 107.7 )

d)

No since sample size is less than 30 , we Could not have computed the confidence intervals