Question

A simple random sample of size n is drawn from a population that is normally distributed. The sample​ mean, x overbarx, is found to be 110​, and the sample standard​ deviation, s, is found to be 10.

(a) Construct a 96​% confidence interval about μ if the sample​ size, n, is 16.

​(b) Construct a 96​% confidence interval about μ if the sample​ size, n, is 24.

​(c) Construct a 98​% confidence interval about μ if the sample​ size, n, is 16.

​(d) Could we have computed the confidence intervals in parts​ (a)-(c) if the population had not been normally​ distributed?

Answer 1

(please take care of number of decimals in the answer as required)

a)

sample mean 'x̄=		110.000
sample size    n=		16
std deviation s=		10.000
std error ='sx=s/√n=10/√16=		2.5000

for 96% CI; and 15 df, value of t=		2.2485
margin of error E=t*std error    =		5.6214
lower bound=sample mean-E =		104.379
Upper bound=sample mean+E =		115.621
from above 96% confidence interval for population mean =(104.38<μ<115.62)

b)

std error ='sx=s/√n=10/√24=

2.0412

for 96% CI; and 23 df, value of t=		2.1770
margin of error E=t*std error    =		4.4437
lower bound=sample mean-E =		105.556
Upper bound=sample mean+E =		114.444
from above 96% confidence interval for population mean =(105.56<μ<114.44)

c)

for 98% CI; and 15 df, value of t=		2.6025
margin of error E=t*std error    =		6.5062
lower bound=sample mean-E =		103.494
Upper bound=sample mean+E =		116.506
from above 98% confidence interval for population mean =(103.49<μ<116.51)

d)

No , since sample size is less than or equal to 30