Question

You want to estimate the mean number of chocolate chips in cookies. The average number of chocolate chips per cookie in a sample of 40 cookies was 23.95, with a sample standard deviation of 2.55. (A) Find a 90% confidence interval for the mean number of chocolate chips using the t-distribution. (B) Find a 90% confidence interval for the mean number of chocolate chips using the standard normal distribution (for this question, assume the population standard deviation is equal to the sample standard deviation). (C) The t-distribution will always give a slightly wider confidence interval than the standard normal distribution. Explain why this is desirable in situations like this, where we don’t know what the population standard deviation is.

Answer 1

A)

sample mean 'x̄=	23.95
sample size   n=	40.00
sample std deviation s=	2.55
std error 'sx=s/√n=	0.4032

for 90% CI; and 39 df, value of t=		1.685
margin of error E=t*std error    =		0.679
lower bound=sample mean-E =		23.27
Upper bound=sample mean+E =		24.63
from above 90% confidence interval for population mean =(23.27,24.63)

B)

for 90 % CI value of z=		1.645
margin of error E=z*std error =		0.663
lower bound=sample mean-E=		23.2868
Upper bound=sample mean+E=		24.6132
from above 90% confidence interval for population mean =(23.29,24.61)

c)

since we do not know population standard deviation, therefore to compensate in the variability of it , it is better to have a little wider interval so that our confidence level does provide the interval which it meant for.