Question

please help with parts a - c!! thanks

please check roundings!!

A random sample of 20 purchases showed the amounts in the table in $). The mean is 547 86 and the standard deviation is 519 60 al Construct a 90% confidence interval for the mean purchases of all customers, assuming that the assumptions and conditions for the confidence interval have been met b) How large is the margin of error? c) How would the confidence interval change if you had assumed that the standard deviation was known to be $200

Answer 1

Solution-A:

xbar=47.86

s=19.60

n=20

alpha=0.10

alpha.2=0.10/2=0.05

df=n-1=20-1=19

tcrit in excel

=T.INV(0.05,19)

=1.729132812

90% confidence interval for mean is

xbar-t*s/sqrt(n),xbar+t*s/sqrt(n)

47.86-1.729132812*19.60/sqrt(20),47.86+1.729132812*19.60/sqrt(20)

40.28174,55.43826

(40.28,55.44)

ANSWER(A)

(40.28,55.44)

Solution-b:

margin of error=t*s/sqrt(n)

=1.729132812*19.60/sqrt(20)

=7.578259

ANSWER:

margin of error=7.58

Solution-c:

90% confidence interval for mean is

xbar-t*s/sqrt(n),xbar+t*s/sqrt(n)

47.86-1.729132812*20/sqrt(20),47.86+1.729132812*20/sqrt(20)

40.12708,55.59292

40.13,55.59

ANSWER

(B)the new confidence interval(40.13,55.59) is narrower than interval from part(A)