Question

Using Excel, find the value of the 95% confidence interval for a single sample using both the z variable and the student's t-distribution where the mean is 10.00 with a standard deviation of 0.50. When calculating the confidence interval for the z-value, assume any value of n is a large sample, and when calculating the confidence interval for the for the student's t-distribution, assume any value of n is a small sample. Compute the confidence intervals for: a. n 50 b. n 100 c. n 1000 (Be sure to list the Excel formulas you used to obtain the z and t-values.)

Answer 1

SolutionA:

with t intervals

margin of error

=CONFIDENCE.T(0.05;0.5;50)

=0.142098

lower limit=sample mean-margin of error

=10-0.142098

95% confidence interval lower limit for mean using t interval =9.857901572

upper limit=sample mean+margin of error

=10+0.142098

  95% confidence interval upper limit for mean using t interval    =10.14209843

For z interval use confidence .norm

margin of error==CONFIDENCE.NORM(0.05;0.5;50)=0.13859

lower limit=sample mean-margin of error

=10-0.13859

95% confidence interval lower limit for mean using z interval

=9.861409618

95% confidence interval upper limit for mean using z interval

upper limit=sample mean+margin of error

10.13859

Solutionb:

margin of error==CONFIDENCE.T(0.05;0.5;100)

=0.099210848

lower limit=sample mean-margin of error	9.900789
upper limit=sample mean+margin of error	10.09921

using Z interval:

margin of error=CONFIDENCE.NORM(0.05;0.5;100)=0.097998
lower limit=sample mean-margin of error	9.902002
upper limit=sample mean+margin of error	10.098

Solutionc:

with t intervals:

excel formula for margin of error=CONFIDENCE.T(0.05;0.5;1000)=0.031027

	n=1000
margin of error	0.031027
lower limit=sample mean-margin of error	9.968973
upper limit=sample mean+margin of error	10.03103

with z intervals:

excel formula to get z interval:

=CONFIDENCE.NORM(0.05;0.5;1000)

=0.03099

	n=1000
margin of error	0.03099
lower limit=sample mean-margin of error	9.96901
upper limit=sample mean+margin of error	10.03099