Question

Excel please I don't know how to do it.

6-13. From a sample of 22 graduate students, the mean number of months of work experience prior to entering an MBA program was 34.86. The national standard deviation is known to be 19 months. What is a 95% confidence interval for the population mean? Compute the confidence interval using the appropriate formula and verify your results using the Excel Confidence Intervals template.

6-15. A survey of 26 college freshmen found that they average 6.85 hours of sleep each night. A 90% confidence interval had a margin of error of 0.497.

      a. What are the lower and upper limits of the confidence interval?
b. What was the standard deviation, assuming that the population standard deviation is known?

Answer 1

Answer 6.13: Here, we will apply the compute the Confidence Interval by applying the following formula:

Confidence Interval = Mean ± [ (Z X Std Dev.) / SQRT (Sample Size)]

Where Sample Size = 22 Students,

Mean = 34.86 months,

Std Dev = 19 months,

Z = Value of Z at 95 % Confidence Level = 1.96 (We derived z value from the z-statistic table)

Hence, the confidence interval = 34.86 ± [ (1.96 X 19). / SQRT (22)]

= 34.86 ± ( 37.24 / 4.6904)

= 34.86 ± 7.9396

Thus, UCL = Upper Control Limit = 34.86 + 7.9396 = 42.7996, and

LCL = Lower Control Limit = 34.86 - 7.9396 = 26.9204

Hence, a 95% confidence interval for the given population mean = (26.9204 , 42.7996) (Rounded to 4 decimal places)

For Verification using Excel:

Use the formula as mentioned below, and then press enter:

=CONFIDENCE.NORM(alpha,std dev,sample size)

Take Alpha = 1- 0.95 = 0.05, Std Dev = 19, and sample size = 22

Hence,

=CONFIDENCE.NORM(0.05,19,22)

= 7.9396 (Tallies with the manually calculated answer)

(Note: The excel formula as calculated above gives the allowable increase or decrease in the mean value)