Question

A sales manager believes that a firms’ sales representatives should spend about 30 percent of their working days traveling. If they are on the road for much less, new orders decline and the service and the news-gathering functions of the representatives are not adequately met. If they travel for much more than 30 percent of the time, expense accounts eat up any incremental profit. A study involving the 25 sales representatives of the firm for the last five months shows a sample mean traveling time of 27 percent with estimated population standard deviation of 5 percent. After a quick preliminary analysis of the data, the sales manager concludes that the discrepancy of 3 percent between the sample mean of 27 percent and the desired mean of 30 percent is within the margin of error and, therefore, it is negligible.   At the 90% level of confidence do you agree with the manager’s conclusion? Would your conclusion change at the 99% level of confidence? Show the necessary steps and interpret your results for each confidence interval.             ​

Answer 1

Here, we have given that,

X: Travelling time

n= Number of sales representatives=25

=sample mean of traveling time=27 %   

= Population standard deviation= 5 %

(A)

Now, we want to find the 90% confidence interval for population mean

Formula is as follows,

Where

E=Margin of error =

Now,

Degrees of freedom = n-1 = 25-1=24

c=confidence level =0.90

=level of significance=1-c=1-0.90=0.10

and we know that confidence interval is always two tailed

Z-critical = 1.645 using excel NOMRSINV(prob=0.10/2)

Now,

  

=1.645

We get the 90% confidence interval for the population mean

Interpretation:

This is the 90% CI which shows that we have 90% confidence that this population mean will fall within this interval.

Here, At the 90% level of confidence we do not agree with manager's conclusion that  the sales manager concludes that the discrepancy of 3 percent between the sample mean of 27 percent and the desired mean of 30 percent is within the margin of error and, therefore, it is negligible

because Here, population mean == 30 % is not contain in this interval.

(B)

Now, we want to find the 99% confidence interval for population mean

Formula is as follows,

Where

E=Margin of error =

Now,

Degrees of freedom = n-1 = 25-1=24

c=confidence level =0.99

=level of significance=1-c=1-0.99=0.01

and we know that confidence interval is always two tailed

Z-critical =2.576 using excel NOMRSINV(prob=0.01/2)

Now,

  

=2.576

We get the 99% confidence interval for the population mean

Interpretation:

This is the 99% CI which shows that we have 99% confidence that this population mean will fall within this interval.

Here, At the 99% level of confidence we do not agree with manager's conclusion that  the sales manager concludes that the discrepancy of 3 percent between the sample mean of 27 percent and the desired mean of 30 percent is within the margin of error and, therefore, it is negligible

because Here, population mean == 30 % is not contain in this interval.