Question

A manager is trying to figure out which of her top 2 salespeople is the best, as measured by average number of sales made per week. Thus, she’s set up the following hypothesis test that they are equally as good, & if she can reject that hypothesis, she will conclude that the salesperson with the highest average weekly sales (Salesperson #2) is the best. She’s looked back at all the weeks worked for both employees to calculate the statistics below.

sales person 1   

n1 = 80 , x¯1 = 104 ,σ1 = 8.4

Sales person 2

n2 = 70

¯x2 = 106

σ2=7.6

a) Estimate the difference between the two population means.

(b) Provide a 90% confidence interval for this mean difference.

(c) Provide a 95% confidence interval for this mean difference.

(d) Provide a 99% confidence interval for this mean difference.

(e) What is the value of the test statistic (the z -value)?

(f) What is the p-value?

(g) With α = 0.10, do you Reject H0 or Fail to reject H0?

(h) With α = 0.05, do you Reject H0 or Fail to reject H0?

Answer 1

(a) Since the estimate of difference between the two population means is the difference between the corresponding sample means.

Therefore, Estimate of difference between the two population means = x^-2 - x^-1= 106-104 = 2.

(b) The Confidence Interval for mean difference is given by

ni o aionicanu ounL

Now, tabulated value of z at 90% level of significance is 1.645.

Therefore, 90% CI for mean difference is

)21-17 , 6)L (106 ㅢoy)+1.by 51 or.t7Dj 80 70 1337 6 1625

(c) Using the same formula above for tabulated value of Z at 95% level which is = 1.96,

The 95% CI is (-0.5608, 4.5608).

(d) Using formula in question (b) for tabulated Z value at 99% level which is = 2.576,

The 99% CI is(-1.36574, 5.365741).

(e) The test statistic is given as

ni hi

(f) P-value is given by

anHo る以 W) ce st

(g) At 0.10 alpha value the tabulated Z score = 1.645

And our calculated Z score is = -1.530715= Z cal.

Since |Z cal | =1.530715 < 1.645, therefore we fail to reject the null hypothesis and conclude that there is no significant difference between the 2 salespersons.

(h) At 0.05 alpha vale the tabulated Z score = 1.96

And our calculated Z score = Z cal = -1.530715

Since |Z cal | =1.530715 < 1.96, therefore we fail to reject the null hypothesis and conclude that there is no significant difference between the 2 salespersons.