Question

Earlier expert noted no sample size given, would the 60 months not be the sample size?

Monthly profits (in million dollars) of two phone companies were collected from the past five years (60 months) and some statistics are given in the following table.

Company Mean SD
Rogers $123.7 $25.5
Bell $242.7 $15.4

(a) (2 pts) Find a point estimate for the difference in the average monthly profits of these two phone companies.

(b) (2 pts) What is the margin of error for a 99% confidence interval for the difference in the average monthly profits of these two phone companies?

(c) (2 pts) Construct a 99% confidence interval based on the margin of error in part (b). Can we conclude that there is a difference in the average monthly profits of these two phone companies?

(d) (7 pts) Given the level of significance α = 1%, test whether Bell’s average monthly profit is more than Rogers’.

Answer 1

Given:

Sample sizes, n1 =n2 =60

Sample means, =123.7 and =242.7

Standard deviations, =25.5 and =15.4

(a)

Point estimate for the difference = = = $123.7 - $242.7 = -($119)

(b)

Margin of error, MoE =Z_crit*SE

SE =Standard Error = = =3.8458

Z- critical value at 99% confidence level for a two-tailed case is Z_crit =2.58

Margin of error, MoE =2.58*3.8458 =9.922

(c)

99% confidence interval for the difference between two population means is: = MoE = -119 9.922 =(-128.922, -109.078).

We can conclude that there is a significant difference between the average monthly profits of these two phone companies because the above interval does not include 0.

(d)

H0:

H1: (one-tailed: right-tailed test)

Test statistic, Z = =119/3.8458 =30.94

Critical value of Z at 0.01 significance level for a right-tailed test is Z_crit =2.33

Conclusion: Since Z: 30.94 > Z_crit : 2.33, there is a strong statistical evidence to reject H0 and claim that the average monthly profit of Bell's company is significantly greater than that of Rogers’.