Two tailed t test for two companies' monthly profits

Question

I have a question revolving around two tailed tests, confidence intervals, etc, and I'm really struggling to figure out how to find the correct answers. For some reason I can't seem to properly wrap my mind around t-tests and two tail tests.
Below you'll find the question and my attempt at some of the answers, which are probably way off.

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

I subtracted the two means (242.7-123.7) and got 119.

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

z = (1 - 0.99)/2

z = 0.01/2

z =0.005

1 - 0.005 = 0.995

p(z < 3.275) = 0.995

From looking online, I found this formula but I couldn't figure out how to fill in the values: ME = t(stdev/√n)

I don't know where to go from here. I know the formulas for the confidence intervals, but I don't understand what I'm supposed to do given that there are two standard devs and means? I don't know whether n is 60 or 2.

(c) Based on the constructed 99% confidence interval in part (b), can we conclude that there is a difference in the average monthly profits of these two phone companies?

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

Answer 1

2-sample pooled t procedure: If you are willing to assume that the two population variances are equal, then use a pooled two-sample t test procedure. Minitab gives the following output for that procedure:

Two-Sample T-Test and CI 

Sample   N   Mean  StDev  SE Mean
1       60  123.7   25.5      3.3
2       60  242.7   15.4      2.0

Difference = μ (1) - μ (2)
Estimate for difference:  -119.00
95% CI for difference:  (-126.62, -111.38)
T-Test of difference = 0 (vs ≠): T-Value = -30.94  
   P-Value = 0.000  DF = 118
Both use Pooled StDev = 21.0643

You can find the formula for the pooled standard deviation (perhaps denoted $s_p)$ in your text.

Notice that Minitab gives a 95% confidence using the value 1.98, which cuts probability 0.025 from the upper tail of Student's t distribution with 118 degrees of freedom. The somewhat longer 99% confidence interval uses the value 2.618, which cuts probability 0.005 from the upper tail of that distribution. These numbers can be found in printed tables or by using a statistical calculator or software.

Also, I believe you are being asked to test $H_0: \mu_1 = \mu_2$ against the one-side alternative $H_a: \mu_1 < \mu_2.$ The output above is for a two-sided test. For a one-sided test you would divide the P-value by 2, but that makes little difference here because the 2-sided P-value is < 0.0005 (indicated as 0.000 in the printout).

Judging from the details you provide, I'm guessing you are supposed to use the pooled 2-sample t procedure. However, in practice many statisticians use the Welch 2-sample procedure discussed below unless they have convincing prior evidence that the population variances are very nearly equal.

Welch 2-sample t procedure: If you are not willing to assume that the two population variances are equal, then you should use the Welch procedure.

When the two sample sizes are equal (as here), the Welch t statistic turns out to be exactly the same as the pooled t statistic (although the formula is a little different).

However, the Welch procedure uses an approximated degrees of freedom, which is generally somewhat smaller than $n_1 + n_2 - 2.$ The (somewhat messy) formula for the degrees of freedom should be shown in your text, if you are supposed to use the Welch procedure.

Minitab output is as follows:

Two-Sample T-Test and CI 

Sample   N   Mean  StDev  SE Mean
1       60  123.7   25.5      3.3
2       60  242.7   15.4      2.0

Difference = μ (1) - μ (2)
Estimate for difference:  -119.00
95% CI for difference:  (-126.63, -111.37)
T-Test of difference = 0 (vs ≠): T-Value = -30.94  
P-Value = 0.000  DF = 96

Notice the smaller degrees of freedom.

Again here, Minitab gives a 95% CI, not a 99% CI. For 96 degrees of freedom the numbers used to make the respective CIs are 1.985 and 2.628.

Answer 2

n is the number of values you have for each mean and in your case they are apparently equal, which

(a) makes comparison of the mean as simple as you have suggested.

(b) the only things you seem to be missing is the value for n, which you state just above your table.

(c) and (d) this should be evident after working out b