Question

A cell phone company offers two plans to its subscribers. At the time new subscribers sign up, they are asked to provide some demographic information. The mean yearly income for a sample of 40 subscribers to Plan A is $50,500 with a standard deviation of $6,700. For a sample of 25 subscribers to Plan B, the mean income is $53,300 with a standard deviation of $6,300.

At the 0.02 significance level, is it reasonable to conclude the mean income of those selecting Plan B is larger? Hint: For the calculations, assume the Plan A as the first sample.

The test statistic is ____. (Negative value should be indicated by a minus sign. Round your answer to 2 decimal places.)

Answer 1

Test and CI for Two Variances

Method

Null hypothesis Sigma(1) / Sigma(2) = 1
Alternative hypothesis Sigma(1) / Sigma(2) not = 1
Significance level Alpha = 0.02

Statistics

Sample N StDev Variance
1 40 81.854 6700.000
2 25 79.373 6300.000

Ratio of standard deviations = 1.031
Ratio of variances = 1.063

98% Confidence Intervals

CI for
Distribution CI for StDev Variance
of Data Ratio Ratio
Normal (0.652, 1.565) (0.426, 2.448)

Tests

Test
Method DF1 DF2 Statistic P-Value
F Test (normal) 39 24 1.06 0.891

From the above test we see that p-value>0.02 hence it is reasonable to assume  that population variances are equal.

Now we apply two sample t tests when population variances are equal.

Two-Sample T-Test and CI

Sample N Mean StDev SE Mean
1 40 50500 6700 1059
2 25 53300 6300 1260

Difference = mu (1) - mu (2)
Estimate for difference: -2800
98% upper bound for difference: 702
T-Test of difference = 0 (vs <): T-Value = -1.68 P-Value = 0.049 DF = 63
Both use Pooled StDev = 6550.4998

Since p-value=0.049>0.02 so it is not reasonable to conclude the mean income of those selecting Plan B is larger.