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 38 subscribers to Plan A is $58,400 with a standard deviation of $9,200. For a sample of 31 subscribers to Plan B, the mean income is $62,500 with a standard deviation of $7,100.

At the 0.010 significance level, is it reasonable to conclude the mean income of those selecting Plan B is larger? Assume unequal population standard deviations. Hint: For the calculations, assume the Plan A as the first sample.

A.) What is the decision rule? (negative amounts should be indicated by a minus sign. round to 3 decimal places)

B.) Compute the value of the test statistic.(negative amounts should be indicated by a minus sign. round to 2 decimal places)

Answer 1

Solution:

Given details of a cell phone company

sample of 38 subscribers to Plan A, the mean income is $58,400 with a standard deviation of $9,200

sample of 31 subscribers to Plan B, the mean income is $62,500 with a standard deviation of $7,100.

0.010 significance level

a) the decision rule:

H₀:

     H₁:

DF = (s1^2/n1 + s2^2/n2)^2/((s1^2/n1)^2/(n1 - 1) + ((s2^2/n2)^2/(n2 - 1))

        = ((9200)^2/38 + (7100)^2/31)^2/(((9200)^2/38)^2/37 + ((7100)^2/31)^2/30)

      = 67

With 67 df and 0.01 significance level the critical value is t_{0.01, 67} = -2.383

Reject H₀, it t is less than -2.383

b) value of the test statistic:

The test statistic t = ()/sqrt(s1^2/n1 + s2^2/n2)

                                 = (58400 - 62500)/sqrt((9200)^2/38 + (7100)^2/31)

                                = -2.09