In 2011, the industries with the most complaints to the Better Business Bureau were banks, cable and satellite television companies, collection agencies, cellular phone providers, and new car dealerships (USA Today, April 16, 2012), The results for a sample of 200 complaints are contained in the DATAfile named B8B. Click on the datafile logo to reference the data. a. Construct a frequency distribution for the number of complaints by industry, Category Bank 26 Cable Car Cell 42 60 Collection 28 Total b. Using α = .01, conduct a hypothesis test to determine if the probability of a complaint is the same for the five industries. The test-statistic is p-value- What is your conclusion?c. Which industry has the most complaints?

Considering this is a categorical variable, and the hypothesis is that all categories have the same probability, you have to apply a Chi-Square Goodness to Fit test.

Observed frequencies per category

1) Bank: 26

2) Cable: 44

3) Car: 42

4) Cell: 60

5) Collection: 28

Total= 200

Statistical hypotheses:

H₀: P₁=P₂=P₃=P₄=P₅=1/5

H₁: At least one of the proposed proportions is different.

α: 0.01

[tex]X^2= sum \frac{(O_i-E_i)^2}{E_i} ~~X^2_{k-1}[/tex]

For this test the formula for the expected frequencies is:

[tex]E_i= n * P_i[/tex]

So the expected values for each category is:

[tex]E_1= n * P_1= 200*1/5= 40[/tex]

[tex]E_2= n * P_2= 200*1/5= 40[/tex]

[tex]E_3= n * P_3= 200*1/5= 40[/tex]

[tex]E_4= n * P_4= 200*1/5= 40[/tex]

[tex]E_5= n * P_5= 200*1/5= 40[/tex]

[tex]X^2_{H_0}= \frac{(26-40)^2}{40} +\frac{(44-40)^2}{40} +\frac{(42-40)^2}{40} +\frac{(60-40)^2}{40} +\frac{(28-40)^2}{40} = 11.5[/tex]

This test is one tailed and so is its p-value, under a Chi-square with 4 degrees of freedom p-value: 0.021484.

The p-value is less than the significance level so the decision is to reject the null hypothesis.

c. The industry with most complaints is the cellular phone providers

I hope this helps!