Question

A vehicle quality survey asked new owners a variety of questions about their recently purchased automobile. One question asked for the owner’s rating of the vehicle using categorical responses of average, outstanding, and exceptional. Another question asked for the owner’s education level with the categorical responses some high school, high school graduate, some college, and college graduate. Assume the sample data below are for 500 owners who had recently purchased an automobile.

	Education
Quality Rating	Some HS	HS Grad	Some College	College Grad
Average	35	30	25	65
Outstanding	45	45	50	95
Exceptional	20	25	25	40

a. Use a .05 level of significance and a test of independence to determine if a new owner's vehicle quality rating is independent of the owner's education.

Compute the value of the x^2 test statistic (to 2 decimals). _________

The p-value is _________

What is your conclusion?

__________(Cannot Conclude/Conclude) that the quality rating is not independent of the education of the owner.

b. Use the overall percentage of average, outstanding, and exceptional ratings to comment upon how new owners rate the quality of their recently purchased automobiles.

Average	__________%
Outstanding	__________%
Exceptional	__________%

New owners ________(do not appear/appear) to be satisfied with the recent purchase of their automobile. _________% of owners rated their automobile as Outstanding or Exceptional.

Answer 1

a) The hypotheses are:

H0: quality rating is independent of the education of the owner

Ha: quality rating is not independent of the education of the owner

Observed frequencies are:

Observed	Some HS	HS Grad	Some College	College Grad	Total
Average	35	30	25	65	155
Outstanding	45	45	50	95	235
Exceptional	20	25	25	40	110
Total	100	100	100	200	500

For expected frequencies, use formula:

E=row sum * column sum/N

For example, for first cell, row sum = 155, column sum = 100, and N = 500.

E=row sum * column sum/N = 155*100/500 = 31.00.

Using same formula, we get:

Expected	Some HS	HS Grad	Some College	College Grad	Total
Average	31.00	31.00	31.00	62.00	155
Outstanding	47.00	47.00	47.00	94.00	235
Exceptional	22.00	22.00	22.00	44.00	110
Total	100	100	100	200	500

Now we will calculate chi square contribution for each cell.

Use formula (O-E)²/E.

For example, for first cell, O=35 and E=31.00.

(O-E)²/E = (35-31.00)²/31.00 = 0.52

For test statistic, add these values.

Using same formula, we get:

Chi square	Some HS	HS Grad	Some College	College Grad	Total
Average	0.52	0.03	1.16	0.15	1.85
Outstanding	0.09	0.09	0.19	0.01	0.37
Exceptional	0.18	0.41	0.41	0.36	1.36
Total	0.78	0.53	1.76	0.52	3.59

Compute the value of the x^2 test statistic (to 2 decimals). 3.59

There are r=3 rows and c=4 columns.

The degree of freedom is:

df = (r-1)(c-1) = (3-1)(4-1) = 6

For p-value, use excel formula =CHIDIST(3.59,6).

The p-value is 0.732

What is your conclusion?

Since p-value = 0.732 > 0.05, we Cannot Conclude that the quality rating is not independent of the education of the owner.

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b)

Average	155/500 = 0.31 = 31%
Outstanding	235/500 = 0.47 = 47%
Exceptional	110/500 = 0.22 = 22%

New owners appear to be satisfied with the recent purchase of their automobile. 47%+22% = 69% of owners rated their automobile as Outstanding or Exceptional.