Question

Is HDTV ownership related to quantity of purchases of other electronics? A Best Buy retail outlet collected the following data for a random sample of its recent customers. At α = 0.10, is the frequency of in-store purchases independent of the number of large-screen HDTVs owned (defined as 50 inches or more)?

In-Store Purchases Last Month
HDTVs Owned	None			One			More Than One			Row Total
None		12			13			14			39
One		17			33			30			80
Two or More		18			45			65			128
Col Total		47			91			109			247

(b) Calculate the chi-square test statistic, degrees of freedom, and the p-value. (Round your test statistic value to 2 decimal places and the p-value to 4 decimal places.)

Test statistic
d.f.
p-value

(c) Find the critical value for chi-Square. Refer to the chi-square Appendix E table. (Round your answer to 2 decimal places.)

Critical value            

2)

Oxnard Kortholt, Ltd., employs 50 workers. Research question: At α = .05, do Oxnard employees differ significantly from the national percent distribution?

Health Care Visits	National Percentage			Oxnard Employees Frequency
No visits		15.6			3
1–3 visits		43.9			19
4–9 visits		25.1			13
10 or more visits		15.4			15
Total		100.0			50

) Calculate the chi-square test statistic, degrees of freedom and the p-value. (Round your test statistic value to 3 decimal places and the p-value to 4 decimal places.)

Test statistic
d.f.
p-value

Answer 1

Question 1

Solution:

Here, we have to use chi square test for independence of two categorical variables.

Null hypothesis: H₀: The frequency of in-store purchases independent of the number of large-screen HDTVs owned.

Alternative hypothesis: H_a: The frequency of in-store purchases dependent of the number of large-screen HDTVs owned.

We are given level of significance = α = 0.10

Test statistic formula is given as below:

Chi square = ∑[(O – E)^2/E]

Where, O is observed frequencies and E is expected frequencies.

E = row total * column total / Grand total

We are given

Number of rows = r = 3

Number of columns = c = 3

Degrees of freedom = df = (r – 1)*(c – 1) = 2*2 = 4

d.f. = 4

α = 0.10

Critical value = 7.78

(by using Chi square table or excel)

Calculation tables for test statistic are given as below:

Observed Frequencies
	In-store purchase
HDTV's owned	None	One	> One	Total
None	12	13	14	39
One	17	33	30	80
Two or more	18	45	65	128
Total	47	91	109	247

Expected Frequencies
	In-store purchase
HDTV's owned	None	One	> One	Total
None	7.421053	14.36842	17.21053	39
One	15.22267	29.47368	35.30364	80
Two or more	24.35628	47.15789	56.48583	128
Total	47	91	109	247

Calculations
(O - E)
4.578947	-1.36842	-3.21053
1.777328	3.526316	-5.30364
-6.35628	-2.15789	8.51417

(O - E)^2/E
2.825308	0.130326	0.598906
0.207512	0.421898	0.796763
1.658802	0.098743	1.28335

Test Statistic = Chi square = ∑[(O – E)^2/E] = 8.021608

χ² test statistic = 8.02

P-value = 0.0908

(By using Chi square table or excel)

P-value < α = 0.10

So, we reject the null hypothesis

There is not sufficient evidence to conclude that the frequency of in-store purchases independent of the number of large-screen HDTVs owned.

Question 2

Solution:

Here, we have to use chi square test for goodness of fit.

Null hypothesis: H₀: The Oxnard employees do not differ significantly from the national percent distribution.

Alternative hypothesis: H_a: The Oxnard employees differ significantly from the national percent distribution.

We are given level of significance = α = 0.05

Test statistic formula is given as below:

Chi square = ∑[(O – E)^2/E]

Where, O is observed frequencies and E is expected frequencies.

α = 0.05

Critical value = 7.814727764

(by using Chi square table or excel)

Calculation tables for test statistic are given as below:

Visits	Proportion	O	E	(O - E)^2/E
No visits	0.156	3	7.8	2.953846154
1 to 3	0.439	19	21.95	0.396469248
4 to 9	0.251	13	12.55	0.016135458
10 or more	0.154	15	7.7	6.920779221
Total	1	50	50	10.28723008

Test Statistic = Chi square = ∑[(O – E)^2/E] = 10.28723008

χ² test statistic = 10.287

We are given

N = 4

Degrees of freedom = df = N – 1 = 3

d.f. = 3

P-value = 0.0163

(By using Chi square table or excel)

P-value < α = 0.05

So, we reject the null hypothesis

There is sufficient evidence to conclude that The Oxnard employees differ significantly from the national percent distribution.