Question

In the Journal of Marketing Research (November 1996), Gupta studied the extent to which the purchase behavior of scanner panels is representative of overall brand preferences. A scanner panel is a sample of households whose purchase data are recorded when a magnetic identification card is presented at a store checkout. The table below gives peanut butter purchase data collected by the A. C. Nielson Company using a panel of 2,500 households in Sioux Falls, South Dakota. The data were collected over 102 weeks. The table also gives the market shares obtained by recording all peanut butter purchases at the same stores during the same period.

Brand	Size	Number of Purchases by Household Panel	Market Shares
Jif	18 oz.	3,193	19.36%
Jif	28	1,876	7.84
Jif	40	792	5.06
Peter Pan	10	4,061	17.64
Skippy	18	6,279	27.16
Skippy	28	1,639	12.54
Skippy	40	1,415	10.40
Total		19,255

Goodness-of-Fit Test
	obs	expected	O – E	(O – E)2/E	% of chisq
	3,193	3,727.768	-534.768	76.715	7.98
	1,876	1,509.592	366.408	88.935	9.25
	792	974.303	-182.303	34.111	3.55
	4,061	3,396.582	664.418	129.969	13.51
	6,279	5,229.658	1,049.342	210.553	21.89
	1,639	2,414.577	-775.577	249.120	25.90
	1,415	2,002.520	-587.520	172.373	17.92
	19,255	19,255.000	.000	961.776	100.00

(a) Show that it is appropriate to carry out a chi-square test.

Each expected value is ≥               

(b) Test to determine whether the purchase behavior of the panel of 2,500 households is consistent with the purchase behavior of the population of all peanut butter purchasers. Assume here that purchase decisions by panel members are reasonably independent, and set α = .05. (Round your answers χ²to 2 decimal places and χ².05 to 3 decimal places.)

χ2χ2
χ2.05χ.052

Answer 1

Solution:-

a)

Each expected value is ≥ 5

	Observed Frequency	Expected Frequency	[(Or,c - Er,c)2/ Er,c]
	3193	3727.768	76.71529286
	1876	1509.592	88.93450844
	792	974.303	34.11093244
	4061	3396.582	129.9692687
	6279	5229.658	210.552704
	1639	2414.577	249.120108
	1415	2002.52	172.3726856
Sum	19255	19255	961.7755001

b)

State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

Null hypothesis: The purchase behavior of the panel of 2,500 households is consistent with the purchase behavior of the population of all peanut butter purchasers.

Alternative hypothesis: The purchase behavior of the panel of 2,500 households is not consistent with the purchase behavior of the population of all peanut butter purchasers.

Formulate an analysis plan. For this analysis, the significance level is 0.05. Using sample data, we will conduct a chi-square goodness of fit test of the null hypothesis.

Analyze sample data. Applying the chi-square goodness of fit test to sample data, we compute the degrees of freedom, the expected frequency counts, and the chi-square test statistic. Based on the chi-square statistic and the degrees of freedom, we determine the P-value.

DF = k - 1 = 8 - 1
D.F = 7

X² = 961.78

X²_Critical = 14.067

Rejection region is X² > 14.067

where DF is the degrees of freedom, k is the number of levels of the categorical variable, n is the number of observations in the sample, E_i is the expected frequency count for level i, O_i is the observed frequency count for level i, and X² is the chi-square test statistic.

Interpret results. Since the X²-value (961.78) lies in the rejection region, hence we have to reject the null hypothesis.