Question

From 104 of its restaurants, Noodles & Company managers collected data on per-person sales and the percent of sales due to "potstickers" (a popular food item). Both numerical variables failed tests for normality, so they tried a chi-square test. Each variable was converted into ordinal categories (low, medium, high) using cutoff points that produced roughly equal group sizes. At α = .10, is per-person spending independent of percent of sales from potstickers? Potsticker % of Sales Per-Person Spending Low Medium High Row Total Low 14 13 8 35 Medium 11 17 5 33 High 10 8 18 36 Col Total 35 38 31 104 You will need to open the Excel file. Then open Minitab. Copy the data (NOT THE TOTALS) into Minitab. Be sure that the 1st number goes into row 1 in Minitab and that you type the column headings (Low, Medium, High) into the grey shaded top header row in Minitab. PictureClick here for the Excel Data File (a) The hypothesis for the given issue is H0: Percentage of Sales and Per-Person Spending are independent. No Yes (b) Calculate the chi-square test statistic, degrees of freedom, and the p-value. (Round your test statistic value to 2 decimal places and p-value to 4 decimal places. Leave no cells blank - be certain to enter "0" wherever required.) Test statistic d.f. p-value (c) We reject the null and find dependence. No Yes

Answer 1

(a) Yes

(b) The output is:

		Low	Medium	High	Total
Low	Observed	14	13	8	35
	Expected	11.78	12.79	10.43	35.00
	O - E	2.22	0.21	-2.43	0.00
	(O - E)² / E	0.42	0.00	0.57	0.99
Medium	Observed	11	17	5	33
	Expected	11.11	12.06	9.84	33.00
	O - E	-0.11	4.94	-4.84	0.00
	(O - E)² / E	0.00	2.03	2.38	4.40
High	Observed	10	8	18	36
	Expected	12.12	13.15	10.73	36.00
	O - E	-2.12	-5.15	7.27	0.00
	(O - E)² / E	0.37	2.02	4.92	7.31
Total	Observed	35	38	31	104
	Expected	35.00	38.00	31.00	104.00
	O - E	0.00	0.00	0.00	0.00
	(O - E)² / E	0.79	4.05	7.87	12.71
		12.71	chi-square
		4	df
		.0128	p-value

chi-square test statistic = 12.71

degrees of freedom = 4

p-value = 0.0128

(c) Yes