Question

From 81 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 = .01, is per-person spending independent of percent of sales from potstickers?

	Potsticker % of Sales
Per-Person Spending	Low			Medium			High			Row Total
Low		13			11			4			28
Medium		5			15			2			22
High		5			12			14			31
Col Total		23			38			20			81

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.

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 =

Answer 1

Null hypothesis , H₀: Per person spending is independent of the percent of sales

Alternative hypothesis,H_a: Per person spending is dependent on the percent of sales

Test statistic :

Where Expexted frequency is Row total * Column total /Grand Total

The value of test statistic is 16.67

Degree of freedom is (r-1)(c-1) = 4

p value is .0022

Since p value is less than .01(the level of significance ) , we reject the null hypothesis and conclude that per person spending and percent of sales from potstickers is not independent of each other