Question

In: Statistics and Probability

From 77 of its restaurants, Noodles & Company managers collected data on per-person sales and the...

From 77 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 11 4 8 23 Medium 6 11 6 23 High 4 13 14 31 Col Total 21 28 28 77 Click 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

Solutions

Expert Solution

using excel>addin>phstat>multiple sample >chi square

we have

Chi-Square Test
Observed Frequencies
per person spending Calculations
percentage of sales Low Medium High Total fo-fe
low 11 6 4 21 4.727273 -0.27273 -4.45455
medium 4 11 13 28 -4.36364 2.636364 1.727273
high 8 6 14 28 -0.36364 -2.36364 2.727273
Total 23 23 31 77
Expected Frequencies
per person spending
percentage of sales Low Medium High Total (fo-fe)^2/fe
6.272727 6.272727 8.454545 21 3.562582 0.011858 2.347019
medium 8.363636 8.363636 11.27273 28 2.27668 0.831028 0.264663
high 8.363636 8.363636 11.27273 28 0.01581 0.667984 0.659824
Total 23 23 31 77
Data
Level of Significance 0.01
Number of Rows 3
Number of Columns 3
Degrees of Freedom 4
Results
Critical Value 13.2767
Chi-Square Test Statistic 10.63745
p-Value 0.030955
Do not reject the null hypothesis
Expected frequency assumption
       is met.

(a) The hypothesis for the given issue is H0: Percentage of Sales and Per-Person Spending are independent yes

(b) the chi-square test statistic=10.64

degrees of freedom = 4

, and the p-value = 0.0310

.(c) since p value is greater tha 0.01 so we don not reject Ho


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