From 74 of its restaurants, Noodles & Company managers collected data on per-per

Question

From 74 of its restaurants, Noodles & Company managers collected data on per-person sales and the percent of sales due to "pot stickers" (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 alpha = .05, is per-person spending independent of percent of sales from pot stickers? The hypothesis for the given issue is Hq: Percentage of Sales and Per Person Spending are independent. Calculate the Chi-Square test statistic, degree of freedom and the row -value.(Round your test statistic value to 2 decimal places and row -value to 4 decimal places. Leave no cell blank -certain to enter "0" wherever required.) Test statistic____________df_____________row-value___________We reject the null and find dependence.

Explanation / Answer

Expected Sale of Potstickers = Row total*column total / Total

Like for low-low = 24*24/74 and so on

Per person Spending

Low

Medium

High

Row Total

Low

7.784

8.432

7.784

24

Medium

9.081

9.838

9.081

28

High

7.135

7.730

7.135

22

Col Total

24

26

24

74

Answer a.

Let the percentage of sale and per person spending are independent under H0.

Answer b.

Test statics = Sum[(A-E)^2/E]

= (14-7.784)^2 / 7.784 + (7-8.432)^2 / 8.432 + ... + (15-7.135)^2/7.135

= 33.848

Df = (m-1)*(n–1) = 2*2 = 4

p-value = <0.0001

So, p is smaller than 0.05.

We have sufficient evidence to reject null hypothesis.

Per person Spending

Low

Medium

High

Row Total

Low

7.784

8.432

7.784

24

Medium

9.081

9.838

9.081

28

High

7.135

7.730

7.135

22

Col Total

24

26

24

74

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