[ PLEASE SHOW ALL WORK WITHOUT THE USE OF EXCEL ] From 78 of its restaurants, No
ID: 3067888 • Letter: #
Question
[ PLEASE SHOW ALL WORK WITHOUT THE USE OF EXCEL ]
From 78 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 = .02, is per-person spending independent of percent of sales from potstickers?
The hypothesis for the given issue is H0: Percentage of Sales and Per-Person Spending are independent.
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.)
From 78 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 = .02, is per-person spending independent of percent of sales from potstickers?
Explanation / Answer
a)
Yes
b)
applying chi square test of independence:
test statistic =6.054
df =(row-1)*(column-1)=(3-1)*(3-1)=4
p value =0.1954 ( please try 0.1951 if this comes wrong)
c)
No as p value is not less than 0.02 level
Expected Ei=row*column/total Low medium High Total Low 5.385 6.667 7.949 20 medium 7.000 8.667 10.333 26 High 8.615 10.667 12.718 32 Total 21 26 31 78 chi square =(Oi-Ei)2/Ei Low medium High Total Low 0.485 0.067 0.113 0.665 medium 0.571 2.167 0.527 3.265 High 0.017 1.260 0.847 2.125 Total 1.073 3.494 1.487 6.054