From 104 of its restaurants, Noodles & Company managers collected data on per-person sales and the...

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From 104 of its restaurants, Noodles & Company managerscollected data on per-person sales and the percent of sales due to\"potstickers\" (a popular food item). Both numerical variablesfailed tests for normality, so they tried a chi-square test. Eachvariable 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 frompotstickers? Potsticker % of Sales Per-Person Spending Low MediumHigh Row Total Low 14 13 8 35 Medium 11 17 5 33 High 10 8 18 36 ColTotal 35 38 31 104 You will need to open the Excel file. Then openMinitab. Copy the data (NOT THE TOTALS) into Minitab. Be sure thatthe 1st number goes into row 1 in Minitab and that you type thecolumn headings (Low, Medium, High) into the grey shaded top headerrow in Minitab. PictureClick here for the Excel Data File (a) Thehypothesis for the given issue is H0: Percentage of Sales andPer-Person Spending are independent. No Yes (b) Calculate thechi-square test statistic, degrees of freedom, and the p-value.(Round your test statistic value to 2 decimal places and p-value to4 decimal places. Leave no cells blank - be certain to enter \"0\"wherever required.) Test statistic d.f. p-value (c) We reject thenull and find dependence. No Yes

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3.8 Ratings (592 Votes)

(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


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