In: Statistics and Probability
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
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