Question

Case Study: Applying a Completely Randomized Design (Detecting Changes in Consumer Comfort)

In a study conducted to investigate browsing activity by shoppers, each shopper was initially classified as a nonbrowser, light browser, or heavy browser. For each shopper, the study obtained a measure to determine how comfortable the shopper was in a store. Higher scores indicated greater comfort. Suppose the following data were collected:

Nonbrowser               Light Browser             Heavy Browser

4                                  5                                  5

5                                  6                                  7

6                                  5                                  5

3                                  4                                  7

3                                  7                                  4

4                                  4                                  6

5                                  6                                  5

4                                  5                                  7

Using a .05 level of significance, conduct a completely randomized design hypothesis test, followed by a post-hoc comparison test if you rejected the null hypothesis. Please provide the 5 steps for both the main effect and the post-hoc test (if required), the Minitab output for each hypothesis test, and state the business implication based upon your analysis. You must use Minitab and the 5 step hypothesis testing process.

Here’s an example of reporting results in APA style:

An one way analysis of variance showed that the effect of noise was significant, F(3,27) = 5.94, p = .007. Post hoc analyses using the Tukey post hoc criterion for significance indicated that the average number of errors was significantly lower in the white noise condition (M = 12.4, SD = 2.26) than in the other two noise conditions (traffic and industrial) combined (M = 13.62, SD = 5.56), F(3, 27) = 7.77, p = .042.

Use this prototype to report results:

A one way analysis of variance showed that the effect of fill in the effect description was/was not significant, F(fill in num and denom df , ) = fill in observed F, p = fill in p-value. Post hoc analyses using the Tukey post hoc criterion for significance indicated that describe any differences and provide the M and SD, F(fill in the df for numerator and denominator) = fill in observed F, p = fill in p-value.

To conduct this test in Minitab, please use the following process:

Stack the data in columns as given above. To begin a one-way ANOVA with Stacked Data, select Stat from the menu bar. Select ANOVA from the pulldown menu. Select One-Way. In the slot Response, list the column containing the observations. In the slot Factor, list the column containing the group identifiers. For multiple comparisons, select Comparisons and make your selection from the dialog box that appears. The multiple comparison options are Tukey’s, Fisher’s, Dunnett’s, or Hsu’s MCB tests. In the multiple comparison dialog box, you can insert the family error rate in the box on the right as a whole number.

Answer 1

First Enter the data in Minitab:

To conduct this test in Minitab, use the following process:

output:

Tukey's Test for Multiple comparsion

Report results:

A one way analysis of variance showed that the effect of fill in the effect description was significant, F(2 , 21 ) = 4.00 ,   p = 0.034 . Post hoc analyses using the Tukey post hoc criterion for significance indicated that the was significantly comfort is Nonbrowser (M = 4.20, SD = 1.035) than the other two Browser.