Homework SourseIn a study conducted to investigate browsing activity by sho
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.
Use = .05 to test for a difference among mean comfort scores for the three types of browsers.
A.) Compute the values identified below (to 2 decimals, if necessary).
B.) Calculate the value of the test statistic (to 2 decimals, if necessary). ______________
C.)The p-value is ?
D.) What is your conclusion?
Use Fisher\'s LSD procedure to compare the comfort levels of nonbrowsers and light browsers. Use = .05.
E.) Compute the LSD critical value (to 2 decimals). _______________
F.) What is your conclusion?
Conclude that nonbrowsers and light browsers have different mean comfort scores OR Cannot conclude that nonbrowsers and light browsers have different mean comfort scores ?
| Light | Heavy | |||
| Nonbrowser | Browser | Browser | ||
| 3 | 4 | 7 | ||
| 4 | 5 | 9 | ||
| 5 | 4 | 7 | ||
| 2 | 3 | 9 | ||
| 2 | 6 | 6 | ||
| 3 | 3 | 8 | ||
| 4 | 5 | 7 | ||
| 3 | 4 | 9 |
Solution
Ho:there is no significant difference among mean comfort scores for the three types of browsers. H1:there is a significant difference among mean comfort scores for the three types of browsers.
Sum of squares within treatments=[((26)2+(34)2+(62)2)/8]-[(122)2/24]=709.5-620.1667=89.33. Total sum of squares=[32+42+...........+72+92]-[(122)2/24]=734-620.1667=113.83. & Error sum of squares=Total sum of squares-Sum of squares within treatments=113.83-89.33=24.5. The table value of F(2,21) at 0.05 is 3.47. Then the ANOVA table is
Fcal>Ftab, so we reject null hypothesis, i.e., we accept alternative hypothesis. Therefore, there is a significant difference among mean comfort scores for the three types of browsers.
(F)By the similar process
The table value of F(1,14) at 0.05 is 4.60, so we accept null hypothesis . Therefore, there is no significant difference between mean comfort scores among nonbrowsers and light browser.
| Source of variation | sum of squares | degrees of freedom | mean sum of squares | F test |
| Treatments | 89.33 | 2 | 89.33/2=44.67 | 44.67/1.17=38.18 |
| Error | 24.5 | 21 | 24.5/21=1.17 | |
| Total | 113.83 | 23 |
