Question

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 Light Heavy Browser Nonbrowser Browser 6 4 8 7 6 6 4 6 5 6 6 8 3 7 5 5 8 7 4 7 6 6 a. Use .05 to test for a difference among mean comfort scores for the three types of browsers. Compute the values identified below (to 2 decimals, if necessary) Sum Squares, Treatment Sum of Squares, Error Mean Squares, Treatment Mean Squares, Error Calculate the value of the test statistic (to 2 decimals, if necessary) The p-value isSelect your answer What is your conclusion? -Select your answer- .05 b. Use Fisher's LSD procedure to compare the comfort levelss of nonbrowsers and light browsers. Use a Compute the LSD critical value (to 2 decimals) What is your conclusion? Select your answer

Answer 1

Answer Date: 24/11/2019 a) The sum of square treatment is, By using Excel software, find sum of square treatment with the help of following steps: Import the data. Choose ANOVA: Single Factor from Data Analysis in Data menu. Choose Input Range and Column in Groups by. Choose Output Range. Click Ok.

Anova: Single Factor SUMMARY Groups Nonbrowser Light Heavy Count 8 8 Sum 58 5 0 38 Average Variance 7.25 1.071429 6.25 1.071429 4.75 1.357143 ANOVA Source of Variation Between Groups Within Groups Total Fcrit 3.4668 SS 25.33333 24.50 49.83333 df MS F P-value 2 12.66667 10.85714 0.000578 2 1 1.166667 23 From the Excel output, the sum of square treatment is 25.33). Te sum of square error is, From Excel output, the sum of square error is 24.50).

The mean square treatment is, From Excel output, the mean square treatment is 12.67|. The mean square error is, From Excel output, the mean square error is 1.17. To test the hypothesis is that there is a difference among mean comfort scores for the three types of browsers at 5% level of significance. The null and alternative hypothesis is, H:%i = 112 = Hz Hq: at leastone u; is unequal. The F-test statistics is, From the Excel output, the F-test statistics is(10.86).

The p-value for this test is, From the Excel output, the p value for this test is 0.0006 less than 0.01. Decision The conclusion is that p-value in this context is less than 0.05 which is 0.0006, so the null hypothesis is rejected at 5% level of significance. There is sufficient evidence to indicate that there is a difference among mean comfort scores for the three types of browsers. The result is statistically significant. b) The Fisher's LSD value is, First, computet critical value then find Fisher's LSD value. The t critical value is,

The sample size is small and two-tailed test. Look in the column headed a = 0.05 and the row headed in the t distribution table by using degree of freedom within is 21 and the t critical value is 2.080. The critical value of Fisher's LSD is, LSD = t , DFWXMS X 11 — nA = 2.080x, 1.17%(6+) = 1.12 The critical value of Fisher's LSD is 1.12). The mean difference table is, Treatments Mean 1 | Nonbrowser-Light | 7.25 Mean 2 6.25 Difference Absolute 1.00 1.00

Decision The statistical decision is that the mean difference is less than critical value of Fisher's LSD which is 1.12, so the null hypothesis is not rejected at 5% level of significance. The mean difference of comfort levels in nonbrowsers and light browsers is not statistically significant.