Question

A mail-order catalog firm designed a factorial experiment to test the effect of the size of a magazine advertisement and the advertisement design on the number of catalog requests received (data in thousands). Three advertising designs and two different size advertisements were considered. The data obtained follow. Size of Advertisement Small Large 8 - 12 Design B 17 Use the ANOVA procedure for factorial designs to test for any significant effects due to type of design, size of advertisement, or interaction. Use a = 0.05. Find the value of the test statistic for type of design. Find the p-value for type of design. (Round your answer to three decimal places.) p-value = State your conclusion about type of design. Because the p-value > a = 0.05, type of design is significant. Because the p-value > a = 0.05, type of design is not significant. Because the p-value sa = 0.05, type of design is significant. Because the p-value sa = 0.05, type of design is not significant. Find the value of the test statistic for size of advertisement. Find the p-value for size of advertisement. (Round your answer to three decimal places.) p-value = State your conclusion about size of advertisement. Because the p-value Sa= 0.05, size of advertisement is not significant. Because the p-value > a = 0.05, size of advertisement is not significant. Because the p-value sa = 0.05, size of advertisement is significant. Because the p-value > a = 0.05, size of advertisement is significant.

Answer 1

We input this data in Excel and by using "Anova: Two-Factor With Replication" option under Data > Data Analysis, we carry out the two-way ANOVA analysis. The data and the output are given below. Using the output, we answer the questions given.

Bookl - Mic Home Insert Page Layout Formulas Data Review View De Connections AL AZ Clear Properties Reapply From From From From Other Existing Refresh + Sort Filter Access Web Text Sources Connections All - So Edit Links . Advanced Get External Data Connections Sort & Filter BSB Text to Remove Data Consolidate What If Columns Duplicates Validation Analysis Data Tools Gro A1 A D F G H B Small C Large 1 Anova: Two-Factor With Replication 2 SUMMARY Small Large Total 40 Count Sum Average Variance 2 20 10 8 2 20 10 8 5.333333333 Count Sum Average Variance 24 42 50 92 21 25 23 32 8 18.66666667 Count Sum Average Variance 2 22 11 32 2 38 60 19 8 34.66666667 Total Count Sum Average Variance 6 6 84 108 14 18 44 50.4 SS df MS ANOVA Source of Variation Sample Columns Interaction Within 2 F P-value F crit 172 10.75 0.010386176 5.14325285 3 0.133974596 5.987377584 1 0.421875 5.14325285 Total 520 11



- Test statistic (design) = f (Sample) = 110.75]. p-value (design) = P-valve (campke) = 10010.10.010). Conclusion and option) Test Statistie ( advertisement) = f(colume) = 3). p-value Cadvertisement) = to p-value (columns) = [0.134]. Conclusion - 2nd option Test vatistic (interaction) = f (interaction) = 0 - p-value = 10.422) Cadcanneohelirien - 2nd option canner