Question

function of advertising expenditures. Historical data for a sample of eight we oss revenue as a Weekly Gross Televison Newspaper Advertising Revenue Advertising ($1000s) ($1000s) ($1000s) 96 6 2.5 90 2 2 95 5 2.5 92 2.5 2.5 95 4 3.3 95 4.5 2.3 95 3.5 4.2 3.5 94 3 istic (to 2 decimals). Use t table. 7 Homework a. Use a 0.01 to test the hypotheses Ho Pr 0 Ha B and/or B2 is not equal to zero for the model y Bo+B11 + Par +e, where 1television advertising ($1000s) 2newspaper advertising ($1000s) Compute the F test statistic (to 2 decimals). Use F table. What is the p-" e? Select your answer-T What is your conclusion? Select your answer- b. Use a 0.05 to test the significance of B. Compute the t test statistic (to 2 decimals). Use t table. What is thep-value? Select your answer- What is your conclusion? Select your answer Should 1 be dropped from the model? - Select your answer c. Use a 0.05 to test the significance of B. Compute the t test statistic (to 2 decimals). Use t table. What is the p-value? - Select your answer- What is your conclusion? Select your answer- Should z be dropped from the model? Select your answer O Icon Key

Answer 1

Date: 09/06/2019 Answer The data represent weekly gross revenue advertising and newspaper advertising as explanatory variables ('s) response variable (y) on television as a) To test the hypothesis is that the overall regression model is significant at 1% level of significance The null and alternative hypothesis is, Ho B0 HBB20 The F-test statistics is By using Excel software, find F-test statistics with the help of following steps: 1) 2) 3) 4) Import the data. Select Data Analysis from Data menu Select Regression and click Ok. Input Y Range and Input X range

Click Ok 5) ANOVA Significance F df SS MS F Regression Residual Total 26.66842464 13.33421 50.06931 0.000493196 5 1.331575361 0.266315 7 28 Coefficients Stan dard Error tStat P-value Lower 95% Upper 95% Intercept Television newspaper 87.85457 85.3333901 0.980781696 87.00549 3.8E-09 82.8122105 0.94998963 1.7027705 1.32638006 0.146422272 9.058595 0.000274 1.26659154 0.263847502 4.800468 0.004882 0.588349946 1.9448331 From the Excel output, the F-test statistics is 50.07 The p-value for this test is, From the Excel output, thep value for this test is |0.0005 Decision

The conclusion is that p-value in this context is less than 0.01 which is 0.0005 hypothesis is rejected at 1% level of significance. There is sufficient evidence to indicate that the overall regression model is significant. The result is statistically significant so the null b) To test the hypothesis is that the slop of television advertising is different from 0 at 5% significance level The null and alternative hypothesis is HaB0 The t-test statistics is, From the Excel output in part (a), the t-test statistics is|9.06 The p-value for this test is From the Excel output in part (a), the p-value for this test is 0.0003

Decision The conclusion is that the p-value in this context is less than 0.05 which is 0.0003, so the null hypothesis is rejected at 5% level of significance. There is sufficient evidence to indicate that the slop of television advertising is different from 0. The result is statistically significant. The slop of television advertising (x1) should not be drop from the model due to the p- value in this context is less than 0.05 which is 0.0003 c) To test the hypothesis is that the slop of newspaper advertising is different from 0 at 5% significance level The null and alternative hypothesis is Ho B2 0 HB0 The t-test statistics is,

From the Excel output in part (a), the t-test statistics is 4.80 The p-value for this test is, From the Excel output in part (a), thep-value for this test is |0.0049 Decision The conclusion is that the p-value in this context is less than 0.05 which is 0.0049, so the null hypothesis is rejected at 5% level of significance. There is sufficient evidence to indicate that the slop of newspaper advertising is different from 0. The result is statistically significant The slop of newspaper advertising (x2) should not be drop from the model due to the p- value in this context is less than 0.05 which is 0.0049.