Question

Supplemental One-way ANOVA problems 1. To see social support enhances recovery from surgery.researchers asked 15 people who had recently had heart surgery how many days they used pain pills after their surgery. Data below are those numbers of days for a group of people who have many friends, a group who have a few friends, and a group who had no friends. Test the hypothesis that social support Influences recovery. Use 4 steps and am .05. No friends Many friends Few friends 2. To see different advertisements can get teens to avoid smoking, researchers asked 15 high school students to view one of three different advertisements. One ad showed a diseased lung. one ad showed happy students who said that they don't need to smoke to be happy, and a third ad showed a doctor who talked about the negative health effects of smoking. Data below reflect the students' desire to smoke. Higher numbers indicate higher desire to smoke. Test to see it there are differences in these advertisements. Use 4 steps and a = 05. doctor Diseased lung H appy students 3. To test to see if people are happier in different towns during the winter researchers found people who lived in either Denver, Chicago, Houston, or Miami and gave them a happiness questionnaire. Their happiness scores are below thigher number indicate more happiness Test to see there are differences between towns. Use 4 steps and am 05. Miami Den Chicago Houston DO

Answer 1

Answer Date: 11/11/2019 To test the hypothesis is that social support influences recovery at 5% level of significance. The null and alternative hypothesis is, H :/4 = 12 = Hq: at least one u; is unequal. The F-test statistics is, By using Excel software, find F-test statistics 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.

df ANOVA Source of Variation Between Groups Within Groups Total SS 128.1333333 50.8 178.9333333 2 12 14 MS F P -value Fcrit 64.06667 15.13386 0.000523642 3.885293835 4.233333 From the Excel output, the F-test statistics is 15.13. The p-value for this test is, From the Excel output, the p value for this test is 0.0005. Decision The conclusion is that p-value in this context is less than 0.05 which is 0.0005, so the null hypothesis is rejected at 5% level of significance. There is sufficient evidence to indicate that social support influences recovery. The result is statistically significant.

To test the hypothesis is that there are differences in advertisements at 5% level of significance. The null and alternative hypothesis is, H :11 = 1l2 = Jlz H,: at least one u; is unequal. The F-test statistics is, By using Excel software, find F-test statistics with the help of following steps: 1) 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. df F ANOVA Source of Variation Between Groups Within Groups Total SS 485.7333333 MS 242.8667 6.666667 P-value 7.99594E-06 Fcrit 3.885293834 2 12 14 36.43 565.7333333

From the Excel output, the F-test statistics is 36.43|. The p-value for this test is, From the Excel output, the p value for this test is 0.0000) Decision The conclusion is that p-value in this context is less than 0.05 which is 0.0000, so the null hypothesis is rejected at 5% level of significance. There is sufficient evidence to indicate that there are differences in advertisements. The result is statistically significant. 3) To test the hypothesis is that there are differences between towns at 5% level of significance. The null and alternative hypothesis is, H : [1] = 1l2 = plz = 1l4 He: at least one pl; is unequal.

The F-test statistics is, By using Excel software, find F-test statistics with the help of following steps: 1) 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. SS ANOVA Source of Variation Between Groups Within Groups Total Fcrit 3.238871522 2.8 36.4 39.2 MS F P -value 3 0.933333 0.410256 0.747824274 16 2.275 19 From the Excel output, the F-test statistics is 0.41]. The p-value for this test is, From the Excel output, the p value for this test is 0.7478.

Decision The conclusion is that p-value in this context is higher than 0.05 which is 0.7478, so the null hypothesis is not rejected at 5% level of significance. There is insufficient evidence to indicate that there are differences between towns. The result is not statistically significant.