In Exercise? ?use analysis of variance for the indicated test. Clancy, Rowling, Tolstoy Readability? Pages were randomly selected by the author from ?The Bear and the Dragon? by Tom Clancy, ?Harry Potter and the Sorcerer’s Stone? by J. K. Rowling, and ?War and Peace? by Leo Tolstoy. The Flesch Reading Ease scores for those pages are listed below. Use a 0.05 significance level to test the claim that the three samples are from populations with the same mean. Do the books appear to have different reading levels of difficulty? C l a n c y R o w l i n g T o l s t o y

Solution 12BSC Step 1 The given problem explain about Clancy, Rowling, Tolstoy Readability pages. We have to check assumptions: 1. The Shapiro-wilk statistic values are First mile is .165, second mile is .212 and third mile is .055 respectively. Then these values are greater than the level of significance 0.05 so the data is found to be normal. Tests of Norm ality Sh Kolm ap ogoro iro v-Smi - rnov a Wi lk Statist Stati df Sig. df Sig. ic stic Clancy .236 12 .065 .901 12 .165 Rowlin .200 .156 12 .910 12 .212 g * Tolsto .201 12 .196 .864 12 .055 y 2. Then we know that mean and standard deviations from below table. Std. N Mean Deviation Clancy 12 70.7333 11.32610 Rowling 12 80.7500 4.68139 Tolstoy 12 66.1500 7.85800 From the above table The standard deviations for the miles first one is 11.3261,second one is 4.6813 and third one is 7.8580. Hence, we can not assume that there is homogeneity of variance among samples only by looking at standard deviations here. Then we have to check this by Levene’s test in SPSS (Analyze-Compare means-one way ANOVA-options-homogeneity of variance test-OK). The SPSS output table is given below : Test of Homogeneity of Variances Levene Statistic df1 df2 Sig. 1.849 2 33 .173 From the above table we know that Significance value of Levene’s statistic. Here,Significance value of Levene’s statistic = .173 We know that the significance level of . Therefore level of significance =0.05 Now, We are comparing Significance value of Levene’s statistic and the significance level of . Hence the Levene’s statistic > the significance level of =0.05 Therefore 0.173 > 0.05 From the information we can say that there exists homogeneity of variance and second assumption is met with. 3.The samples are randomly selected it is mentioned in the question. 4. The samples are independent of each other . 5. The samples are categorized according to one factor difficulty level . Step 2 From the given problem we can assume that the three samples are from populations with the same mean . The null hypothesis is that the three samples are from populations with the same mean = = . 1 2 3 Then ANOVA for the data set table in below. ANOV A Sum of d Mean F Sig. Squares f Square Betwee 1338.00 9.46 n 2 669.001 .001 2 9 Groups Within 2331.38 3 70.648 Groups 7 3 3669.38 3 Total 9 5 The above table is SPSS output for one-way ANOVA table.