ANOVA Solutions: Testing Differences in Population Means, Assignments of Statistics

Download ANOVA Solutions: Testing Differences in Population Means and more Assignments Statistics in PDF only on Docsity! STA 6127: Solutions of Exercises for ANOVA (Chap. 12) 1. a) Let µ1, µ2, ..., µ12 be the population mean number of true friends of individuals belonging to the 12 astrological signs. The Null hypothesis is given by H0 : µ1 = µ2 = ... = µ12 and the alternative hypothesis is Ha: at least two of the population means are different. b) The value of the F statistic is pretty small, keeping in mind that the mean of the F distribution is about 1. So, there seems to be only weak evidence against the Null hypothesis. c) Since the P-value is large, there is not much evidence against the null. If the null were true, the probability would be 0.82 of getting an F statistic of .61 or larger. 4. a)(i) The Null hypothesis is given by H0 : µ1 = µ2 = µ3 and the alternative hypothesis is Ha: at least two population means differ. Here µi is the population mean number of good friends of individuals who belong to the ith class of the variable measuring the frequency of visits to a bar or tavern, i =1, 2, 3. (ii) The test statistic is F = 3.03. (iii) The P value is 0.049, giving fairly strong evidence against the null hypothesis. b) The standard deviations are much larger than the means, which suggests the response distributions may be very highly skewed. The test is robust to non-normality, but with extremely skewed distributions the mean is not the must useful measure of center. It may be better to compare the medians. 6. To do this enter the data as two columns viz. “scores” and “Groups”. The first column contain the scores of the students i.e 4, 6, ... , 5. The second column contains indicators of the groups these scores belong to. So the first three elements of this column are 1, the next two are 2 and the last three are 3. Now go to “Analyze”, then “Compare Means”, then “One Way ANOVA”. The dependent variable would be “Scores” and the independent one would be “Groups”. Then click on “Options” and then on “Descriptives”. Click O.K. You should get the result shown in the text. 7. You can do this either by hand (or calculator) or using software. a) The sample means are respectively : 2, 3 and 13. b) The within-group sum of squares is 12. The variance estimate is 12/3 = 4. c) The between-group sum of squares is 148. The variance estimate is 148/2 = 74. d) The Null hypothesis is H0 : µ1 = µ2 = µ3. The Alternative hypothesis is Ha: at least two population means differ. Here, µi denotes the mean repair cost of the ith bumper. The test statistic is F = 74/4 = 18.5 with df1 = 2, df2 = 3. From the F table, the P-value lies between 0.01 and 0.05. There is fairly strong evidence against then null, and it seems that there do exist differences between the mean repair costs of at least one pair of bumpers. e) The required ANOVA table is given by: Table 1: Source Sum of Squares df Mean Square F P Between 148 2 74 18.5 < 0.05 Within 12 3 4 Total 160 5 8. a) The required 95% confidence interval is (2.0 − 3.0) ± 3.182 √ 4.0(1/2 + 1/2) = −1.0 ± 6.4 = (- 7.4, 5.4). Since this C.I. contains 0, there is no significant difference between the mean repair costs of bumpers A and B. b) Here σ̂ is 2, ȳ1 = 2, ȳ2 = 3 and ȳ3 = 13. To achieve overall confidence level of at least 0.95, the Bonferroni method uses error probability for each interval of 0.05/3 = 0.0167. Thus, we use a t-score with two tail probability of 0.0167, or single tail probability 0.0083. From the table, with df=3, we have t0.0083 ' 4.7. So, the 95% Bonferroni simultaneous CI for: (i) µ1 − µ2 is (2 − 3) ± 4.7(2) √ 1/2 + 1/2 = (-10.4, 8.4). (ii) µ1 − µ3 is (2 − 13) ± 4.7(2) √ 1/2 + 1/2 = (-20.4, -1.6). (iii) µ2 − µ3 is (3 − 13) ± 4.7(2) √ 1/2 + 1/2 = (-0.6, -19.4). We can interpret the C.I.s as follows: For (ii) we infer that the mean repair cost of bumper 1 is less than that of bumper 3 by at least 160 dollars and at most 2040 dollars. The other two C.I.s can be interpreted similarly. The 95% confidence applies to the entire set of three C.I.s rather than to each one. Since the last two C.I.s do not contain 0 (and have negative values), we conclude that there is significant evidence that the mean repair cost of bumper 3 is more than that of both bumper 1 and 2. There is no evidence though that the mean repair costs of bumpers 1 and 2 differ. 9. a) Since the standard deviations are the same, the within-groups variability would remain the same. The between-groups variability would decrease, and there is less evidence of differences among the means, so the F test statistic would be smaller. (So, the P-value would be larger.) b) Since the standard deviations have decreased, the within-groups variability would decrease. So, the F-statistic value would increase (and thus the P-value would decrease). c) With larger sample sizes, effects of a particular size are more highly significant (with any significance test). The F statistic would be larger. d) Larger in (a), smaller in (b) and (c). 13. a) Since the P-value is small, there is strong evidence that there exist differences between the mean number of friends (of individuals belonging to the three groups). The F-test is just a global test of inde- pendence between the response and explanatory variables. The conclusion of this test does not specify which means are different or how different are they. So, it is not possible to know the exact nature of dependence between the explanatory variable (nature of happiness) and response variable (number of friends) just by using the F test. b) This implies that we are 95% confident that the population mean number of friends of “very happy” individuals is between 0.7 and 5.3 higher than that of “pretty happy” individuals. c) The Bonferroni intervals are generally wider than the ordinary C.I. because the multiple comparison approach uses a higher confidence level for each separate interval to ensure achieving the simultaneous confidence level. d) Since only the first interval does not contain 0, we conclude that only the very happy and pretty happy groups are significantly different in the population mean number of friends. 20. a) The required regression model is given by : E(Y ) = α + β1Z1 + β2Z2, where Z1 = 1 for bumper A and 0 otherwise while Z2 = 1 for bumper B and 0 otherwise. b) In ANOVA, H0 is µ1 = µ2 = µ3. In regression, it is β1 = β2 = 0. c) The prediction equation ŷ = 13 − 11z1 − 10z2 would give the three sample means of 2, 3, and 13 (when plugging in the dummy variable values). 23. a) Let’s use the 0.05 significance level to make decisions. (i) Since the p-value corresponding to Gender is greater than 0.05, we conclude that there is no significant evidence of a Gender effect. (ii)Since the p-value corresponding to Race is 0.000, we conclude that there is significant evidence of a Race effect. b) Controlling for race, the sample mean number of hours of TV watched by males and females are very similar (2.79 and 2.71 for whites, 4.16 and 4.13 for blacks). Thus, the sample does not show any strong evidence of a gender effect. Controlling for gender, there is a large difference between the sample mean number of hours of TV watched by whites and blacks (2.79 and 4.16 for males, 2.71 and 4.13 for females). So, the sample does seem to have strong evidence of a race effect. Table 4: Analysis Variable : Abortion CHURCH = freq RELIGION = fund N Mean Variance 9 1.5555556 2.2777778 CHURCH = freq RELIGION = nonfund N Mean Variance 6 2.6666667 1.8666667 CHURCH = infreq RELIGION = fund N Mean Variance 3 3.6666667 0.333333 CHURCH = infreq RELIGION = nonfund N Mean Variance 8 5.500000 2.8571429 Model with interaction: Source DF Sum of Squares F Value Pr > F Model 3 68.89316239 10.48 0.0002 Error 22 48.22222222 Corrected Total 25 117.11538462 Source DF Sum of Squares F Value Pr > F CHURCH 1 33.21174004 15.15 0.0008 RELIGION 1 11.77777778 5.37 0.0301 CHURCH*RELIGION 1 0.70859539 0.32 0.5754 Table 5: Source DF Sum of Squares F Value Pr > F Model 2 68.18456701 16.03 0.0001 Error 23 48.93081761 Corrected Total 25 117.11538462 Source DF Sum of Squares F Value Pr > F CHURCH 1 36.84299191 17.32 0.0004 RELIGION 1 11.06918239 5.20 0.0321 Table 6: Model Summary R R-squared Adj. R-squared Standard Error 0.763 0.582 0.546 1.459 ANOVA S.S df M.S F Sig. Regression 68.185 2 34.092 16.025 0 Residual 48.931 23 2.127 Total 117.115 25 Coefficients Unstandardized Standardized 95% C.I for B B Std.Error Beta t Sig L.B U.B Const. 5.377 0.470 11.445 0.000 4.405 6.349 Religion -1.384 0.607 -0.325 -2.281 0.032 -2.638 -0.129 Church -2.547 0.612 -0.593 -4.162 0.000 -3.813 -1.281 Attd.