assignment6 anova answer · for example, q1.3 asks you to use s2(n −1) to get sst. s is the...

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IUPUI/SPEA (Fall 2007) K300 (4392) Statistical Techniques 1 K300 (4392) Statistical Techniques (Fall 2007) Assignment 6: ANOVA and Chi-Square test (190 points, Due November 14) Instructor: Hun Myoung Park [email protected], (317) 274-0573 Instructor’s Comment The emphasis of assignments 5 through 8 is given to understanding of both theoretical and practical aspects. You are asked to read SPSS output and analyze using it, and then to replicate the output manually. During the first stage, you should be able to read the SPSS output correctly to draw information. This is what public managers are supposed to do. Otherwise, you do not know what you are doing. In the second stage, you should be able to understand how SPSS produces the output. Question 1 Several of you did not read questions carefully. For example, Q1.3 asks you to use ) 1 ( 2 n s to get SST. s is the overall standard deviation of 15.582, while n is the number of overall observations 19. Variance is standard deviation squared. Overall variance is not 15.582, but 15.582 2 . Q1.9 asks you to conduct hypothesis test. 3 2 1 μ μ μ IS NOT the denial of 3 2 1 0 : μ μ μ = = H ; it is only one case of the denial. You need to use narrative alternative hypothesis; for example, at least one group mean is different from others. (the same in the chi square test) You should read F score (critical value) from the table on page 639 (not p. 640) as specified in the question. This is a two tailed test. I do not understand why many of you ignored the instruction; please read and follow instructions carefully. A p- value is, simply say, a probability of risk taking; again, therefore, a p-value SHOULD NOT be smaller than zero and larger than 1. Also you may not ignore step 4 and 5. You have to say either “Reject the null hypothesis at the .05 level” or “Do not reject the null hypothesis at the .05 level” in step 4. The final step is substantive interpretation. You need to go back to step 1 and re-paraphrase the null or alternative hypothesis. DO NOT say “reject the null hypothesis” again as substantive interpretation; the phrase does not have any substance. Who can know what you are testing? If you are not sure what is going on hypothesis testing, PLEASE read through the lecture note 1 (hypothesis testing) or come and see me. For Q1.10, you need to copy all 19 observations from question 28 on page 610. Do not ignore “…” If you cannot replicate SPSS output, you should ask yourself, “what is wrong in my computation?” As mentioned, x is the overall sample mean and j x is the sample mean of group j. So you have to compute four means (one overall mean and three group means).

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Page 1: Assignment6 ANOVA Answer · For example, Q1.3 asks you to use s2(n −1) to get SST. s is the overall standard deviation of 15.582, while n is the number of overall observations 19

IUPUI/SPEA (Fall 2007) K300 (4392) Statistical Techniques

1

K300 (4392) Statistical Techniques (Fall 2007) Assignment 6: ANOVA and Chi-Square test (190 points, Due November 14)

Instructor: Hun Myoung Park [email protected], (317) 274-0573

Instructor’s Comment The emphasis of assignments 5 through 8 is given to understanding of both theoretical and practical aspects. You are asked to read SPSS output and analyze using it, and then to replicate the output manually. During the first stage, you should be able to read the SPSS output correctly to draw information. This is what public managers are supposed to do. Otherwise, you do not know what you are doing. In the second stage, you should be able to understand how SPSS produces the output. Question 1 Several of you did not read questions carefully. For example, Q1.3 asks you to use

)1(2 −ns to get SST. s is the overall standard deviation of 15.582, while n is the number of overall observations 19. Variance is standard deviation squared. Overall variance is not 15.582, but 15.5822. Q1.9 asks you to conduct hypothesis test. 321 μμμ ≠≠ IS NOT the denial of

3210 : μμμ ==H ; it is only one case of the denial. You need to use narrative alternative hypothesis; for example, at least one group mean is different from others. (the same in the chi square test) You should read F score (critical value) from the table on page 639 (not p. 640) as specified in the question. This is a two tailed test. I do not understand why many of you ignored the instruction; please read and follow instructions carefully. A p-value is, simply say, a probability of risk taking; again, therefore, a p-value SHOULD NOT be smaller than zero and larger than 1. Also you may not ignore step 4 and 5. You have to say either “Reject the null hypothesis at the .05 level” or “Do not reject the null hypothesis at the .05 level” in step 4. The final step is substantive interpretation. You need to go back to step 1 and re-paraphrase the null or alternative hypothesis. DO NOT say “reject the null hypothesis” again as substantive interpretation; the phrase does not have any substance. Who can know what you are testing? If you are not sure what is going on hypothesis testing, PLEASE read through the lecture note 1 (hypothesis testing) or come and see me. For Q1.10, you need to copy all 19 observations from question 28 on page 610. Do not ignore “…” If you cannot replicate SPSS output, you should ask yourself, “what is wrong in my computation?” As mentioned, x is the overall sample mean and jx is the sample mean of group j. So you have to compute four means (one overall mean and three group means).

Page 2: Assignment6 ANOVA Answer · For example, Q1.3 asks you to use s2(n −1) to get SST. s is the overall standard deviation of 15.582, while n is the number of overall observations 19

IUPUI/SPEA (Fall 2007) K300 (4392) Statistical Techniques

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Question 2 In question 2.1, you have to read a t statistic depending on the result of F-test for equal variance. If you do not reject the null hypothesis of equal variance, report the t statistic on the first line; otherwise, report the second t statistic. The F statistic in the t test is not comparable to F statistic in ANOVA. The former examines if two independent variables have the equal variance, while the latter examines if all group means are same. When the first degrees of freedom is 1 in one-way ANOVA (comparing two group means), the F statistic is the squared t statistic of two independent t-test. Question 3 In Question 3, chi square 2χ is not X2; all expected frequencies should be larger than 5; expected frequencies are used to compute chi square; observed frequencies should be used to compute gamma and lambda. Some of you appear confused with interpretation of p-values for gamma and lambda. The null hypothesis, in most cases, is the statistic in question is not discernable from zero; not significant or no effect. Therefore, if p-value is small (say smaller than .05), you can reject the null hypothesis and conclude that there is some significant effect. For example, the gamma is .288 with a p-value of .030. You just take only 3 percent risk when rejecting the null hypothesis of no effect. Obviously, you should reject the null hypothesis at the .05 level (but not at the .01 level): a significant positive relationship. Lambda is interpreted as the proportional reduction in error. Lambda zero means that knowing information of volunteering reduces error by zero percent (or does not reduce the error) when predicting IQ. Others - You should write question numbers before solving questions. Please follow the questions step by step. - You should solve questions in order. Q1.1 comes before Q1.2, which comes before Q1.3, and so forth. DO NOT jump around. - If your computation produces results different from SPSS output, you should go back and double-check your computation. - Manual computation does not allow you to use computer software packages such as Microsoft Excel and SAS (although the instructor uses Excel for answer key; definitely, the instructor is not a student). You should do by hand (calculator). Note: handwrite this comment on sheets and hand in Wednesday, November 28. Submission is used to check your attendance. Do not use a wordprocessor or other computer software packages.

Page 3: Assignment6 ANOVA Answer · For example, Q1.3 asks you to use s2(n −1) to get SST. s is the overall standard deviation of 15.582, while n is the number of overall observations 19
Page 4: Assignment6 ANOVA Answer · For example, Q1.3 asks you to use s2(n −1) to get SST. s is the overall standard deviation of 15.582, while n is the number of overall observations 19
Page 5: Assignment6 ANOVA Answer · For example, Q1.3 asks you to use s2(n −1) to get SST. s is the overall standard deviation of 15.582, while n is the number of overall observations 19
Page 6: Assignment6 ANOVA Answer · For example, Q1.3 asks you to use s2(n −1) to get SST. s is the overall standard deviation of 15.582, while n is the number of overall observations 19
Page 7: Assignment6 ANOVA Answer · For example, Q1.3 asks you to use s2(n −1) to get SST. s is the overall standard deviation of 15.582, while n is the number of overall observations 19
Page 8: Assignment6 ANOVA Answer · For example, Q1.3 asks you to use s2(n −1) to get SST. s is the overall standard deviation of 15.582, while n is the number of overall observations 19
Page 9: Assignment6 ANOVA Answer · For example, Q1.3 asks you to use s2(n −1) to get SST. s is the overall standard deviation of 15.582, while n is the number of overall observations 19
Page 10: Assignment6 ANOVA Answer · For example, Q1.3 asks you to use s2(n −1) to get SST. s is the overall standard deviation of 15.582, while n is the number of overall observations 19
Page 11: Assignment6 ANOVA Answer · For example, Q1.3 asks you to use s2(n −1) to get SST. s is the overall standard deviation of 15.582, while n is the number of overall observations 19