Statistics Homework Solutions 211

Statistics Homework Solutions 211 - 3. (a) MSE 2 9659, J i...

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SECTION 9.2 211 17. (a) Source DF SS MS F P Grade 3 1721.4 573.81 9.4431 0.000 Error 96 5833.4 60.765 Total 99 7554.9 (b) Yes, F 3 96 9 4431, P 0 001 ( P 0). 19. (a) Source DF SS MS F P Soil 2 2.1615 1.0808 5.6099 0.0104 Error 23 4.4309 0.19265 Total 25 6.5924 (b) Yes. F 2 23 5 6099, 0 01 P 0 05 ( P 0 0104). Section 9.2 1. (a) Yes, F 5 6 46 64, P 0. (b) q 6 6 05 5 63. The value of MSE is 0.00508. The 5% critical value is therefore 5 63 0 00508 2 0 284. Any pair that differs by more than 0.284 can be concluded to be different. The following pairs meet this criterion: A and B, A and C, A and D, A and E, B and C, B and D, B and E, B and F, D and F. (c) t 6 025 15 4 698 (the value obtained by interpolating is 4.958). The value of MSE is 0.00508. The 5% critical value is therefore 4 698 2 0 00508 2 0 335. Any pair that differs by more than 0.335 may be concluded to be different. The following pairs meet this criterion: A and B, A and C, A and D, A and E, B and C, B and D, B and E, B and F, D and F. (d) The Tukey-Kramer method is more powerful, since its critical value is smaller (0.284 vs. 0.335). (e) Either the Bonferroni or the Tukey-Kramer method can be used.
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Unformatted text preview: 3. (a) MSE 2 9659, J i 12 for all i . There are 7 comparisons to be made. Now t 88 025 7 2 754, so the 5% critical value is 2 754 2 9659 1 12 1 12 1 936. All the sample means of the non-control formulations differ from the sample mean of the control formulation by more than this amount. Therefore we conclude at the 5% level that all the non-control formulations differ from the control formulation. (b) There are 7 comparisons to be made. We should use the Studentized range value q 7 88 05 . This value is not in the table, so we will use q 7 60 05 4 31, which is only slightly larger. The 5% critical value is 4 31 2 9659 12 2 14. All of the non-control formulations differ from the sample mean of the control formulation by more than this amount. Therefore we conclude at the 5% level that all the non-control formulations differ from the control formulation....
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This note was uploaded on 12/20/2011 for the course STA 3163 taught by Professor Mattgilg during the Fall '11 term at UNF.

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