Statistics Homework Solutions 213

Statistics Homework Solutions 213 - 213 SECTION 9.3 (b) The...

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Unformatted text preview: 213 SECTION 9.3 (b) The sample means are X 1 1 998, X 2 3 0000, X 3 5 300. The sample sizes are J1 5, J2 J3 3. The upper 5% point of the Studentized range is q3 8 05 4 04. The 5% critical value for X 1 X 2 and for X 1 X 3 is 4 04 1 3718 2 1 5 1 3 2 44, and the 5% critical value for X 2 X 3 is 4 04 1 3718 2 1 3 1 3 2 73. Therefore means 1 and 3 differ at the 5% level. ¢ £ ¤¥ £ ¡ ¨ ¡ ¤ ¨ ¡ ¨ ¨ ¨ ¡ ¡ ¡ ¨ ¡ ¡ ¡ ¨ ¨ 0 05 ¨ ¡ ¡¥ £ ¢ ¡ ¡¥ ¤£ ¥ ¡ ¤ ¡ ¡ ¨ 5, MSTr £ ¡ £ ¦¥ ¤ ¨ ¡ ¨ £ ¨ ¨ ¡ 4, J ¡¥ ¡ ¡ £ ¨ ¤ ¡ 88 04, I ¨ ¨ 19 554. ∑I 1 J X i X 2 I 1 i F MSTr MSE 19 554 3 85 5 08. There are 3 and 16 degrees of freedom, so 0 01 P (a computer package gives P 0 012). The null hypothesis of no difference is rejected at the 5% level. ¡ ¨ £ ¡ 13. (a) X ¡ ¡ (b) q4 16 05 4 05, so catalysts whose means differ by more than 4 05 3 85 5 3 55 are significantly different at the 5% level. Catalyst 1 and Catalyst 2 both differ significantly from Catalyst 4. ¨ ¡ £ ¨ ¨ ¨ ¡ ¡ The value of the F statistic is F MSTr MSE 19 554 MSE. The upper 5% point of the F3 16 distribution is 3.24. Therefore the F test will reject at the 5% level if 19 554 MSE 3 24, or, equivalently, if MSE 6 035. £ ¡ ¨ § ¨ ¨ £ ¨ ¡ £ ¡ 15. ¨ The largest difference between the sample means is 89 88 85 79 4 09. The upper 5% point of the Studentized range distribution is q4 16 05 4 05. Therefore the Tukey-Kramer test will fail to find any differences significant at the 5% level if 4 09 4 05 MSE 5, or equivalently, if MSE 5 099. ¨ ¨ ¡ ¨ ¨ ¡ ¡ £ ¨ ¡ ¨ ¡ ¡ ¨ ¡ Therefore the F test will reject the null hypothesis that all the means are equal, but the Tukey-Kramer test will not find any pair of means to differ at the 5% level, for any value of MSE satisfying 5 099 MSE 6 035. ¨ ¡ ¡ ¨ Section 9.3 Let I be the number of levels of oil type, let J be the number of levels of piston ring type, and let K be the number of replications. Then I 4, J 3, and K 3. ¡ 3. ¤ ¦¥ ¡¥ ¡ 1 ¡ ¤ ¡ (d) The number of degrees of freedom for error is IJ K 1J 2. 1 ¡¥ (c) The number of degrees of freedom for interaction is I 1 ¡ (b) The number of degrees of freedom for piston ring type is J ¡ 1 ¡ ¡ (a) The number of degrees of freedom for oil type is I ¡ ¡ ¤ 1. 6. 24. (e) The mean squares are found by dividing the sums of squares by their respective degrees of freedom. The F statistics are found by dividing each mean square by the mean square for error. The number of degrees of freedom for the numerator of an F statistic is the number of degrees of freedom for its effect, and the number of degrees of freedom for the denominator is the number of degrees of freedom for error. P-values may be obtained from the F table, or from a computer software package. ...
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