Unformatted text preview: Section 154 1518 a) 1. The parameters of interest are the mean current (note: set circuit 1 equal to sample 2 so that Table X can be used. Therefore 1=mean of circuit 2 and 2=mean of circuit 1) 2. H 0 : 1 = 2 3. H 1 : 1 > 2
4. = 0.025 5. The test statistic is w = ( n1 + n 2 )( n1 + n 2 + 1)  w 2 1 2 6. We reject H0 if w2 * w0.025 = 51, because =0.025 and n1=8 and n2=9, Appendix A, Table X gives the critical value. 7. w1 = 78and w2 = 75 and because 75 is less than 51, do not reject H0 8. Conclusion, do not reject H0. There is not enough evidence to conclude that the mean of circuit 1 exceeds the mean of circuit 2. b) 1. The parameters of interest are the mean image brightness of the two tubes 2. H 0 : 1 = 2 3. H 1 : 1 > 2
4. = 0.025 5. The test statistic is z = W1  w1 0 w 1 157 6. We reject H0 if Z0 > Z0.025 = 1.96 7. w1 = 78, w 1 =72 and 2 w 1 =108 78  72 = 0 . 58 10 . 39 Because Z0 < 1.96, cannot reject H0 8. Conclusion, fail to reject H0. There is not a significant difference in the heat gain for the heating units at = 0.05. Pvalue =2[1  P( Z < 0.58 )] = 0.5619 z0 = 1519 1. The parameters of interest are the mean flight delays 2. H 0 : 1 = 2 3. H 1 : 1 2
4. =0.01 5. The test statistic is w = ( n1 + n 2 )( n1 + n 2 + 1)  w 2 1 2 6. We reject H0 if w * w0.01 = 23, because =0.01 and n1=6 and n2=6, Appendix A, Table X gives the critical value. 7. w1 = 40 and w2 = 38 and because 40 and 38 are greater than 23, we cannot reject H0 8. Conclusion, do not reject H0. There is no significant difference in the flight delays at = 0.01. 1520 a) 1. The parameters of interest are the mean heat gains for heating units 2. H 0 : 1 = 2 3. H 1 : 1 2
4. = 0.05 5. The test statistic is w = ( n1 + n 2 )( n1 + n 2 + 1)  w 2 1 2 6. We reject H0 if w * w0.01 = 78, because = 0.01 and n1 = 10 and n2 = 10, Appendix A, Table X gives the critical value. 7. w1 = 77 and w2 = 133 and because 77 is less than 78, we can reject H0 8. Conclusion, reject H0 and conclude that there is a significant difference in the heating units at = 0.05. b) 1. The parameters of interest are the mean heat gain for heating units 2. H 0 : 1 = 2 3. H 1 : 1 2
4. =0.05 5. The test statistic is z = W1  w1 0 w 1 6. We reject H0 if Z0 > Z0.025 = 1.96 7. w1 = 77, w 1 =105 and 2 w 1 =175 77  105 =  2 . 12 13 . 23 Because Z0  > 1.96, reject H0 8. Conclusion, reject H0 and conclude that there is a difference in the heat gain for the heating units at =0.05. Pvalue =2[1  P( Z < 2.19 )] = 0.034 z0 = 1521 a) 1. The parameters of interest are the mean etch rates 2. H 0 : 1 = 2 3. H 1 : 1 2
4. = 0.05 158 5. The test statistic is w = ( n1 + n 2 )( n1 + n 2 + 1)  w 2 1 2 6. We reject H0 if w * w0.05 = 78, because = 0.05 and n1 = 10 and n2 = 10, Appendix A, Table X gives the critical value. 7. w1 = 73 and w2 = 137 and because 73 is less than 78, we reject H0 8. Conclusion, reject H0 and conclude that there is a significant difference in the mean etch rate at = 0.05. b) 1. The parameters of interest are the mean temperatures 2. H 0 : 1 = 2 3. H 1 : 1 2
4. = 0.05 5. The test statistic is z = W1  w1 0 w 1 6. We reject H0 if Z0 > Z0.025 = 1.96 7. w1 = 55 , w 1 =232.5 and 2 w 1 =581.25 258  232 . 5 = 1 . 06 24 . 11 Because Z0 < 1.96, do not reject H0 8. Conclusion, do not reject H0. There is not a significant difference in the pipe deflection temperatures at = 0.05. Pvalue =2[1  P( Z < 1.06 )] = 0.2891 z0 = 1522 a) 1. The parameters of interest are the mean temperatures 2. H 0 : 1 = 2 3. H 1 : 1 2
4. = 0.05 5. The test statistic is w = ( n1 + n 2 )( n1 + n 2 + 1)  w 2 1 2 6. We reject H0 if w * w0.05 = 185, because = 0.05 and n1 = 15 and n2 = 15, Appendix A, Table X gives the critical value. 7. w1 = 258 and w2 = 207 and because both 258 and 207 are greater than 185, we cannot reject H0 8. Conclusion, do not reject H0. There is not significant difference in the pipe deflection temperature at = 0.05. b) 1. The parameters of interest are the mean etch rates 2. H 0 : 1 = 2 3. H 1 : 1 2
4. =0.05 5. The test statistic is z = W1  w1 0 w 1 6. We reject H0 if Z0 > Z0.025 = 1.96 7. w1 = 73, w 1 =105 and 2 w 1 =175 73  105 =  2 . 42 13 . 23 Because Z0 > 1.96, reject H0 8. Conclusion, reject H0 and conclude that there is a significant difference between the mean etch rates. Pvalue = 0.0155 z0 = 159 1523 a) The data are analyzed in ascending order and ranked as follows: Group 2 2 2 1 2 2 1 2 2 1 1 2 2 1 1 1 2 1 1 1 Distance 244 258 260 263 263 265 267 268 270 271 271 271 273 275 275 279 281 283 286 287 Rank 1 2 3 4.5 4.5 6 7 8 9 11 11 11 13 14.5 14.5 16 17 18 19 20 The sum of the ranks for group 1 is w1=135.5 and for group 2, w2=74.5 Since w2 is less than w0.05 = 78 , we reject the null hypothesis that both groups have the same mean. b) When the sample sizes are equal it does not matter which group we select for w1 10(10 + 10 + 1) = 105 2 10 * 10(10 + 10 + 1) 2 W1 = = 175 12 135.5  105 Z0 = = 2.31 175 W =
1 Because z0 > z0.025 = 1.96, we reject H0 and conclude that the sample means are different for the two groups. When z0 = 2.31 the Pvalue = 0.021 ...
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 Spring '09
 KAILASHKAPUR
 Statistics, Statistical hypothesis testing, 2 w, 2.31 175 W

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