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Unformatted text preview: Section 153 Note to Instructors: When an observation equals the hypothesized mean the observation is dropped and not considered in the analysis. 1512 a) 1. Parameter of interest is the mean titanium content 2 . H 0 : = 8 .5 3 . H 1 : 8 .5
4. =0.05 5. The test statistic is w = min( w + , w ). 6. We reject H0 if w * w0.05 = 52, because = 0.05 and n = 20, Appendix A, Table IX gives the critical value. 7. w+ = 80.5 and w  = 109.5 and w = min(80.5,109.5) = 80.5 8. Conclusion, because 80.5 > 52, we cannot reject H0. The mean titanium content is not significantly different from 8.5 at = 0.05. b) 1. Parameter of interest is the mean titanium content 2 . H 0 : = 8 .5 3 . H 1 : 8 .5
4. =0.05 W +  n ( n + 1) / 4 n ( n + 1)( 2 n + 1) / 24 6. We reject H0 if the Z0 > Z0.025 = 1.96, at = 0.05 7. w+ = 80.5 and w  = 109.5 and
5. The test statistic is Z = 0
Z0 = 80 . 5  19 ( 20 ) / 4 19 ( 20 )( 39 ) / 24 =  0 . 58 8. Conclusion, because 0.58 < 1.96, we cannot reject H0. The mean titanium content is not significantly different from 8.5 at =0.05. 1513 1. Parameter of interest is the mean pH 2. H 0 : = 7. 0 3 . H 1 : 7 .0
4. = 0.05 5. The test statistic is w = min( w + , w ). 6. We reject H0 if w * w0.05 = 8, because = 0.05 and n = 10, Appendix A, Table IX gives the critical value. 7. w+ = 50.5 and w  = 4.5 and w = min(50.5, 4.5) = 4.5 8. Conclusion, because 4.5 < 8, we reject H0 and conclude that the mean pH is not equal to 7. 1514 1. Parameter of interest is the difference in the mean caliper measurements 2. H 0 : D = 0 3. H 1 : D 0
4. =0.05 5. The test statistic is w = min( w + , w ). 6. We reject H0 if w * w0.05 = 3 because = 0.05 and n = 8 Appendix A, Table IX gives the critical value. 7. w+ = 21.5 and w  = 14.5 and w = min(21.5,14.5) = 14.5 8. Conclusion, do not reject H0 because 14.5 > 13. There is a not a significant difference in the mean measurements of the two calipers at = 0.05. 156 1515 1. Parameter of interest is the mean impurity level 2. H 0 : = 2. 5 3. H 1 : < 2. 5
4. = 0.05 5. The test statistic is w+ 6. We reject H0 if w+ * w0.05 = 60 because = 0.05 and n = 20 Appendix A, Table IX gives the critical value. 7. w+ = 5 and w  = 205 8. Conclusion, because 5 < 60, we reject H0 and conclude that the mean impurity level is less than 2.5 ppm. 1516 1. Parameter of interest is the difference in mean drying times 2. H 0 : D = 0 3. H 1 : D 0
4. =0.01 5. The test statistic is w = min( w + , w ). 6. We reject H0 if w * w0.05 = 27 because = 0.01 and n = 18 Appendix A, Table IX gives the critical value. 7. w+ = 141.5 and w  = 29.5 and w = min(141.5, 29.5) = 29.5 8. Conclusion, not enough evidence to reject H0. 1517 1. Parameter of interest is the difference in mean hardness readings for the two tips 2. H 0 : D = 0 3. H 1 : D 0
4. = 0.05 5. The test statistic is w = min( w + , w ). 6 We reject H0 if w * w0.05 = 3, because = 0.05 and n = 8 Appendix A, Table IX gives the critical value. 7. w+ = 24.5 and w  = 11.5 and w = min(24.5, 11.5) = 11.5 8. Conclusion, because 11.5 > 3 cannot reject H0. There is not a significant difference in the tips. ...
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 Spring '09
 KAILASHKAPUR

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