252y0772 11 26 07 Page layout view ECO252 QBA2 THIRD EXAM November 29 2007 Version 2 Name Key Student number Class Day and hour I 8 points Do all the following 2 points each unless noted otherwise Make Diagrams Show your work All probabilities must be between zero and positive 1 x N 27 14 75 27 46 27 P 1 36 z 3 43 z 1 P 46 x 77 P 14 14 P 0 z 3 57 P 0 z 1 36 4998 4131 0867 For z make a diagram Draw a Normal curve with a mean at 0 Indicate the mean by a vertical line Shade the area between 1 36 and 3 43 Because this is on one side of zero we must subtract the area between zero and 1 36 from the larger area between zero and 3 43 If you wish make a completely separate diagram for x Draw a Normal curve with a mean at 27 Indicate the mean by a vertical line Shade the area between 46 and 77 This area is on one side of the mean 27 so we subtract to get our answer 39 27 21 27 z P 0 43 z 0 86 2 P 21 x 39 P 14 14 P 0 43 z 0 P 0 z 0 86 1664 3051 4715 For z make a diagram Draw a Normal curve with a mean at 0 Indicate the mean by a vertical line Shade the entire area between 0 43 and 0 86 Because this is on both sides of zero we must add the area between 0 43 and zero to the area between zero and 0 86 If you wish make a completely separate diagram for x Draw a Normal curve with a mean at 27 Indicate the mean by a vertical line Shade the entire area between 21 and 39 This area is on both sides of the mean 27 so we add to get our answer 0 27 P z 1 93 P 1 93 z 0 P z 0 3 P x 0 P z 14 4732 5 9732 For z make a diagram Draw a Normal curve with a mean at 0 Indicate the mean by a vertical line Shade the entire area above 1 86 Because this is on both sides of zero we must add the area between 1 86 and zero to the area above zero If you wish make a completely separate diagram for x Draw a Normal curve with a mean at 27 Indicate the mean by a vertical line Shade the entire area above zero remembering that zero is below the mean This area is on both sides of the mean 27 so we add to get our answer 4 x 08 Do not try to use the t table to get this For z make a diagram Draw a Normal curve with a mean at 0 z 08 is the value of z with 8 of the distribution above it Since 100 8 92 it is also the 92nd percentile Since 50 of the standardized Normal distribution is below zero your diagram should show that the probability between z 08 and zero is 92 50 42 or P 0 z z 08 4200 The closest we can come to this is P 0 z 1 41 4203 1 40 is also pretty close So z 08 1 41 or something slightly smaller To get from z 08 to x 08 use the formula x z x x 27 1 41 14 46 74 If you wish make a completely Draw a Normal curve with a mean at 27 Show that 50 of the distribution is which is the opposite of z separate diagram for x 1 252y0772 11 26 07 Page layout view below the mean 27 If 8 of the distribution is above x 08 it must be above the mean and have 42 of the distribution between it and the mean 46 74 27 Check P x 46 74 P z 14 P z 1 41 P z 0 P 0 z 1 41 5 4207 0793 08 2 252y0772 11 26 07 Page layout view II 22 points Do all the following 2 points each unless noted otherwise Do not answer a question yes or no without giving reasons Show your work when appropriate Use a 5 significance level except where indicated otherwise Note that this is extremely long and that no one will do all the problems so look them over 1 Turn in your computer problems 2 and 3 marked as requested in the Take home 5 points 2 point penalty for not doing 2 In an ordinary 1 way ANOVA if the computed F statistic exceeds the value from the F table at the given significance level we can a Reject the null hypothesis because the difference between the means is not significant b Reject the null hypothesis because there is evidence of a significant difference between some of the means c Not reject the null hypothesis because the difference between the means is not significant d Not reject the null hypothesis because the difference between the means is significant c Not reject the null hypothesis because the difference between the variances is not significant d Not reject the null hypothesis because the difference between the variances is significant e None of the above 7 3 After an analysis if variance you would use the Tukey Kramer procedure or similar confidence intervals to check a For Normality b For equality of variances c For independence of error terms d For pairwise differences in means e For all of the above f For none of the above 4 If an ordinary one way ANOVA has 25 columns 17 rows and 17 25 425 the degrees of freedom for the F test are a 400 and 24 b 408 and 16 c 24 and 400 d 16 and 408 e 400 and 424 f 408 and 424 g 424 and 400 h 424 and 408 i 16 and 24 j None of the above The correct answer is Explanation This is a one way ANOVA The total number of observations is n 425 and the number of columns is m 25 This means there are 425 1 424 total degrees of freedom and that between the columns there are 25 1 24 degrees of freedom This leaves 424 24 400 degrees of freedom for the error within term Numbers are filled in below F Source SS DF MS F 24 400 1 40 Between SSB m 1 SSB MSB F 01 MSB F m 1 MSW 24 n m 400 Within SSW SSW MSW n m Total SST n 1 424 3 252y0772 11 26 07 Page layout view 5 Assuming that your answer to 4 is correct and that the significance level is 1 the correct value of F from the table is 1 84 This may have to be approximate If so what did you use 1 12 Note I will check your answer against what you said in the previous question The answer above is 24 …
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