252y0122 3 26 01 ECO252 QBA2 Name SECOND HOUR EXAM Hour of Class Registered Circle March 20 2001 MWF 10 11 TR 12 30 2 00 I 14 points Do all the following Make diagrams x N 7 9 16 7 3 7 z P 1 11 z 1 00 1 P 3 x 16 P 9 9 P 1 11 z 0 P 0 z 1 00 3665 3413 7078 6 7 0 7 z P 0 78 z 0 11 2 P 0 x 6 P 9 9 P 0 78 z 0 P 0 11 z 0 2823 0438 2385 14 7 24 7 z P 3 44 z 0 78 3 P 24 x 14 P 9 9 P 3 44 z 0 P 0 z 0 78 4997 2823 7820 5 7 2 7 z P 1 00 z 0 22 4 P 2 x 5 P 9 9 P 1 00 z 0 P 0 22 z 0 3413 0871 2542 2 7 5 F 2 The Cumulative probability up to 2 P x 2 P z 9 P z 0 56 P z 0 P 0 56 z 0 5 2123 2877 6 A symmetrical interval about the mean with 81 probability We want two points x 905 and x 095 so that P x 905 x x 095 8100 Make a diagram showing 6 in the middle at the center of a 81 region split into two areas with probabilities of 4050 From the diagram if we replace x by z P 0 z z 095 4050 The closest we can come is P 0 z 1 31 4049 Use z 095 1 31 and x z 105 7 1 31 9 7 11 79 or 4 79 to 18 79 To check this note 18 795 7 4 79 7 z that P 4 79 x 18 79 P 9 9 P 1 31 z 1 31 2 P 0 z 1 31 2 4049 8098 8100 7 x 006 We want a point x 006 so that P x x 006 006 Make a diagram of z showing zero in the middle 4940 between 0 and z 006 and 006 above z 006 From the diagram if we replace x by z P 0 z z 006 4940 The Normal table says P 0 z 2 51 4940 So z 006 2 51 and x z 006 7 2 51 9 29 59 29 59 7 To check this note P x 29 59 P z P z 2 51 9 P z 0 P 0 z 2 51 5 4940 006 252y0122 3 20 01 II 6 points 2 point penalty for not trying part a Show your work A test shopper goes to Miller s Supermarkets 8 times and to Albert s Supermarkets 8 times The amount that was spent using a standardized shopping list is shown below You can regard these data as two independent random samples from populations with a normal distribution For Albert s the sample mean is 115 25 and the sample standard deviation is 1 75255 Obs 1 2 3 4 5 6 7 8 xm xa Miller s 118 119 120 118 120 122 120 120 Albert s 112 115 115 117 117 117 115 114 a Compute s m the standard deviation for Miller s 3 b Compute a 99 confidence interval for the difference between the population means 1 and 2 assuming that the variances are equal for the two parent populations According to your confidence interval is there a significant difference between the population means You must tell why 3 c Extra Credit i Redo the confidence interval on the assumption that the variances are not equal 6 ii Use an F test to tell whether you should have assumed that the variances were or were not equal 2 Solution a Use x1 for Miller s and x 2 for Albert s x 2 115 25 and s 2 1 75255 so s 22 3 07143 Line 1 2 3 4 5 6 7 8 x1 s12 x1 118 119 120 118 120 122 120 120 957 x 1 n1 x 2 1 x12 13924 14161 14400 13924 14400 14884 14400 14400 114493 957 119 625 8 n1x12 n1 1 114493 8 119 625 2 11 875 1 696429 s1 1 30247 7 7 b From Table 3 of the Syllabus Supplement Interval for Confidence Hypotheses Interval Difference H 0 0 d t 2 sd between Two H 1 0 Means 1 1 sd s p 1 2 unknown n1 n 2 variances DF n1 n 2 2 assumed equal Test Ratio t d 0 sd s 2p Critical Value d cv 0 t 2 sd n1 1 s12 n2 1 s22 n1 n 2 2 2 3 252y0122 3 20 01 x1 119 625 s12 1 696429 s1 1 30247 x 2 115 25 s 22 3 0714 s 2 1 75255 DF n1 n 2 2 8 8 2 14 d x1 x 2 119 625 115 25 4 375 01 s 2p n1 1 s12 n2 1 s22 n1 n2 2 7 1 696429 7 3 0714 1 696429 3 0714 2 38391 14 2 sd s p 1 1 n1 n 2 2 38391 1 1 8 8 2 38391 25 14 t 005 2 977 0 595979 0 77200 Confidence Interval d t 2 sd 4 375 2 977 0 77200 4 375 2 2982 or 2 076 to 6 673 The interval does not include 0 so there is a significant difference between the means H 0 0 H 0 1 2 H 0 1 2 0 Formally our hypotheses are H 0 or or We reject H 1 H1 1 2 H1 1 2 0 1 2 c i From the formula table Interval for Confidence Interval Difference Between Two Means Unknown Variances Assumed Unequal Hypotheses H 0 0 H 1 0 s12 s22 1 2 sd n1 n2 Same as H2 s12 s22 0 1 2 H 1 1 2 n n if 0 2 1 0 d t 2 sd DF Test Ratio t d 0 sd 0 Critical Value d cv 0 t 2 sd s12 2 n1 n1 1 s 22 2 n2 n2 1 x1 119 625 s12 1 696429 s1 1 30247 x 2 115 25 s 22 3 0714 s 2 1 75255 d x1 x 2 119 625 115 25 4 375 01 4 s12 1 69643 0 212054 n1 8 s 22 3 0714 0 383929 n2 8 sd s12 s 22 0 595982 0 77200 n1 n 2 s12 s 22 0 595982 n1 n 2 DF s12 s 22 n1 n 2 2 2 2 s12 s 22 n1 n2 n1 1 n2 1 595982 2 0 212054 2 0 383929 2 7 12 925 so use 12 degrees of freedom 7 5 252y0122 3 20 01 12 t 005 3 055 Confidence Interval The 2 sided interval is d t 2 sd 4 375 3 055 0 77200 4 375 2 359 or 2 016 to 6 734 c ii …
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