252y0121 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 6 9 16 6 3 6 z P 1 00 z 1 11 1 P 3 x 16 P 9 9 P 1 00 z 0 P 0 z 1 11 3413 3665 7078 5 6 0 6 z P 0 67 z 0 11 2 P 0 x 5 P 9 9 P 0 67 z 0 P 0 11 z 0 2486 0438 2048 14 6 24 6 z P 3 33 z 0 89 3 P 24 x 14 P 9 9 P 3 33 z 0 P 0 z 0 89 4996 3133 8129 4 6 2 6 z P 0 89 z 0 22 4 P 2 x 4 P 9 9 P 0 89 z 0 P 0 22 z 0 3133 0871 2262 2 6 5 F 2 The Cumulative probability up to 2 P x 2 P z 9 P z 0 44 P z 0 P 0 44 z 0 5 1700 3300 6 A symmetrical interval about the mean with 79 probability We want two points x 895 and x 105 so that P x 895 x x105 7900 Make a diagram showing 6 in the middle at the center of a 79 region split into two areas with probabilities of 3950 From the diagram if we replace x by z P 0 z z 105 3950 The closest we can come is P 0 z 1 25 3944 or P 0 z 1 26 3962 Since neither of these is much closer than the other use z 105 1 255 and x z 105 6 1 255 9 6 11 295 or 5 295 to 17 295 To check this note that 17 295 6 5 295 6 P 5 295 x 17 295 P z 9 9 P 1 295 z 1 2558 2 P 0 z 1 255 2 3950 7900 7 x 008 We want a point x 008 so that P x x 008 008 Make a diagram of showing zero in the middle 4920 between 0 and z 008 and 008 above z 008 From the diagram if we replace x by z P 0 z z 008 4920 The Normal table says z P 0 z 2 41 4920 So z 008 2 41 and x z 008 6 2 41 9 27 69 To 27 69 6 check this note P x 27 69 P z P z 2 41 9 P z 0 P 0 z 2 41 5 4920 008 252y0121 3 19 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 Miller s the sample mean is 115 25 and the sample standard deviation is 1 75255 xm Obs 1 2 3 4 5 6 7 8 Miller s 112 115 115 117 117 117 115 114 xa Albert s 119 121 122 120 122 124 122 122 a Compute s a the standard deviation for Albert 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 x 2 for Albert s and x1 for Miller 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 x1 119 121 122 120 122 124 122 122 972 x 1 n1 x12 14161 14641 14884 14400 14884 15376 14884 14884 118114 972 2 121 50 s1 8 x 2 1 n1x12 n1 1 118114 8 121 50 2 16 2 2857 7 7 s1 1 51186 b From Table 3 of the Syllabus Supplement Interval for Confidence Hypotheses Interval Test Ratio Critical Value 2 Difference between Two Means unknown variances assumed equal d t 2 sd sd s p 1 1 n1 n 2 DF n1 n 2 2 H 0 0 H 1 0 1 2 t d 0 sd s 2p d cv 0 t 2 sd n1 1 s12 n2 1 s22 n1 n 2 2 3 252y0121 3 19 01 x1 121 50 s12 2 2857 s1 1 51186 x 2 115 25 s 22 3 0714 s 2 1 75255 DF n1 n 2 2 8 8 2 14 d x1 x 2 121 50 115 25 6 25 01 s 2p n1 1 s12 n2 1 s22 n1 n2 2 7 2 2857 7 3 0714 2 2857 3 0714 2 67955 14 2 2 67955 1 1 8 8 2 67955 25 14 t 005 2 977 sd s p 1 1 n1 n 2 0 6698875 0 818467 Confidence Interval d t 2 sd 6 25 2 977 0 81847 6 25 2 4366 or 3 81 to 8 69 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 Solution b From the formula table Interval for Confidence Hypotheses Interval Difference between Two Means unknown variances assumed unequal d t 2 s d sd s12 s 22 n1 n 2 s12 n 1 DF 2 H 0 0 H 1 0 1 2 Same as H2 0 1 2 s22 H 1 1 2 n if 0 0 2 Test Ratio t d 0 sd 0 Critical Value d cv 0 t 2 s d s12 n1 n1 1 s 22 2 n2 n2 1 x1 121 50 s12 2 2857 s1 1 51186 x 2 115 25 s 22 3 0714 s 2 1 75255 d x1 x 2 121 50 115 25 6 25 01 4 s12 2 2857 0 285714 n1 8 s 22 3 0714 0 383929 n2 8 sd s12 s 22 0 669643 0 8183172 n1 n 2 s12 s 22 0 669643 n1 n 2 DF s12 s 22 n1 n 2 2 2 2 s12 s 22 n1 n2 n1 1 n2 1 669643 2 2 28571 2 383929 2 4 13 7052 so use 13 degrees of freedom 7 5 252y0121 3 19 01 13 t 005 3 012 Confidence Interval The 2 sided interval is d t 2 sd …
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