252y0752 10 19 07 Open in Print Layout format ECO252 QBA2 FIRST EXAM October 4 and 8 2007 Version 2 Name KEY Class hour Student number Show your work Make Diagrams Include a vertical line in the middle Exam is normed on 50 points Answers without reasons are not usually acceptable I 8 points Do all the following x N 3 13 0 3 P z 0 23 0 23 z 0 P z 0 0910 5 5910 1 P x 0 P z 13 2 3 40 3 z P 3 31 z 0 08 P 3 31 z 0 2 P 40 x 2 P 13 13 P 0 08 z 0 4995 0319 4676 1 252y0752 10 19 07 Open in Print Layout format 3 3 3 3 z P 0 46 z 0 1772 3 P 3 x 3 P 13 13 4 x 125 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 125 is the value of z with 12 5 of the distribution above it Since 100 12 5 87 5 it is also the 875 fractile Since 50 of the standardized Normal distribution is below zero your diagram should show that the probability between z 125 and zero is 87 5 50 37 5 or P 0 z z 125 3750 If we check this against the Normal table the closest we can come to 3750 is P 0 z 1 15 3749 1 16 is also acceptable here but clearly worse So z 125 1 15 This is the value of z that you need for a 75 confidence interval To get from z 125 to x 085 use the x formula x z which is the opposite of z x 3 1 15 13 17 95 If you wish make a completely separate diagram for x Draw a Normal curve with a mean at 3 Show that 50 of the distribution is below the mean 3 If 12 5 of the distribution is above x 125 it must be above the mean and have 37 5 of the distribution between it and the mean 17 95 3 Check P x 17 95 P z P z 1 15 P z 0 P 0 z 1 15 13 5 3749 1251 125 This is identical to the way you normally get a p value for a rightsided test 2 252y0752 10 19 07 Open in Print Layout format II 9 points 2 point penalty for not trying part a Our sales of microwave ovens in five randomly picked months appear below 122 126 140 142 150 a Compute the sample standard deviation s of expenditures Show your work 2 b Assuming that the underlying distribution is Normal compute a 99 confidence interval for the mean 2 c Redo b when you find out that there were only 14 months to pick the data from 2 d Assume that the population standard deviation is 10 and create a 75 two sided confidence interval for the mean 2 e Use your results in a to test the hypothesis that the mean is below 140 at the 99 level 3 State your hypotheses clearly f Extra Credit Given the data test the hypothesis that the population standard deviation is below 15 3 Solution a Compute the sample standard deviation s of expenditures The first two columns are needed for the Row x x x x x 2 x2 computational shortcut method The first third 1 122 14884 14 196 and fourth are needed for the definitional 2 126 15876 10 100 method Using both methods or the 3 140 19600 4 16 4 142 20164 6 36 definitional method wastes time 5 150 22500 14 196 x 680 x 2 93024 680 93024 0 544 x and x x 680 136 00 s n 5 2 x x 2 nx 2 n 1 x 0 a check n 5 x x 2 544 93024 5 136 00 2 544 136 4 4 s x 136 11 6619 b Assuming that the underlying distribution is Normal compute a 99 confidence interval for the mean 2 x t n 1 s x 136 4 604 5 2154 136 00 24 01 or 111 99 to 160 01 s x 2 sx n 11 6619 136 4 t n 1 t 005 4 604 27 20 5 2154 2 5 5 c Redo b when you find out that there were only 14 months to pick the data from 2 x t n 1 s x 136 00 4 604 4 3304 136 00 19 98 or 116 02 to 155 98 2 sx sx n 11 6619 N n N 1 5 14 5 14 1 136 00 9 9 27 20 18 8307 4 3394 5 13 13 d Assume that the population standard deviation is 10 and create a 75 two sided confidence interval for the mean 2 We found z 125 1 15 on the last page We have x x n 10 5 10 n 5 x 136 00 and 100 20 4 4721 5 x z 3 x1 136 000 1 15 4 4721 136 00 5 14 or 130 86 to 141 14 e Use your results in a to test the hypothesis that the mean is below 140 at the 99 level 3 State your hypotheses clearly The statement that the mean is below 140 does not contain an equality so it must be an 3 252y0752 10 19 07 Open in Print Layout format alternate hypothesis We have the following information 01 x 136 00 n 5 and s x sx n 4 3 747 Needless to say 5 2154 Since this is a one sided hypothesis we will use t n 1 t 01 because of the small sample size we are assuming that the parent distribution is Normal Our hypotheses are H 0 140 so H 1 140 0 140 Since we are worrying about the mean being too small this is a left sided test There are three ways to do this Do only one of them i Test Ratio t x 0 136 00 140 0 7670 This is a left sided test the smaller the sx 5 2154 sample mean is the more negative will be this ratio We will reject the null hypothesis if the ratio is smaller than t n 1 t 4 3 747 Make a diagram showing a Normal curve with a mean at 0 and a 01 shaded reject zone below 3 747 Since the test ratio is not below 3 747 we cannot reject H 0 If you wish to find a p value for your hypothesis note that the t ratio is 0 7670 The p value will be the probability that t is below 0 7670 The line of the t table for 4 degrees of freedom is below 40 35 30 25 20 15 10 05 025 01 005 001 df 45 4 0 134 0 271 0 414 0 569 0 741 0 941 1 190 …
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