Alternate formulas for this section include those below.252y0752 10/19/07 (Open in ‘Print Layout’ format) ECO252 QBA2 Name: ____KEY________ FIRST EXAM Class hour: _____________ October 4 and 8, 2007 Version 2 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. 13,3~ Nx1. 0xP 23.01330 zPzP 0023.0 zPz = .0910 + .5 = .59102. 240 xP 08.031.3133213340 zPzP 031.3 zP 008.0 zP0319.4995. = .46761252y0752 10/19/07 (Open in ‘Print Layout’ format)3. 33 xP 046.013331333 zPzP = .17724. 125.x(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. 125.z 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 125.z and zero is 87.5% - 50% = 37.5% or .3750.0125. zzP If we check this against the Normal table, the closest we can come to .3750 is .3749.15.10 zP (1.16 is also acceptable here, but clearly worse.) So .15.1125.z This is the value of z that you need for a 75% confidence interval. To get from 125.z to 085.x, use the formula zx , which is the opposite of xz . 95.171315.13 x. 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 125.x, it must be above the mean and have 37.5% of the distribution between it and the mean.Check: 95.17xP 15.113395.17 zPzP 15.100 zPzP.125.1251.3749.5. This is identical to the way you normally get a p-value for a right-sided test.2252y0752 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 150a. 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 themean. (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 below15 (3)? Solution: a) Compute the sample standard deviation,s, of expenditures.Row x 2x xx 2xx 1 122 14884 -14 196 2 126 15876 -10 100 3 140 19600 4 16 4 142 20164 6 36 5 150 22500 14 196 680 93024 0 544The first two columns are needed for the computational (shortcut) method. The first, third and fourth are needed for the definitional method. Using (both methods or) the definitional method wastes time.680x, ,930242x 0xx(a check), 5442xx and 5n.00.1365680nxx 1364544400.13659302412222nxnxsx6619.11136 xsb) Assuming that the underlying distribution is Normal, compute a 99% confidence interval for the mean. (2) xnstx12 01.2400.1362154.5604.4136 or 111.99 to 160.01 nssxx2154.520.27513656619.11 604.44005.12ttnc) Redo b) when you find out that there were only 14 months to pick the data from. (2) xnstx12 98.1900.1363304.4604.400.136 or 116.02 to 155.981NnNnssxx 11451456619.113394.48307.1813920.27139500.136d) Assume that the population standard deviation is 10 and create a 75% two-sided confidence interval for the mean. (2) We found 15.1125.z on the last page. We have 10, 5n, 00.136x and4721.4205100510nxx 14.500.1364721.415.1000.13613xzx 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 3252y0752 10/19/07 (Open in ‘Print Layout’ format)alternate hypothesis. We have the following information. 01., 00.136x, 5n and nssxx2154.5. Since this is a one-sided hypothesis we will use 747.3401.1ttn. Needless to say, because of the small sample size, we are assuming that the parent distribution is Normal. Our hypotheses are 140:140:10HH so 1400. 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: 7670.02154.514000.1360xsxt . This is a left-sided test - the smaller the sample mean is, the more negative will be this ratio. We will reject the null hypothesis if the ratio is smallerthan 747.3401.1ttn. Make a diagram showing a Normal curve with a mean at 0 and a shaded 'reject' zone below -3.747. Since the test ratio is not below -3.747, we cannot reject 0H.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.df .45 .40 .35 .30 .25 .20 .15 .10 .05 .025 .01 .005 .0014 0.134 0.271 0.414 0.569 0.741 0.941 1.190 1.533 2.132 2.776 3.747 4.604 7.173What this tells us, among other things, is that 20.941.0 tPand 25.741.0 tP. Since-0.7670 lies between -0.941 and -0.741, the probability that t lies below -0.7670 must
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