252y0411 2 19 04 ECO252 QBA2 Name KEY FIRST HOUR EXAM Hour of class registered February 25 2004 Class attended if different Show your work Make Diagrams Exam is normed on 50 points Answers without reasons are not usually acceptable I 8 points Do all the following x N 1 7 Do not make diagrams of x with zero in the middle Make up your mind If you are diagramming x put the mean in the middle if you are diagramming z put zero in the middle Copies x of the diagrams are available on a separate sheet Remember z and that this equation implies that if we have z and need a value of x x as in parts j n we use z 25 5 1 1 1 z P 0 z 3 50 4998 or 49977 Make a 1 P 1 x 25 5 P 7 7 diagram Your diagram for z shows a Normal curve with zero in the middle and the area shaded from zero to 3 50 Because this area starts at zero you do not need to add or subtract 16 1 0 1 z P 0 14 z 2 14 0557 4838 5395 2 P 0 x 16 P 7 7 Make a diagram Your diagram for z shows a Normal curve with zero in the middle and the area shaded from 0 14 to 2 14 Because this area is on both sides of zero add 3 F 25 5 The cumulative probability up to 25 5 F 25 5 25 5 1 P z P z 3 50 P z 0 P 0 z 3 50 5 4998 9998 or 7 99977 Make a diagram Your diagram for z shows a Normal curve with zero in the middle and the area shaded from the extreme left to 3 50 Because this area is on both sides of zero add x 015 I m quoting from the take home part of the exam but you really should make a diagram showing a Normal curve centered at zero with 1 5 above z 015 48 5 between z 015 and zero and 4 50 below zero To find z 03 2 z 015 recall that since z 015 is 1 5 from the top of the distribution 48 5 from zero So P 0 z z 15 4850 According to the Normal table P 0 z 2 17 4850 So z 015 2 17 Now if we use x z we get and 50 1 5 x 015 1 2 17 7 16 19 252y0411 2 19 04 II 5 points 2 point penalty for not trying part a A random sample is taken of the time spent waiting in line at a bank The following data is found Recomputing what I ve done for you is a great way to waste time x x2 1 2 3 4 5 6 7 8 9 10 11 Sum 5 3 6 2 7 1 3 4 2 10 2 45 25 9 36 4 49 1 9 16 4 100 4 257 a Compute the sample standard deviation Solution x x 45 4 090909 s of s 2 the waiting times Show your work 2 x 2 nx 2 n 11 n 1 72 90909 7 290909 s 7 290909 2 700 10 b Compute a 90 confidence interval for the mean 3 Given x 4 0909 s 2 70 n 11 and 10 So s x 10 DF n 1 10 and t 05 1 812 s n 257 11 4 090909 2 10 2 700 11 0 8141 x t 2 s x 4 0909 1 812 0 8141 4 09 1 48 or 2 61 to 5 57 2 252y0411 2 19 04 III Do all of the following Problems 18 points Show your work except in multiple choice questions 1 Lee We want to test whether the weight of an average bag of cat food is above 50 pounds at a 5 significance level What are our null and alternative hypotheses a H 0 50 and H 1 50 b H 0 50 and H 1 50 H 0 50 and H 1 50 d H 0 50 and H 1 50 e H 0 50 and H 1 50 f H 0 50 and H 1 50 c 2 Lee We want to test whether the weight of an average bag of cat food is above 50 pounds at a 5 significance level We take a sample of 144 and find a sample mean of 49 and a sample standard deviation of 5 What is the value of our t or z test ratio a 2 0 b 2 5 c 2 4 d 2 8 e 3 0 Solution 3 t x 0 49 50 1 12 2 4 5 sx 5 5 144 12 Lee A company wishes to test the volume of fuel in a 50 gallon drum Since the company assumes that it is being cheated it uses H 0 50 Assume 01 and that from a sample of 31 it gets a sample mean and a sample standard deviation of 1 1 gallons From the results of x 50 the sample it computes the ratio 1 1 It should do the following 31 a Reject H 0 if the ratio is below 2 576 b Reject H 0 if the ratio is below 2 327 c Reject H 0 if the ratio is below 2 750 d Reject H 0 if the ratio is below 2 457 e Reject H 0 if the ratio is below 2 576 or above 2 576 f Reject H 0 if the ratio is above one of the numbers in a d g None of the above supply correct value or values Solution The alternate hypothesis is H 0 50 We thus do a left sided test of t x 50 30 1 1 against t n 1 t 01 2 457 31 3 252y0411 2 19 04 4 Lee The price paid by students at an WCU for a new textbook is believed to be Normally distributed with a population standard deviation of 15 75 A random sample of 50 students has a sample mean of 47 80 A 90 confidence interval for the population mean is a 44 14 to 51 46 b 43 04 to 52 56 c 46 34 to 53 66 d 44 10 to 51 50 e 45 06 to 50 54 Solution x 47 80 15 75 n 50 and 10 So s 15 75 x 2 22739 and z 05 1 645 n 50 x z 2 s x 47 80 1 645 2 22739 47 80 3 66 or 44 04 to 51 46 5 In problem 4 is the price paid by students in the sample significantly different from 44 12 1 Solution Yes since 44 12 is not on the correct confidence interval Answer depends in the interval that you selected in problem 4 6 The head of an accounting department wants to test to see if the proportion of bad invoices is below 0 1 …
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