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. 7,1~ Nx 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 of the diagrams are available on a separate sheet. Remember xz and that this equation implies that if we have z and need a value of x, as in parts j-n, we use zx .1. 5.251 xP 4998.50.30715.25711 zPzP (or .49977) Make a 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. 2. 160 xP 5395.4838.0557.14.214.07116710 zPzP 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. 5.25F (The cumulative probability up to 25.5) 5.25F 50.3715.25 zPzP 9998.4998.5.50.300 zPzP (or .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. 4. 015.x 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 015.z, 48.5% between 015.z and zero and 50% below zero. “To find 015.203.zz , recall that since 015.z is 1.5% from the top of the distribution and 50% - 1.5% = 48.5% from zero. So .4850.015. zzP According to the Normal table .4850.17.20 zP So .17.2015.z” Now, if we use zx , we get .19.16717.21015.x252y0411 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 2x 1 5 25 2 3 9 3 6 36 4 2 4 5 7 49 6 1 1 7 3 9 8 4 16 9 2 4 10 10 100 11 2 4 Sum 45 257 a. Compute the sample standard deviation, s, of the waiting times. Show your work! (2) Solution: 090909.41145nxx 10090909.41125712222nxnxs290909.71090909.72 700.2290909.7 sb. Compute a 90% confidence interval for the mean , .(3)Given: ,0909.4x ,70.2s 11n and .10. So ,8141.011700.2nssx101 nDF and 812.11005.t 48.109.48141.0812.10909.42xstxor 2.61 to 5.57.2252y0411 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)50:0Hand 50:1Hb)50:0Hand 50:1Hc)50:0Hand 50:1Hd) *50:0Hand 50:1He)50:0Hand 50:1Hf)50:0Hand 50:1H2. (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 ztest ratio?a) -2.0b) -2.5c) *-2.4d) -2.8e) -3.0Solution: 4.25121251144550490xsxt3. (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 .50:0H 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 the sample it computes the ratio 311.150x. It should do the following:a) Reject 0H if the ratio is below -2.576.b) Reject 0H if the ratio is below -2.327.c) Reject 0H if the ratio is below -2.750.d) *Reject 0H if the ratio is below -2.457.e) Reject 0H if the ratio is below -2.576 or above 2.576.f) Reject 0H 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 .50:0H We thus do a left-sided test of311.150xt against 457.23001.1ttn3252y0411 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 asample mean of $47.80. A 90% confidence interval for the population mean is:a) *$44.14 to $51.46b) $43.04 to $52.56c) $46.34 to $53.66d) $44.10 to $51.50e) $45.06 to $50.54Solution: ,80.47x ,75.15 50n and .10. So,22739.25075.15nsx and 645.105.z 66.380.4722739.2645.180.472xszx 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%. She examines 10000 invoices (, finds 4 bad ones) and tries to test the null hypothesis with a confidence interval 05.. The appropriate confidence interval is:a) 10000999.001.960.10004. pb) 100009996.0004.960.10004. pc) * 100009996.0004.645.10004. pd) 100009996.0004.645.10004. pe) None of the above – write in correct interval From the formula table we have: Interval for Confidence IntervalHypotheses Test Ratio Critical ValueProportionpszpp2nqpsppq 10100:H:Hpppppppz0pcvzpp20nqpp00001 pq The Hypotheses are 001.:001.:10pHpH4252y0411 2/19/04 7. We
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