Computer Question10/23/98 252z98514. A consumer advocacy group compares the proportions of automobiles produced by two different manufacturers that need major repairs in the first three years of ownership. The results are as below:Manufacturer 1 Manufacturer 2 10004511nx 10002722nxa. Do a 2-sided 99% confidence interval for the proportion produced by manufacturer 1 that need major repairs. (2)b. Do a 2-sided 91% confidence interval for the proportion produced by manufacturer 1 that need major repairs. (2)c. An automobile columnist states that over 5% of automobiles produced by the first manufacturer need repairs in the first three years. State the null and alternative hypotheses to be tested in this problem..(1)d. Do a test of the hypothesis in c) at a 95% confidence level. (3)e. Repeat d) at the 91% confidence level.(2)f. If the true proportion that need repairs is 6%, what is the power of the test? (3)g. If I claim that the proportion needing repairs is 6%, on how large a sample would this have to be based to be accurate within 0.5%? (2)Solution: From page 10 of the Syllabus Supplement:Interval for Confidence IntervalHypotheses Test Ratio Critical ValueProportionp p z sspqnq ppp 21HH0::p pp p01 0zp pp0p p zp qncv pp 00 02a) 955.1 and .04510004511 pqnxp 00655553.1000955.045. nqpsp 017.045.00655.576.2045.005.pszppb) From page 1 70.1045.z. So 011.045.00655.70.1045.005.pszppc)05.:H05.:H10 pp d) 00689.100095.05.00nqpp 645.105.zz (i) Test Ratio: 726.0.00689.05.045.0pppz. This is below 645.1so accept 0H. or (ii) Critical Value: 06134.00689.645.105.0pcvzpp . .045pis below this, so accept 0H. or (iii) Confidence Interval: .0342.00655.645.1045.05.pszpp. This does not contradict 0H.e) We want a point 09.z, so that 09.09.zzP. From the diagram, P z z0 4109 ... The closest we can come is P z0 134 4099 . .. So z..09134. Redo one of the three ways of doing d) with 1.645 replaced by 1.34. .10/23/98 252z9851f) The power of a test is the probability of rejecting the null hypothesis when it is false. It is the opposite of the probability of wrongly accepting the null hypothesis. According to d) above, if05., we will accept 0H if p is less than or equal to .06134. So 0714.5.118.01100094.06.06.06134.106.1 zPzPpzPpowerpcv4286.5714.1 g) From the outline, if we assume a 95% confidence level so that 960.12z, 8667005.96.194.06.2222epqzn5. A consumer advocacy group compares the proportions of automobiles producer by two different manufacturers that need major repairs in the first three years of ownership. The results are as below:Manufacturer 1 Manufacturer 2 10004511nx 10002722nxa. Test the hypothesis that the proportions are equal at the 90% level. (5)b. Do a confidence interval for the difference at the 95% level. (4)c. Test the hypothesis that the proportion that need repairs is greater for the first manufacturer than the second at the 90% level. (2)d. Find p-values for your results in a) and c). (2)Those who do not learn from the past are doomed to repeat it.210/23/98 252z98516. A city sets a requirement that a sewer pipe must have a mean breaking strength exceeding 2500 pounds per linear foot. A test of 200 sections of sewer pipe yields an average strength of 2530 pounds per linear foot and a standard deviation of 200. Assume a confidence level of 95%.a. State null and alternative hypotheses for this problem and a critical value for the sample mean.(2)b. Given the sample results above, do you accept or reject the null hypotheses?(1)c. What is the power of the test if the true mean is 2535 pounds per square inch? (3)d. Do a power curve for this test. (4)e. If we claim that a population has a Poisson distribution with a mean of 9 , and the actual value foundis 15, do a hypothesis test at the 95% level. (3)f. Find the lowest value above 9 and the highest value below 9 that would lead to a rejection of the null hypothesis in f). Using these, find the power of the test if the mean is actually 12.5.(5)Solution: Interval for Confidence IntervalHypotheses Test Ratio Critical ValueMean ( known) x zx2HH0 01 0:: zxx0x zcv x 02a)2500:H2500:H10. 200 and 2530,200,645.1,05.05. sxnz. Since the above can be used for large samples, replace x with 142.14200200nssx. If we use the critical value method, 26.2523142.14645.125000xcvszx.b) We accept 0H if 26.2523x, so reject 0H.c) We accept 0H if 26.2523x, so we need the probability that x is less than 2523.26 when the true mean is 2535. 2005.2995.5.84.0142.142535252325352523 zPzPxP. This is the probability of a type II error, so that the power is 1 - .2005 = .7995.d) Try the following points: At 25001the power is 5% (the significance level) At 25111the power is 142.14251125231251125231 zPxP 2005.2995.5.184.01 zP At 25231(the critical value) the power is 50%. At 25351the power is .7995. At 25461the power is 142.14254625231254625231 zPxP 9484.44845.163.11 zP310/23/98 252z9851e)9:H9:H10 mm. From the Poisson table for a mean of 9, 152 xPvaluep 08824.95853.121412 xP. Since this is above .05, accept 0H.f) Since the number in e) is pretty close to .05, try 16 (Look for probabilities above 97.5%!). Also try numbers close to zero (Look for probabilities below 2.5%!). For 16, 162 xPvaluep 02214.97786.121512 xPbelow .05 For 3, 32 xPvaluep 04246.02123.2 below .05.These tell us our reject region, so our accept region is 4
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