12/16/98 252zz98714. A consumer advocacy group compares the proportions of automobiles producer by three different manufacturers that need major repairs in the first three years of ownership. The results are as below:Manufacturer 1 Manufacturer 2 Manufacturer 3 10004511nx 10002722nx 5001833nxa. Test the hypothesis that the proportions for the first two manufacturers are equal at the 90% level. (5)b. Do a confidence interval for the difference between the first two manufacturers 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)e. Test the hypothesis that the proportions for all three manufacturers are equal at the 90% level.(6)Solution: From page 10 of the Syllabus Supplement:Interval for ConfidenceIntervalHypotheses Test Ratio Critical ValueDifference between proportions q p 1 p p z sp 2p p p 1 2sp qnp qnp 1 112 22HH :01: p pp p00p p p0 01 020 or p0zp pp 0If pp qnp qnp 001 01102 022 Or use sp p p zcv p 02 If pp qpn n00 01 101 2 2122110nnpnpnp0100:H:H pppp or 211210:H:H pppp 645.110.2z 045.100045111nxp 027.100027122nxp 018.027.045.21 pppNote: 036.500183p, a fact that we will not use in this form) 036.1000100027452122110nnpnpnp 00833.1000110001964.036.1121000nnqpp 00832.1000973.027.1000955.045.222111nqpnqpspa) (i)161.200833.018.0pppz. This is outside the interval 645.1 so reject 0H. or (ii) 0137.00833.645.1020pcvzpp. This interval does not include .018 so reject 0H. or (ii) 0137.018.00832.645.1018.2pszpp This interval does not include 0 so reject 0H.b) 0163018.00832.96.1018.2pszpp.8c)211210:H:H pppp . this is the same as 0:H0:H10 pp 282.110. zUse thematerial in a) (i) 161.2zThis is above 1.282 so reject 0H. Or (ii) 011.00833.282.100pcvzpp . Since this is below .018, reject 0H. Or (iii) .007.011.018.00832.282.1018.2pszpp this is above 0, so reject 0H.d) For a), 0308.48465.216.22 zP. For b), 0308.48465.16.2 zP912/16/98 252zz9871e)UniformityH :0 or 3210: pppH . OManuf. 1 Manuf. 2 Manuf. 3 Total rpRepaired 45 27 18 90 .036Not repaired 955 973 482 2410 .964Total 1000 1000 500 2500 1.000We get the items in E by multiplying the column totals by rp. For example 36.00 = .036(1000).EManuf. 1 Manuf. 2 Manuf. 3 Total rpRepaired 36.00 36.00 18.00 90 .036Not repaired 964.00 964.00 482.00 2410 .964Total 1000.00 1000.00 500.00 2500 1.000 O E EO2 45 36 56.250 27 36 20.250 18 18 18.000 955 964 946.084 973 964 982.084 482 482 482.000 2500 2500 2504.668 We now compute 668.42500668.250422nEO . Since our degrees of freedom are 2131211 cr (ris number of rows and c is number of columns), we test this against 9915.52205.. Since our computed 2 is less than 2 from the table, we accept 0H.1012/16/98 252zz98715.a. A restaurant owner suspects that there is a pattern to when couples enter her restaurant. Older couples seem to come in earlier and younger later. One night she records the apparent age of couples from 5:30 to 10:00 PM (O means ‘older’ and Y means ‘younger’) and gets the following pattern:OOOOOOYOOOOOOYYYOYYYYYYOYYYOYYYIs this random? Assume that the significance level is 5%. (5) b. Use a Kolmogorov-Smirnov test to find if the number of arrivals over 100 days follows a Poisson distribution with a mean of 3. (5)Arrivals 0 1 2 3 4 5 6Number of Days 6 18 32 24 11 2 7 c. Suppose that in b) the null hypothesis had been ‘Poisson’ with no indication of the mean. Choose a method and do the problem with this new hypothesis. (6)Solution:a)15,15,1021 nnr. From the table, the critical values of rare 10 and 16, so we reject RandomnessH :0.b) A table showing computations for a) and b) appears below. For the K-S computations, we find the cumulative expected distribution eFfrom the Poisson(3) table. To get the cumulative observed distribution oF, we divide O by nand add the results. x OxOeFnO oFD xP E 0 6 0 .04874 .06 .06 .0102 .082085 8.208 1 18 18 .19915 .18 .24 .0408 .205213 20.521 2 32 64 .42319 .32 .56 .1368 .256516 25.652 3 24 72 64723 .24 .80 .1527 .213763 21.376 4 11 44 .81526 .11 .91 .0947 .133602 13.360 5 2 10 .91608 .02 .93 .0132 .066801 6.680 6 7 42 .96649 .06 1.00 .0335 .027834 2.783Total 100 250 1.00 Since the maximum difference is .1527 and the 5% critical value from the table is 136.10036.136.1n, where On, we reject 3:0PoissonH.c) The only possible method if parameters are unknown (except in the case of the normal distribution) is the Chi-squared method. We computed the mean by summing xO and dividing by n. In this case, our mean is 250 divided by 100 or 2.5. E is computed by looking up probabilities under the Poisson(2.5) distribution and multiplying them by n. O E EO2 6 8.208 4.3860 18 20.521 15.7887 32 25.652 39.9189 24 21.376 26.9461 11 13.360 9.0569 2 6.680 0.5988 7 2.783 17.6069 100 98.580 114.3023 We now compute 2023.41002023.11422nEO . Since our degrees of freedomare equal to the number of lines minus 1 minus the number of parameters estimated from the data = 7 - 1 -111 = 5, we test this against 0705.115205.. Since our computed 2 is less than 2 from the table, we accept PoissonH
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