12 16 98 252zz9871 4 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 x1 45 x2 27 x 3 18 n1 1000 n2 1000 n 3 500 a 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 Confidence Hypotheses Interval Difference p p z 2 s p H 0 p p0 between p p1 p2 H 1 p p0 proportions q 1 p s p pq pq 1 1 2 2 n1 n2 p0 p01 p02 or p 0 0 Test Ratio z p p0 p If p 0 p p01q 01 p02 q 02 n1 n2 Or use H 0 p p 0 p1 H 1 p p 0 or H 0 p1 p 2 x1 45 045 n1 1000 Note p 3 p0 If p0 0 p p 0 q 0 p0 1 n1 1 n2 n1 p1 n2 p2 n1 n2 10 z 2 1 645 x2 27 027 p p1 p 2 045 027 018 n1 1000 n1 p1 n2 p2 45 27 036 n1 n2 1000 1000 s p p1q1 p q 2 2 n1 n2 i z or ii H 1 p1 p 2 s p pcv p0 z 2 p 18 036 a fact that we will not use in this form 500 1 1 p p0 q 0 0 n1 n2 a p2 Critical Value 1 1 00833 1000 1000 036 964 045 955 027 973 1000 1000 00832 p p 0 018 2 161 This is outside the interval 1 645 so reject H 0 p 00833 p cv p 0 z 2 p 0 1 645 00833 0137 This interval does not include 018 so reject H 0 or ii p p z 2 s p 018 1 645 00832 018 0137 This interval does not include 0 so reject H 0 b p p z 2 s p 018 1 96 00832 018 0163 8 c H 0 p1 p2 H1 p1 p 2 this is the same as H 0 p 0 H1 p 0 10 z 1 282 Use the material in a i z 2 161 This is above 1 282 so reject H 0 Or ii p cv p 0 z p 0 1 282 00833 011 Since this is below 018 reject H 0 Or iii p p z 2 s p 018 1 282 00832 018 011 007 this is above 0 so reject H 0 d For a 2 P z 2 16 2 5 4846 0308 For b P z 2 16 5 4846 0308 9 12 16 98 252zz9871 e H 0 Uniformity or H 0 p1 p 2 p 3 Manuf 1 Manuf 2 O Repaired Not repaired Total We get the items in 036 1000 45 955 1000 Repaired Not repaired Total E O2 18 482 500 Total pr 90 2410 2500 036 964 1 000 E by multiplying the column totals by p r For example 36 00 E O 27 973 1000 Manuf 3 Manuf 1 Manuf 2 Manuf 3 Total pr 36 00 964 00 1000 00 36 00 964 00 1000 00 18 00 482 00 500 00 90 2410 2500 036 964 1 000 E 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 O2 n 2504 668 2500 4 668 Since our degrees of freedom are E r 1 c 1 2 1 3 1 2 r is number of rows and c is number of columns we test this 2 2 against 05 5 9915 Since our computed 2 is less than 2 from the table we accept H 0 2 We now compute 10 12 16 98 252zz9871 5 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 OOOOOOYOOOOOOYYYOYYYYYYOYYYOYYY Is 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 6 Number 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 r 10 n1 15 n2 15 From the table the critical values of r are 10 and 16 so we reject H 0 Randomness b A table showing computations for a and b appears below For the K S computations we find the cumulative expected distribution Fe from the Poisson 3 table To get the cumulative observed n and add the results Fe Fo O n distribution F o we divide O by x O xO D P x 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 2 783 Total 100 250 1 00 Since the maximum difference is 1527 and the 5 critical value from the table is 1 36 1 36 136 where n O we reject H 0 Poisson 3 n 100 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 6 18 32 24 11 2 7 100 O2 E 8 208 20 521 25 652 21 376 13 360 6 680 2 783 98 580 E 4 3860 15 7887 39 9189 26 9461 9 0569 0 5988 17 6069 114 3023 2 We now compute O2 n 114 2023 100 4 2023 Since our degrees of freedom E …
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