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09 STAT HWA a

# 09 STAT HWA a - Dr Valerie R Bencivenga Economics 329...

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Dr. Valerie R. Bencivenga Economics 329 PRACTICE ASSIGNMENT #9A: RIGHT ANSWERS 1.a. 0 12 12 12 1 11 11 12 2 10 10 12 3 9 9 12 5 . 5 . C 5 . 5 . C 5 . 5 . C 5 . 5 . C ) 5 . p | 9 X ( P max 073 . 000244 . 0029 . 0161 . 0537 . b. 0 12 12 12 1 11 11 12 2 10 10 12 3 9 9 12 6 . 4 . C 6 . 4 . C 6 . 4 . C 6 . 4 . C ) 4 . p | 9 X ( P 015 . 000017 . 0003 . 0025 . 0125 . c. 0 12 12 12 1 11 11 12 2 10 10 12 3 9 9 12 4 . 6 . C 4 . 6 . C 4 . 6 . C 4 . 6 . C ) 6 . p | 9 X ( P power 2254 . 0022 . 0174 . 0639 . 1419 . 7746 . r powe 1 d. ) 75 . p | 9 X ( P power 0 12 12 12 1 11 11 12 2 10 10 12 3 9 9 12 25 . 75 . C 25 . 75 . C 25 . 75 . C 25 . 75 . C 351 . e. .1 .2 .3 .4 .5 .6 .7 Probability of rejecting H 0 .4 .5 .6 .75 p f. Use the terms in the answer to part a. We know that for the rejection region , 9 X . 05 . max For the rejection region , 10 X . 0192 . max For the rejection region , 11 X . 0031 . max Therefore, the rejection region 10 X is the largest with , 05 . max and the rejection region 11 X is the largest with . 01 . max (We want the largest rejection region subject to max not exceeding the specified level, because the larger the rejection region, the greater is the power of the test.) g. Even if there is no discrimination in hiring, there is a positive probability that the firm will end up with all men. Therefore, we cannot say with certainty that there has been discrimination.

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2.a. Accept 0 H if cv X . Reject 0 H if cv X . The smallest rejection region is 1 X (in other words, 0 X ). b. First, keeping the rejection region and the number of trials constant, we know that is at its largest when p is at the boundary between 0 H and A H , here 2 . p . Since we are interested in maximum , this tells us that we should do our calculation using 2 . p . Second, holding p and the number of trials constant, the smaller the rejection region, the smaller is . Since we want to see if it is possible to find a rejection region where the maximum value of does not exceed 5%, we should do our calculation for the smallest rejection region, here 0 X . Since , 05 . 1074 . 8 . 2 . C ) 2 . p | 0 X ( P 10 0 0 10 it is not possible to control maximum at 5%. c. A third result (in addition to the two in part b) is that increasing sample size, here the number of trials, allows both and to be reduced. We know from part b that 10 n is too small to bring maximum below 5%. Therefore, we must increase n, using the rejection region 0 X and 2 . p , for the same reasons as in part b. This yields 05 . 0859 . 8 . 2 . C ) 11 n , 2 . p | 0 X ( P 11 0 0 11 05 . 0687 . 8 . 2 . C ) 12 n , 2 . p | 0 X ( P 12 0 0 12 05 . 0550 . 8 . 2 . C ) 13 n , 2 . p | 0 X ( P 13 0 0 13 05 . 0440 . 8 . 2 . C ) 14 n , 2 . p | 0 X ( P 14 0 0 14 Only when n is increased to 14 does maximum for the smallest rejection region drop below 5%.
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09 STAT HWA a - Dr Valerie R Bencivenga Economics 329...

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