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solutions1 - Stat 665 (Spring 2011) Kaizar Solutions 1...

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Unformatted text preview: Stat 665 (Spring 2011) Kaizar Solutions 1 Exercises 40 points total 1. (6 points) Agresti problem 1.8 (use both the score and exact methods dont use Wald!) (a) pi = y n = 344 1170 = 0 . 294. (b) The wording of the question is unclear as to the null hypothesis, so I am choosing the hypotheses: H : . 5 , H a : < . 5 Score test: z = radicalBig o (1- ) n = . 294 . 5 radicalBig . 5(1- . 5) 1170 = 14 . 09 The p-value is P ( Z 14 . 09), where Z is N(0,1). Using R, I find this to be 2 . 1 10- 45 . I reject the null hypothesis and conclude that a minority would claim to be willing to accept cuts in their standard of living to protect the environment. Exact test: Using R, I find a p-value < 2 . 2 10- 16 . Again, I reject the null hypothesis and conclude that a minority would claim to be willing to accept cuts in their standard of living to protect the environment. (c) Score interval: From R, I find the interval: (0.26, 0.33) I am approximately 99% confident that the true proportion of people would claim to be willing to accept cuts in their standard of living to protect the environment. Exact interval: From R, I find the interval: (0.26, 0.33) I am at least 99% confident that the true proportion of people would claim to be willing to accept cuts in their standard of living to protect the environment. Note that for a true proportion near 0.3, the sample size (1170) appears to be large enough that the two tests seem to agree well. 2. (8 points) Agresti problem 1.12 In this case, pi = 0. Therefore, the estimated standard error is also 0, which makes the test statistic z infinite. Obviously this is not a useful test statistic. Since the estimated standard error is zero, the 95% interval is simply the estimate: 0. That is, according to the Wald procedure, I am 95% confident that the true proportion of intro stats students who are vegetarian is exactly zero. The score test is: z = radicalBig o (1- ) n = . 5 radicalBig . 5(1- . 5) 25 = 5 Using R, I find the p-value to be 5 . 7 10- 7 . 1 Based on this test, I reject the null hypothesis that half of intro stats students are vege- tarian. Note that a 95% confidence interval is an inversion of a test. That is, the interval is the collection of values of for which the score test would not reject the null hypothesis. I would reject the test whenever z > 1 . 96 and whenever z < 1 . 96. Notice that as I increase the hypothesized , the value of z decreases (since Im subtracting a bigger and bigger value). This is not linear, since the denominator is also changing. Ive plotted the relationship between and z below, where the dotted lines are at the edges of the rejec- tion region....
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This note was uploaded on 07/26/2011 for the course STA 665 taught by Professor Staff during the Spring '10 term at Ohio State.

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solutions1 - Stat 665 (Spring 2011) Kaizar Solutions 1...

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