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

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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 – don’t use Wald!) (a) ˆ 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 0 : π 0 . 5 , H a : π < 0 . 5 Score test: z = ˆ π π 0 radicalBig π o (1 - π 0 ) n = 0 . 294 0 . 5 radicalBig 0 . 5(1 - 0 . 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, ˆ 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 = ˆ π π 0 radicalBig π o (1 - π 0 ) n = 0 0 . 5 radicalBig 0 . 5(1 - 0 . 5) 25 = 5 Using R, I find the p-value to be 5 . 7 × 10 - 7 . 1

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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 π 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 π 0 , the value of z decreases (since I’m subtracting a bigger and bigger value). This is not linear, since the denominator is also changing. I’ve plotted the relationship between π 0 and z below, where the dotted lines are at the edges of the rejec- tion region. 0.0 0.4 0.8 -60 -40 -20 0 all possible pi pi_0 z 0.00 0.10 0.20 -3 -2 -1 0 1 2 interesting range of pi pi_0[1:50] z[1:50] Visually, you can see that the confidence interval should go from the smallest possible value (zero) to a value at about 0.13. When I actually calculate the value of the score statistic (z) when π 0 = 0 . 133, I get: z = ˆ π π 0 radicalBig π o (1 - π 0 ) n = 0 0 . 133 radicalBig 0 . 133(1 - 0 . 133) 25 = 1 . 96 which confirms that 0.133 is the edge of the rejection region.
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• Spring '10
• Staff
• Null hypothesis, Statistical hypothesis testing, Statistical tests, Likelihood function, Likelihood-ratio test

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

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