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2. To test the null hypothesis pi = P2 with two independent samples of sizes nį and n2, we require that n1 < N1/20, that n2 < N2/20, that n1 * pi * (1 – p1) > 10, and that n2 * p2 * (1– p2) > 10. Our test statistic TS is pi – p2 Va•(1 – p) • (+ + +) where p is the pooled estimate p = (x1 + x2)/(n1 + n2). (a) For a two-tailed test, we may compute +za/2 = ±InvNorm(a/2,0, 1) and check that the test-statistic falls in a tail. Or we may instead compute P = 2 * normalcdf (|TS|, 0,0, 1) and check that P < a. Suppose our null hypothesis is pi p2, that our sample sizes are ni = 1500 and n2 = 1600, and that we find x1 = 506 and x2 = 692. Assume that N1 and N2 are very large. i. If we use a two-tailed test, what is our alternative hypothesis: pi + p2 ? pi < P2 ? or pi > P2 ? ii. What would we conclude if we rejected the null hypothesis with a two-tailed test? iii. Can we reject the null hypothesis at level of significance a = 0.05, using a two-tailed test? iv. Can we reject the null hypothesis at level of significance a = 0.01, using a two-tailed test?

2. To test the null hypothesis pi = P2 with two independent samples of sizes nį and n2, we require that n1 < N1/20, that n2 < N2/20, that n1 * pi * (1 – p1) > 10, and that n2 * p2 * (1– p2) > 10. Our test statistic TS is pi – p2 Va•(1 – p) • (+ + +) where p is the pooled estimate p = (x1 + x2)/(n1 + n2). (a) For a two-tailed test, we may compute +za/2 = ±InvNorm(a/2,0, 1) and check that the test-statistic falls in a tail. Or we may instead compute P = 2 * normalcdf (|TS|, 0,0, 1) and check that P < a. Suppose our null hypothesis is pi p2, that our sample sizes are ni = 1500 and n2 = 1600, and that we find x1 = 506 and x2 = 692. Assume that N1 and N2 are very large. i. If we use a two-tailed test, what is our alternative hypothesis: pi + p2 ? pi < P2 ? or pi > P2 ? ii. What would we conclude if we rejected the null hypothesis with a two-tailed test? iii. Can we reject the null hypothesis at level of significance a = 0.05, using a two-tailed test? iv. Can we reject the null hypothesis at level of significance a = 0.01, using a two-tailed test?

Algebra & Trigonometry with Analytic Geometry
Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter10: Sequences, Series, And Probability
Section10.8: Probability
Problem 31E
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Solve questions 1 through 4 please
2. To test the null hypothesis p1 = P2 with two independent samples of sizes ni and n2, we require that ni < N1/20, that n2 < N2/20, that n1 * p1 * (1 – p1) > 10, and that n2 p2 * (1 – p2) > 10. Our test statistic TS is pi – p2 p• (1 – p) • (+ + ) where p is the pooled estimate p = (x1 + x2)/(n1 + n2). (a) For a two-tailed test, we may compute ±z,/2 = ±InvNorm(a/2,0, 1) and check that the test-statistic falls in a tail. Or we may instead compute P = 2 * normalcdf (|TS\, 0, 0, 1) and check that P < a. Suppose our null hypothesis is pi = P2, that our sample sizes are ni = 1500 and n2 = 1600, and that we find x1 = 506 and x2 = 692. Assume that N1 and N2 are very large. i. If we use a two-tailed test, what is our alternative hypothesis: p1 # p2 ? p1 < p2 ? or pi > p2 ? ii. What would we conclude if we rejected the null hypothesis with a two-tailed test? iii. Can we reject the null hypothesis at level of significance a = 0.05, using a two-tailed test? iv. Can we reject the null hypothesis at level of significance a = 0.01, using a two-tailed test?
Transcribed Image Text:2. To test the null hypothesis p1 = P2 with two independent samples of sizes ni and n2, we require that ni < N1/20, that n2 < N2/20, that n1 * p1 * (1 – p1) > 10, and that n2 p2 * (1 – p2) > 10. Our test statistic TS is pi – p2 p• (1 – p) • (+ + ) where p is the pooled estimate p = (x1 + x2)/(n1 + n2). (a) For a two-tailed test, we may compute ±z,/2 = ±InvNorm(a/2,0, 1) and check that the test-statistic falls in a tail. Or we may instead compute P = 2 * normalcdf (|TS\, 0, 0, 1) and check that P < a. Suppose our null hypothesis is pi = P2, that our sample sizes are ni = 1500 and n2 = 1600, and that we find x1 = 506 and x2 = 692. Assume that N1 and N2 are very large. i. If we use a two-tailed test, what is our alternative hypothesis: p1 # p2 ? p1 < p2 ? or pi > p2 ? ii. What would we conclude if we rejected the null hypothesis with a two-tailed test? iii. Can we reject the null hypothesis at level of significance a = 0.05, using a two-tailed test? iv. Can we reject the null hypothesis at level of significance a = 0.01, using a two-tailed test?
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