Question

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You wish to test the following claim (H) at a significance level of a = 0.05. HP = P2 HP: <P2 You obtain 90 successes in a sample of size na = 787 from the first population. You obtain 75 successes in a sample of size 12 = 583 from the second population. For this test, you should NOT use the continuity correction, and you should use the normal distribution as an approximation for the binomial distribution. What is the test statistic for this sample? (Report answer accurate to three decimal places.) test statistic What is the p-valu for this sample? (Report answer accurate to four decimal places.) p-value

Answer 1

ket random. det p, and fz be proportion of some Charecterestic (let some characterestic A) in two populations samples of sizes n, and my are drawn from the first and secord population. det f, and I denotes the number of success (that is let number of members with characterestica) the random samples. Then, fi~ bin (n., p.) and independently. Then, haf I over, hen, f = 90, n. = 787 B = { $z=75, ny = 583 in fan bin (M2355 - 90 787 ng 75 1 583

Under Ho a let p = P2 =p we have Ê (f1 + f2) =(2, 2) þ > (Hx52) = 1 þ is estimated by h = fitta 90+ 75 0-12044 si nith, 787 +583 Test statistic Z = P-P2 a No 1) under Ho P (1-P) & to 2 75/583 787 I 0.12044 (1-0./2041) { + 787 5833 0.01429 .0.803 494 0.0003163 13 0.803

Now, p-value + z = ] Since, it is a left tailed test 3o = observed value of 2. Now, we got, zo=-0.8035 Ho ² p-value Zrunco) PHz < - 0.80 35 • (-0.8036) 12 = 1 - 0 (0.8035 1) 0.789616 0.2108