The Seattle Seahawks have played a total of 702 games and have won 362 of them. The New York Jets have played a total of 927 games and have won 408 of them. a. Setup a hypothesis test to test the claim that the Seattle Seahawks are a better team than the New York Jets.b. Even though we will use a normal model to conduct our hypothesis test, briefly explain why we may not have met the required conditions.c. Find the standardized test statistic.d. State the conclusion of the hypothesis test at a level of significance of 0.05.

Name: The Seattle Seahawks have played a total of 702 games and have won 362
Uploaded: 2024-03-20T07:07:12.000Z
Channel: Bartleby
Description: The Seattle Seahawks have played a total of 702 games and have won 362 of them. The New York Jets have played a total of 927 games and have won 408 of them. a. Setup a hypothesis test to test the claim that the Seattle Seahawks are a better team than

A z-test for two proportions is a hypothesis test that attempts to make a claim about the population proportions p1 and p2. Specifically, we are interested in assessing whether or not it is reasonable to claim that p1 = p2, using sample information. The Z-test for two proportions has two non-overlapping hypotheses, the null and the alternative hypothesis. Given Information : The Seattle Seahawks have played a total of 702 games and have won 362 of them. The New York Jets have played a total of 927 games and have won 408 of them. p = x/n Let, p1 be the proportion of number of matches won by Seattle Seahawks p1 = 362/702 = 0.54 p2 be the proportion of number of matches won by New York Jets p2 = 408/927 = 0.44 where , n1 = 702 & n2 = 362 are the sample sizes . State the hypothesis : H0 : Seattle Seahawks are a similar team than the New York Jets p1 = p2 H1 : Seattle Seahawks are a better team than the New York Jets p1 > p2 Test Statistic : Z=p1-p2p^(1-p^)1n1+1n2=0.52-0.44(0.48)(1-0.48)1702+1927Z = 3.15 where , p^=n1p1+n2p2n1+n2=702(0.52)+927(0.44)702+927=365.04+407.881629p^=0.4745 The test statistic Z equals 3.151999, which is not in the 95% region of acceptance: [-∞ : 1.644854].p̂1-p̂2=0.08, is not in the 95% region of acceptance: [-∞ : 0.04235]. Decision Rule : The p-value equals 0.000810783, ( p(x≤Z) = 0.999189 ). It means that the chance of type I error (rejecting a correct H0) is small: 0.0008108 (0.081%).The smaller the p-value the more it supports H1. Since p-value < α, H0 is rejected.The proportion of Group 1's population is considered to be greater than the proportion of Group 2's population. Conclusion : There is enough evidence to claim that Seattle Seahawks are a better team than the New York Jets .