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Thank you!! Migrants through Mexico - 1, NUMERICAL ANSWER There are 7 questions in this set. As you proceed through these questions, use necessary information from previous questions. The Washington Post and the Mexican newspaper Reforma surveyed 1.200 Mexican adults in July 2019. One question asked "What should Mexico do with the migrants from Central America that cross through the country trying to reach the United States?" In the sample, 660 (the proportion 0.55) answered that these migrants should be deported to their country of origin. Assume that the Mexican adults surveyed were a random sample of adults. We want to test the null hypothesis that the population proportion who thought migrants should be deported to their country of origin was 0.6, against the alternative hypothesis that it was not: Ho:p=0.6 vs. HA:P +0.6. We'll use the sample proportion as the test statistic. The significance level is 0.05. Let's begin with the rejection region approach. The form of the rejection region is: Reject Ho if <cv1 or if > CV2 Calculate the value of cv1. Three decimals - Calculate the value of cv2. Three decimals 2 What is the conclusion of the test? FAIL TO REJECT Ho (accept Ho) REJECT Ho in favor of HA 3 REJECT FAIL TO REJECT Now let's estimate the population proportion, and use the confidence interval to test the hypothesis. Calculate the lower (smaller) endpoint of the 95% confidence interval for p. Three decimals 4

Answer 1

VS We are to test H. P=0.6 H. Pto.6 P = sample proportion of individuals who arsuared that these migrants should be deported to their country. Here P = 0.55 with n=1200 (sample size) The test statistic for testing Ho vs Hy is given by In (Å - 0.6) under Ho Tu N(0,1) 50.6(1-0.6) reject to at level & (0.05) if ITI > Zala Ta So we =20.05 here Zodile 20.025 Z 1.2 [where P(*> Zasa) 2 where Xu N(0,1) a l-p(xC Zarz) = 2 1-8 (2a/2) = 2 I (22/2) = 1 - 2 / 2 Zo.025 = 3^(-0.025) = 3- (0.975) = 1-46 obtained from standard normal table reject to if T> 1.96 722 here T<-1.96 we ie 11200 (- 0.6) $1200 (P-0.6) if > 1.96 02 <-1.96 ✓ 0.6(1.0.6) 0.6(-0.6) 70. 7107 Ⓡ-0-6>1.96 or 70.7107. (0-0.6) (-1.96 > P >0.62771858 Ř <0. 5722814 <cv, 30.572 È > CV₂ (c = 0.628

Here hence we Noul have obtained È = 0.55 & pccvo Preject Ho (occept Ha) confidence internal for p is given by Ē £1.96 (to here p =0.55 n1200 .. CI is 0.55 I (1.96 x 0.55(1-0.55) 1200 0.55 ± 0.028148 0.521852 0.522 Here, [0.521852 ,0.578148] : lower endpoint upper endpoint 0.578148 y 0.578 the sample in proportion falls within confidence interual limits hence we accept Ho according to confidence interuals Two: The conclusions are different in this problem. In some problems the conclusions are the same and in some they are different. Now they want {8 + 1.96 J3(-6)}-{P=1-26JB(HE)} È (1-P) <0.03 3.92 so.03 02 È (1-22 2 3 392 P (1-P) 9 153664 n PCI-E) > 153664 153669 9 or n> 7(1-P) 153664 Nou max value of $(1-P) = -2 127, 4268.4499 Ang 4269 Þ(1-P) <1 1 aleways #p : max value is