The 7 Best Technologies Of The Colour Television

Statistics – violence on television?

Inference on proportions of two populations A survey of 1,000 Canadians revealed I think 625 is too much violence on television as well. In a survey of 1,500 Americans, 780 believe there is too much violence on television. A) The data presented sufficient evidence to conclude that there is a difference in the proportions of Canadians and Americans who believe it is too much violence on television? Use a shout = 0.01, the null and alternative hypotheses, the observed test statistic, the p-value and state your conclusion in the context of the question. b) Review the assumptions required for the purposes of this test. c) To establish a confidence interval 99% believe the difference in the proportion of Canada and the U.S. who believe there is too much violence on television. d) Interpret the confidence interval. Any guidance would be greatly appreciated!

(A). That p1 population (true) proportion of Canadians who believe that is too much violence on television and let the population is p2 (true) the proportion of Americans who believe the same shit (ie, say it is too violent on TV!). We want to test Ho: P1-P2 = Do = 0 (ie-p1-p2 = p1 = 0 or p2). You Versus: P1-P2 ≠ (D0 = 0). (IE-P1-P2 P1 ≠ ≠ 0 or P2). Our sample the proportion of Canadians believe there is much violence on TV ^ = p1 = (625/1000) = 0.625 Our sample proportion of Americans believe there is too much violence on television = P2 = ^ (780/1500) 0.52 Our observed test statistic = z = [(P1-P2 ^ ^) - D0] / √ ([^ p1 * (1 - ^ p1) / (n1-1) +] [^ p2 * (1 - ^ p2) / (n2-1)]) = [(0.625 to 0.52) -0] / √ ( [0.625 * (1-0.625) / (1000-1 )]+[^ 0.52 * (1-0.52) / (1500-1)]) = 5.24 ie | | z = 24.05 alpha = 0.05 level of significance, from our point of discharge = z (alpha / 2) = z (0 , 05 / 2) = 1.96 p = z0.025 Our value twice in the right area of the test statistic-| Z | from the standard normal curve = 2 * [P (z> 0)-P (0 = 2*(0.5-0.4990)
= 2*0.001
= 0.002

Conclusion: Debido a que nuestra estadística de prueba de | z | = 5,24 es mucho mayor que that we reject the points z (alpha / 2) = 1.96 z0.025, then we reject Ho for alpha = 0.05 has a significance level, based on our two samples survey. In addition, since the two-tailed p value of 0.002 is less than 0.05 and even less than 0.001, while there is strong evidence to conclude that there is a difference significantly between proportions of Canadians and Americans who believe there is too much violence on television. (B). Since ^ n * p1 = 1.000 * 0.625 = 625 ^ n * p2 = 1500 * 780 = 0.52 are less than 5, then n is considered important, therefore, the sampling distribution of (^ ^ P1-P2) – the probability distribution of all possible differences between the two sample proportions of Canadians and Americans believe is too much violence on television after distribution from A to Z (C). Our confidence level of 99% = 100 (1-alpha)% = 99% 1-alpha = 0.99 Alpha = 0.01 Therefore, a 99% confidence interval estimate the difference in the proportion of Canadians and Americans who believe that television is too violent. = [(P1-P2 ^ ^) (+ / -) z (alpha / 2) * √ ([^ P1 * (1 - ^ p1) / (n1-1 )]+[^ P2 * (1 - ^ p2) / (n2-1)])] = [(0652-52) (+ -) z (0.01 / 2) * √ ([^ .625 * (1-0625) / (1000-1 )]+[^ 0.52 * (1 - 0.52) / (1500-1)])] = [0.132 (+ -) z0.005 * √ ([^ .625 * (1-0625) / 999] + [^ .52 * (1-.52) / 1499])] = [0132 (+ -) 2.575 * (√ [0.234375/999] + [0.2496/1499])] = [0.08,0.18] (d). The previous interval indicates that we are convinced that 99% difference between their actual proportions in Canada and the Americans who believe there is too much violence on television should not be the value is between 0.08 and 0.18. Hope this helps.

Shouldn’t You Kill Your Television? (Part 1)