A poll asked for people\'s opinion on whether closing local newspapers would hur
ID: 3055162 • Letter: A
Question
A poll asked for people's opinion on whether closing local newspapers would hurt civic life, 432 of 1005 respondents said it would hurt civic life a lot Complete parts a through d below. a. Find the proportion of the respondents who said that closing local papers would hurt civic life a lot. (Round to three decimal places as needed.) b. Find a 95% confidence interval the population pro or on who believed closing ne spapers would hurt c c 1 e a ot Assume the o use a simple and m san e RS). In ac sed random samo n buta mo e complex et d than SR Round to three decimal places as needed.) c. Find an 80% confidence interval for the population proportion who believed closing newspapers would hurt civic life a lot. Round to three decimal places as needed.) d. Which interval is wider and why? @) A. The 95% interval is wider. To get a higher degree of certainty, the interval needs to be widened. O B. The 80% interval is wider. To get a higher degree of certainty, the interval needs to be widened. d C. The 95% interval is wider. To get a higher degree of certainty the interval needs to be widened just on the right side ? D. The 95% interval is wider. To get a higher degree of certainty, the interval needs to be widened just on the left side.Explanation / Answer
Solution : Given that p = 0.430 , q = 0.570 , n = 1005
a. the proportion of the respondents is = 432/1005 = 0.430
b. 95% confidence interval for z = 1.96
E = z*sqrt((p*q)/n)
= 1.96*sqrt((0.430*0.570)/1005)
= 0.0306
95% CI = 0.43 - 0.0306 < p < 0.43 + 0.0306
= 0.3994 < p < 0.4606
= (0.399,0.461)
c. 80% confidence interval for z = 1.2820
E = 1.2820*((0.430*0.570)/1005)
= 0.00031
80% CI = 0.430 - 0.00031 < p < 0.430 + 0.00031
= 0.4296 < p < 0.4303
= (0.430,0.430)