Question

A random sample of 1028 adults in a certain large country was asked "Do you pretty much think televisions are a necessity or a luxury you could do without?" Of the 1028 adults surveyed, 527 indicated that televisions are a luxury they could do without. Complete parts (a) through (e) below. Click here to view the standard normal distribution table (page 1). Click here to view the standard normal distribution table (page 2). (a) Obtain a point estimate for the population proportion of adults in the country who believe that televisions are a luxury they could do without p= (Round to three decimal places as needed.) (b) Verify that the requirements for constructing a confidence interval about p are satisfied. is The sample a simple random sample, the value of which is 10, and the less than or equal to 5% of the (Round to three decimal places as needed.) (c) Construct and interpret a 95% confidence interval for the population proportion of adults in the country who believe that televisions are a luxury they could do without. Select the correct choice below and fill in any answer boxes within your choice. (Type integers or decimals rounded to three decimal places as needed. Use ascending order.) confident the proportion of adults in the country who believe that televisions are a luxury they could do without is between and O A. We are OB. There is a % probability the proportion of adults in the country who believe that televisions are a luxury they could do without is between and (d) Is it possible that a supermajority (more than 60%) of adults in the country believe that television is a luxury they could do without? Is it likely? It is that a supermajority of adults in the country believe that television is a luxury they could do without because the 95% confidence interval (Type an integer or a decimal. Do not round.) (e) Use the results of part (c) to construct a 95% confidence interval for the population proportion of adults in the country who believe that televisions are a necessity. Upper bound Lower bound: (Round to three decimal places as needed.)

Answer 1

n = 1028 X=527 point estimate of the population proportion P = x 527 -0.5126 n 1028 a 6 The sample is simple random sample the Value of np (1-P) = 1028 (0-5126) (1–0.5126) 2256-84 than or is is which 256.84 equal greater can to 10 and be the sample size 5% of the population size assumed to be 95% confidence interval with critical z assosciated 957 score can confidena interval be calculated by using the left tail area of Indl 2=1-as 20.05 795 100 al dq=00/= Z=-1.96 Z=1.96 -0.025 CS Scanded with left tail area the lives in 0.975 camzanner

table Z 1.9 of Z row starting with and the Column of 0.06 value 722 - 11.96 Pt kalu P (1-P) 95% confidence interval. وا n 0- 5126 +1.96 0.5126(1-0.520 1028 (6-5126 10.03) 0-5126 +0.03 0-03) = (0.517 5126 -0.03, 0-462, 1.5426 ra 5426 Confidence interval = 0.482 LP 20-5426 that the proportion 95% confident we are believe that of who adults in the the country do televisions are luxury they could 0.5426 without is 0.482 and between @ It is not possible that a super majority believe that if adults in the country could televisions do without ore luxury they CS because with the 95% confidenca interval does not contain CamScanner 0-6

r 0.482 LP C0.5426 Confidence interval 95% believe aym population, proportion of of adults that are televisions necessary = (1- Lower bound) (i- upper bound CI-0.482), (-0.54 5426) 0- 518. 0.4574 Lower bound = 0-4 574 Upper bound -0-518 CS Scanned with Camscanner