Question

A random sample of 1016 adults in a certain large country was asked? "Do you pretty...

A random sample of 1016 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 1016 adults? surveyed, 533 indicated that televisions are a luxury they could do without. Complete parts? (a) through? (c) below. Click here to view the standard normal distribution table (page 1). LOADING... Click here to view the standard normal distribution table (page 2). LOADING... ?(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. ModifyingAbove p with caretequals nothing ?(Round to three decimal places as? needed.) ?(b)??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.) A. We are nothing?% confident the proportion of adults in the country who believe that televisions are a luxury they could do without is between nothing and nothing. B. There is a nothing?% chance the proportion of adults in the country who believe that televisions are a luxury they could do without is between nothing and nothing. ?(c) 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 ? not possible possible, but not likely likely that a supermajority of adults in the country believe that television is a luxury they could do without because the 95?% confidence interval ? contains does not contain nothing. ?(Type an integer or a decimal. Do not? round.)

Homework Answers

Answer #1

CI = p ± Z a/2 Sqrt(p*(1-p)/n)))
x = Mean
n = Sample Size
a = 1 - (Confidence Level/100)
Za/2 = Z-table value
CI = Confidence Interval
Mean(x)=533
Sample Size(n)=1016
a)
Sample proportion = x/n =0.5246
b)
Confidence Interval
= [ 0.5246 ± 1.96* ( Sqrt ( 0.52460 *(1-0.52460) /1016)
=
(0.493899, 0.555314)

We are 95 % confident the proportion of adults in the country who believe that televisions are a luxury they could do without is between (0.493899 and 0.555314)
c)

it is not likely
because the 95?% confidence interval does not contain 60. ?

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