Question

A random sample of 1005 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
1005 adults surveyed, 522 indicated that televisions are a luxury they could do without Complete parts (a) through (d) below.
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(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
pe
(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)
O A. We are
% confident the proportion of adults in the country who believe that televisions are a luxury they could do without is between and
B. There is a
% chance the proportion of adults in the country who believe that televishans are a luxury they could do without is between
(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
that a supermajority of adults in the country believe that television is a luxury they could do without because the 95% confidence
interval
and
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Answer 1

From the information given in the exercise, we build the confidence interval and solve this question. First, we have to find the point estimate for the population proportion, then using this point estimate, and sample size, we build the confidence interval. According to the built confidence interval, question c is answered.

Item a:

522 out of 1005 indicated that television is a luxury that they could do without, so:

$\pi = \frac{522}{1005} = 0.5194$

Thus, the point estimate for the population proportion of adults in the country who believe that televisions are a luxury they could do without is 0.5194.

Item b:

In a sample with a number n of people surveyed with a probability of a success of $\pi$, and a confidence level of $1-\alpha$, we have the following confidence interval of proportions.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

z is the z-score that has a p-value of .

For this problem, we have that:

$n = 1005,\pi = 0.5194$

95% confidence level

So $\alpha = 0.05$, z is the value of Z that has a p-value of $1 - \frac{0.05}{2} = 0.975$, so $Z = 1.96$.  

The lower limit of this interval is:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.5194 - 1.96\sqrt{\frac{0.5194*0.4806}{1005}} = 0.4885$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.5194 + 1.96\sqrt{\frac{0.5194*0.4806}{1005}} = 0.5503$

Thus, the 95% confidence interval for the population proportion of adults in the country who believe that televisions are a luxury they could do without is (0.4885,0.5503). The interpretation is that:

We are  95% confident the proportion of adults in the country who believe that televisions are a luxury they could do without is between 0.4885 and 0.5503.

Item c:

It is possible, but unlikely that a supermajority of adults in the country believe that television is a luxury they could do without because the 95% confidence  interval does not contain 60%.

For another example of a confidence interval for a proportion, you can check https://brainly.com/question/16807970