From the information given in the exercise, we build the confidence interval and solve this question.
Item a:
522 out of 1005 indicated that television is a luxury that they could do without, so:
![\pi = \frac{522}{1005} = 0.5194](https://tex.z-dn.net/?f=%5Cpi%20%3D%20%5Cfrac%7B522%7D%7B1005%7D%20%3D%200.5194)
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
, and a confidence level of
, we have the following confidence interval of proportions.
In which
z is the z-score that has a p-value of
.
For this problem, we have that:
95% confidence level
So
, z is the value of Z that has a p-value of
, so
.
The lower limit of this interval is:
The upper limit of this interval is:
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 brainly.com/question/16807970