A random sample of 1007 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 1007 adults surveyed, 522 indicated that televisions are a luxury they could do without.
1) Obtain a point estimate for the population proportion of adults in the country who believe that televisions are a luxury they could do without.
2) Verity that the requirements for constructing a confidence interval about p are satisfied.
3) Construct and interpret a 95% confidence inter a for the population proportion of adults in the country who believe a televisions are a luuy ey ou within your choice.
A) We are_% confident proportion of adults in the country who believe that televisions are a luxury they could do without is betweenand_.
B. There is a% chance the proportion of adults in the country who believe that televisions are a luxury the could do without is betweenand____.
4) Is it possible that a supermaiority (more than 60%) of adults in the country believe that television is a luxury they could do without? Is it likale?

c. A) We are 95% confident proportion of adults in the country who believe that televisions are a luxury they could do without is between 0.485 and 0.5493.

Step-by-step explanation:

A point estimate for the population proportion of adults in the country who believe that televisions are a luxury they could do without can be calculated from the sample proportion.

The sample proportion is p=0.5184.

[tex]p=X/n=522/1007=0.5184[/tex]

The requirements for constructing a confidence interval about p are:

Random: the sample is random, so it is satisfied.
Normal: the values for np and n(1-p) have to be equal or greater than 10. This is satisfied (np=522, n(1-p)=485).
Independent: as the sample is less than 10% of the population and the sampling is random, this is also satisfied.

We have to calculate a 95% confidence interval for the proportion.

The critical z-value for a 95% confidence interval is z=1.96.

The standard error of the proportion is:

[tex]\sigma_p=\sqrt{\dfrac{p(1-p)}{n}}=\sqrt{\dfrac{0.5184*0.4816}{1007}}\\\\\\ \sigma_p=\sqrt{0.000248}=0.0157[/tex]

The margin of error (MOE) can be calculated as:

[tex]MOE=z\cdot \sigma_p=1.96 \cdot 0.0157=0.0309[/tex]

Then, the lower and upper bounds of the confidence interval are:

[tex]LL=p-z \cdot \sigma_p = 0.5184-0.0309=0.4875\\\\UL=p+z \cdot \sigma_p = 0.5184+0.0309=0.5493[/tex]

The 95% confidence interval for the population proportion is (0.4875, 0.5493).