Mathematics, 24.12.2019 05:31, misst002

Arandom sample of 1014 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 1014 adults surveyed, 538 indicated that televisions are a luxury they could do without. complete pans (a) through (e) below. p =

(b) verify that the requirements for constructing a confidence interval about p are satisfied. the a simple random sample, the value is, which is , and the less than or equal to 5% of

(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.

a. there is a % chance the proportion of adults in the country who believe that televisions are a luxury they could do without is between and
b. we are % confident 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? is that a supermajority of adults in the country believe that television is a luxury they could do without because the 95% confidence interval
(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. the 95% confidence interval is (, ).

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a) $\hat p=\frac{538}{1014}=0.531$ estimated proportion of people indicated that televisions are a luxury they could do without

b) Verify that the requirements for constructing a confidence interval about p are satisfied. The sample is a simple random sample, the value of np(1-p)=1014*0.531*(1-0.531)=252.53 is, which is > 10, and the less than or equal to 5% of the population.

c) The 95% confidence interval would be given by (0.500;0.562)

B. We are 95% confident the proportion of adults in the country who believe that televisions are a luxury they could do without is between 0.500 and 0.562

d) We can say that is possible that more than 60% of adult Americans believe that television is a luxury they could do without because we need to remember that the confidence interval is just a point of estimation and iits probable that the true proportion would be not contained on the interval. But we can say that is not that likely that the true proportion is greater than 60%

e) The 95% confidence interval would be given by (0.438;0.500)

Step-by-step explanation:

Notation and definitions

X=538 number of people indicated that televisions are a luxury they could do without

n=1014 random sample taken

$\hat p=\frac{538}{1014}=0.531$ estimated proportion of people indicated that televisions are a luxury they could do without

true population proportion of people indicated that televisions are a luxury they could do without

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The population proportion have the following distribution

$p \sim N(p,\sqrt{\frac{p(1-p)}{n}})$

a) $\hat p=\frac{538}{1014}=0.531$ estimated proportion of people indicated that televisions are a luxury they could do without.

c) In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 95% of confidence, our significance level would be given by $\alpha=1-0.95=0.05$ and $\alpha/2 =0.025$ . And the critical value would be given by:

$z_{\alpha/2}=-1.96, z_{1-\alpha/2}=1.96$

The confidence interval for the mean is given by the following formula:

$\hat p \pm z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}$

If we replace the values obtained we got:

$0.531 - 1.96\sqrt{\frac{0.531(1-0.531)}{1014}}=0.500$

$0.531 + 1.96\sqrt{\frac{0.531(1-0.531)}{1014}}=0.562$

The 95% confidence interval would be given by (0.500;0.562)

B. We are 95% confident the proportion of adults in the country who believe that televisions are a luxury they could do without is between 0.500 and 0.562

e) The new propotion on this case would be

$\hat p =1-0.531=0.469$