Which statement could be used in step 2 when proving

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timurjin [86]

Answer:

Step-by-step explanation:

its 9

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3 years ago

A random sample of 1014 adults in a certain large country was asked "Do you pretty much think televisions are a necessity or a l

arlik [135]

Answer:

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

$p$ 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.

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) 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

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 new propotion on this case would be

$\hat p =1-0.531=0.469$