Question

In a survey of US adults, twenty percent say they never exercise. This is the highest level seen in five years.

Answer 1

Answer:

$0.20 - 1.96\sqrt{\frac{0.2(1-0.2)}{1000}}=0.175$

$0.20 + 1.96\sqrt{\frac{0.2(1-0.2)}{1000}}=0.225$

The 95% confidence interval would be given by (0.175;0.225)

Step-by-step explanation:

Assuming this question: In a survey of 1000 US adults, twenty percent say they never exercise. This is the highest level seen in five years.1 Find a 95% confidence interval for the proportion of US adults who say they never exercise. Round your answers to three decimal places.

Previous concepts

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

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

Solution to the problem

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:

$0.20 - 1.96\sqrt{\frac{0.2(1-0.2)}{1000}}=0.175$

$0.20 + 1.96\sqrt{\frac{0.2(1-0.2)}{1000}}=0.225$

The 95% confidence interval would be given by (0.175;0.225)