Question

A news organization conducted a survey about preferred methods for obtaining the news. A random sample of 1,605 adults living in a certain state was selected, and 16.2 percent of the adults in the sample reported that television was their preferred method. Which of the following is an appropriate margin of error for a 90 percent confidence interval to estimate the population proportion of all adults living in the state who would report that television is their preferred method for obtaining the news?

Answer 1

Answer:

$ME= 1.64 *\sqrt{\frac{0.162*(1-0.162)}{1605}}=0.015$

Step-by-step explanation:

Notation and definitions  

$n=1605$ random sample taken  

$\hat p=0.162$ estimated proportion of adults in the sample reported that television was their preferred method

$p$ true population proportion of adults in the sample reported that television was their preferred method

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  

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

In order to find the critical values 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 90% of confidence, our significance level would be given by $\alpha=1-0.9=0.1$ and $\alpha/2 =0.05$. And the critical values would be given by:  

$z_{\alpha/2}=-1.64, z_{1-\alpha/2}=1.64$  

The margin of error for the proportion interval is given by this formula:  

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

And if we replace the values we got:

$ME= 1.64 *\sqrt{\frac{0.162*(1-0.162)}{1605}}=0.015$