Question

In a recent poll, 370 people were asked if they liked dogs, and 7% said they did. Find the margin of error of this poll, at the 90% confidence level. Give your answer to three decimals.

Answer 1

Answer:

$ME=1.64\sqrt{\frac{0.07(1-0.07)}{370}}=0.0218$

Step-by-step explanation:

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".  

$p$ represent the real population proportion of interest

$\hat p=0.07$ represent the estimated proportion for the sample

n=370 is the sample size required (variable of interest)

$z$ represent the critical value for the margin of error

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 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 90% of confidence, our significance level would be given by $\alpha=1-0.90=0.10$ and $\alpha/2 =0.05$. And the critical value 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 replacing into formula (a) the values provided we got:

$ME=1.64\sqrt{\frac{0.07(1-0.07)}{370}}=0.0218$

The margin of error on this case would be ME=0.0218