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

`ME=1.64\sqrt{(0.07(1-0.07))/(370)}=0.0218`

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{(\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{(\hat p (1-\hat p))/(n)}` (a)

And replacing into formula (a) the values provided we got:

`ME=1.64\sqrt{(0.07(1-0.07))/(370)}=0.0218`

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