Question

A survey of 480 high school students found that 37% had a pet. Find the margin of error. Round to the nearest percent. Use the margin of error to find an interval that is likely to contain the true population proportion.

Answer 1

Answer:

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

And replacing we got:

$ME= 1.96 *\sqrt{\frac{0.37 (1-0.37)}{480}}= 0.0432$

And replacing into the confidence interval formula we got:

$0.37 - 1.96 *\sqrt{\frac{0.37 (1-0.37)}{480}}=0.327$

$0.37 + 1.96 *\sqrt{\frac{0.37 (1-0.37)}{480}}=0.413$

And the 95% confidence interval would be given (0.327;0.413).

Step-by-step explanation:

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".

Solution to the problem

The confidence interval would be given by this formula

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

Assuming a 95% of confidence. For the 95% confidence interval the value of $\alpha=1-0.95=0.05$ and $\alpha/2=0.025$, with that value we can find the quantile required for the interval in the normal standard distribution.

$z_{\alpha/2}=1.96$

The margin of error would be:

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

And replacing we got:

$ME= 1.96 *\sqrt{\frac{0.37 (1-0.37)}{480}}= 0.0432$

And replacing into the confidence interval formula we got:

$0.37 - 1.96 *\sqrt{\frac{0.37 (1-0.37)}{480}}=0.327$

$0.37 + 1.96 *\sqrt{\frac{0.37 (1-0.37)}{480}}=0.413$

And the 95% confidence interval would be given (0.327;0.413).