Question

The National Center for Education Statistics reported that 47% of college students work to pay for tuition and living expenses. Assume that a sample of 450 colleges students was used in this study. 1. Provide a 95% confidence interval for the population proportion of college students who work to pay for tuition and living expenses.

Answer 1

`0.47 - 1.96\sqrt{(0.47(1-0.47))/(450)}=0.424`

`0.47 + 1.96\sqrt{(0.47(1-0.47))/(450)}=0.516`

The 95% confidence interval would be given by (0.424;0.516)

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

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

If we replace the values obtained we got:

`0.47 - 1.96\sqrt{(0.47(1-0.47))/(450)}=0.424`

`0.47 + 1.96\sqrt{(0.47(1-0.47))/(450)}=0.516`

The 95% confidence interval would be given by (0.424;0.516)