Question

1. Find the estimated margin of error when estimating the proportion of red chips in a mystery bag if 18 red chips were drawn from the bag in a random sample of 50 chips.

Answer 1

`ME=1.96 \sqrt{(0.36 (1-0.36))/(50)} =0.133`

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".

The population proportion have the following distribution

`p \sim N(p,\sqrt{(p(1-p))/(n)})`

Solution to the problem

We assume for this case a confidence level of 95%. 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 margin of error for the proportion interval is given by this formula:

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

The estimated proportion of red chips is given by:

`\hat p = (18)/(50)= 0.36`

And if we replace into formula (a) we got this:

`ME=1.96 \sqrt{(0.36 (1-0.36))/(50)} =0.133`