Question

Assume the population proportion is to be estimated from the sample described. Find the approximate margin of error and the 95% confidence interval for the population proportion. Sample size = 225, sample proportion = 0.28 The margin of error is 0.0587. (Round to four decimal places as needed.) Find the 95% confidence interval. (Round to the three decimal places as needed.)

Answer 1

We can conclude that the true population proportion at 95% of confidence is between (0.221;0.339)

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 =0.28` represent the estimated proportion for the sample

n=225 is the sample size required

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

The population proportion have the following distribution

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

The confidence interval would be given by this formula

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

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 is given by :

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

`Me=1.96 \sqrt{(0.28(1-0.28))/(225)}=0.0587`

And replacing into the confidence interval formula we got:

`0.28 - 1.96 \sqrt{(0.28(1-0.28))/(225)}=0.221`

`0.28 + 1.96 \sqrt{(0.108(1-0.28))/(225)}=0.339`

And the 95% confidence interval would be given (0.221;0.339).

We can conclude that the true population proportion at 95% of confidence is between (0.221;0.339)