Question

Assuming that all necessary conditions are met, what needs to be changed in the formula so that we can use it to constuct a confidence interval estimate for the population proportion p.

Answer 1

Answer:

The confidence interval for the mean is given by the following formula:

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

Step-by-step explanation:

1) Notation and definitions

$X$ number of people or things with a specific characteristic.

$n$ random sample taken

$\hat p$ estimated proportion of people or things with a specific characteristic.

$p$ true population proportion of  people or things with a specific characteristic.

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 population proportion have the following distribution

$p \sim N(p,\sqrt{\frac{p(1-p)}{n}})$

If we have all the conditions required in order to construct the confidence interval:

1) Random sample selected

2) np >10

3) n(1-p)>10

Then the confidence interval for the mean is given by the following formula:

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