Question

A police officer would like to estimate the proportion of petty theft cases that have gone unsolved in a large city. He takes a random sample of 81 petty theft cases and finds 52 of these have gone unsolved. Set up the equation for the 95% confidence interval for p.

Format: Estimate x Crit. value x standard error

Answer 1

Answer:

$0.642 - 1.96\sqrt{\frac{0.642(1-0.642)}{81}}=0.538$

$0.642 + 1.96\sqrt{\frac{0.642(1-0.642)}{81}}=0.746$

The 95% confidence interval would be given by (0.538;0.746)

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".  

The population proportion have the following distribution

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

Solution to the problem

For this case the estimated proportion of interest would be:

$\hat p = \frac{52}{81}= 0.642$

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:  

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

If we replace the values obtained we got:

$0.642 - 1.96\sqrt{\frac{0.642(1-0.642)}{81}}=0.538$

$0.642 + 1.96\sqrt{\frac{0.642(1-0.642)}{81}}=0.746$

The 95% confidence interval would be given by (0.538;0.746)