Question

For a project, a student randomly sampled 182 other students at UTSA to determine if the majority of students were in favor of a proposal to sell alcohol at Chili’s. He found that 75 were in favor of the proposal. Use this information to calculate a 95% confidence interval to estimate the proportion of UTSA students that are in favor of a proposal to sell alcohol at Chili’s.

Answer 1

Answer:

The 95% confidence interval would be given by (0.340;0.484)

Step-by-step explanation:

Notation and definitions

$X=75$ number of students that were in favor of the proposal

$n=182$ random sample taken

$\hat p=\frac{75}{182}=0.412$ estimated proportion of students that were in favor of the proposal

$p$ true population proportion of  students that were in favor of the proposal

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

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

Confidence interval

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:

$0.412 - 1.96\sqrt{\frac{0.412(1-0.412)}{182}}=0.340$

$0.412 + 1.96\sqrt{\frac{0.412(1-0.412)}{182}}=0.484$

The 95% confidence interval would be given by (0.340;0.484)