Question

An article a newspaper reported on the topics that teenagers most want to discuss with their parents. The​ findings, the results of a​ poll, showed that​ 46% would like more discussion about the​ family's financial​ situation, 37% would like to talk about​ school, and​ 30% would like to talk about religion. These and other percentages were based on a national sampling of 529 teenagers. Estimate the proportion of all teenagers who want more family discussions about school. Use a 95 ​% confidence level. Express the answer in the form ModifyingAbove p with caret plus or minusE and round to the nearest thousandth.

Answer 1

Answer:

$\hat p=0.37$ estimated proportion of teenagers who want more family discussions about school

$0.37 - 1.96\sqrt{\frac{0.37(1-0.37)}{529}}=0.329$

$0.37 + 1.96\sqrt{\frac{0.37(1-0.37)}{529}}=0.411$

The 95% confidence interval would be given by (0.329;0.411)

Step-by-step explanation:

Notation and definitions

$n=529$ random sample taken

Estimate the proportion of all teenagers who want more family discussions about school.

$\hat p=0.37$ estimated proportion of teenagers who want more family discussions about school

$p$ true population proportion of teenagers who want more family discussions about school

Confidence interval

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

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

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:

$0.37 - 1.96\sqrt{\frac{0.37(1-0.37)}{529}}=0.329$

$0.37 + 1.96\sqrt{\frac{0.37(1-0.37)}{529}}=0.411$

The 95% confidence interval would be given by (0.329;0.411)