Question

Suppose that administrators at a large urban high school want to gain a better understanding of the prevalence of bullying within their school. They select a random sample of 200 students who are asked to complete a survey anonymously. Of these students, 63 report that they have experienced bullying. Use this information to find a 90% z-confidence interval for p, the true proportion of students at this school who have experienced bullying. Give the limits (bounds) of the confidence interval as a proportion, precise to three decimal places.
lower limit =______________

upper limit =___________

Answer 1

Answer:

lower limit =0.261

upper limit =0.369

Step-by-step 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".  

Description in words of the parameter p

$p$ represent the real population proportion of people that have experienced bullying

X= 63 people in the random sample that have experienced bullying

n=200 is the sample size required  

$\hat p=\frac{63}{200}=0.315$ represent the estimated proportion of people that have experienced bullying

$z_{\alpha/2}$ represent the critical value for the margin of error  

The population proportion have the following distribution  

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

Numerical estimate for p

In order to estimate a proportion we use this formula:

$\hat p =\frac{X}{n}$ where X represent the number of people with a characteristic and n the total sample size selected.

$\hat p=\frac{63}{200}=0.315$ represent the estimated proportion of people that they were planning to pursue a graduate degree

Confidence interval

The confidence interval for a proportion is given by this formula  

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

For the 90% confidence interval the value of $\alpha=1-0.90=0.1$ and $\alpha/2=0.05$, with that value we can find the quantile required for the interval in the normal standard distribution.  

$z_{\alpha/2}=1.64$  

And replacing into the confidence interval formula we got:  

$0.315 - 1.64 \sqrt{\frac{0.315(1-0.315)}{200}}=0.261$  

$0.315 - 1.64 \sqrt{\frac{0.315(1-0.315)}{200}}=0.369$  

And the 90% confidence interval would be given (0.261;0.369).  

We are confident at 90% that the true proportion of people that they were planning to pursue a graduate degree is between (0.261;0.369).  

lower limit =0.261

upper limit =0.369