Question

2. The school newspaper at a large high school reported that 120 out of 200 randomly selected students favor assigned parking spaces. Compute the margin of error.
Interpret the resulting interval in context.

Answer 1

Answer:

$ME= 1.96\sqrt{\frac{0.6 (1-0.6)}{200}}=0.0679$

$0.6 - 1.96\sqrt{\frac{0.6 (1-0.6)}{200}}=0.532$

$0.6 + 1.96\sqrt{\frac{0.6 (1-0.6)}{200}}=0.668$

The 95% confidence interval would be given by (0.532;0.668)

We are confident at 95% that the true proportion of students in favor of the assigned parking spaces is between 0.532 and 0.668.

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

The estimated proportion on this case is given by:

$\hat p =\frac{120}{200}=0.6$

We assume for this case a confidence level of 95%

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:  

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

The margin of error is given by:

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

And if we replace we got:

$ME= 1.96\sqrt{\frac{0.6 (1-0.6)}{200}}=0.0679$

If we replace the values obtained we got:

$0.6 - 1.96\sqrt{\frac{0.6 (1-0.6)}{200}}=0.532$

$0.6 + 1.96\sqrt{\frac{0.6 (1-0.6)}{200}}=0.668$

The 95% confidence interval would be given by (0.532;0.668)

We are confident at 95% that the true proportion of students in favor of the assigned parking spaces is between 0.532 and 0.668.