Question

Suppose that administrators of a large school district wish to estimate the proportion of children in the district enrolling in kindergarten who attended preschool. They took a simple random sample of children in the district who are enrolling in kindergarten. Out of 50 children sampled, 34 had attended preschool.
Construct a large-sample 95% z-confidence interval for p, the proportion of all children enrolled in kindergarten who attended preschool. Give the limits of the confidence interval as decimals, precise to at least three decimal places.

Construct a plus four 95% z-confidence interval for p. Give the limits of the confidence interval as decimals, precise to at least three decimal places.

Answer 1

Answer:

a) The 95% confidence interval would be given by (0.551;0.809)

b) $(\hat p -Me ,\hat p +ME)=(0.667-0.126, 0.667+0.126)=(0.541,0.792)$

Step-by-step explanation:

Part a

Notation and definitions

$X=34$ number of children that had attended preschool

$n=50$ random sample taken

$\hat p=\frac{34}{50}=0.68$ estimated proportion of children that had attended preschool

$p$ true population proportion of children that had attended preschool

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:

$0.68 - 1.96\sqrt{\frac{0.68(1-0.68)}{50}}=0.551$

$0.68 + 1.96\sqrt{\frac{0.68(1-0.68)}{50}}=0.809$

The 95% confidence interval would be given by (0.551;0.809)

Part b

For this case the estimation for the proportion is given by:

$\hat p=\frac{X+2}{n+4}=\frac{34+2}{50+4}= 0.667$

And the margin of error is given by:

$ME= Z_{\alpha/2} \sqrt{\frac{\hat p (1-\hat p)}{n+4}}$

And if we replace we got:

$ME= 1.96 \sqrt{\frac{0.667 (1-0.667)}{50+4}}=0.126$

And the confidence interval is given by:

$(\hat p -Me ,\hat p +ME)=(0.667-0.126, 0.667+0.126)=(0.541,0.792)$