Question

In a sample of 500 new websites registered on the Internet, 24 were anonymous (i.e., they shielded their name and contact information). (a) Construct a 95% confidence interval for the proportion of all new websites that were anonymous. (Do not round the intermediate answers. Round your answers to 4 decimal places.) The 95% confidence interval is from to (b) May normality of p be assumed? No

Answer 1

Answer:

$0.048 - 1.96\sqrt{\frac{0.048(1-0.048)}{500}}=0.0293$

$0.048 + 1.96\sqrt{\frac{0.048(1-0.048)}{500}}=0.0667$

The 95% confidence interval would be given by (0.0293;0.0667)

For the normal approximation we need to check the following conditions:

$np =500* 0.048= 24>10$

$n(1-p) = 500*(1-0.48) = 260>10$

So then we satisfy both conditions and we can use the normal approximation and for this reason we can construct the confidence interval

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

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 estimated proportion is given by:

$\hat p =\frac{24}{500}= 0.048$

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:

$0.048 - 1.96\sqrt{\frac{0.048(1-0.048)}{500}}=0.0293$

$0.048 + 1.96\sqrt{\frac{0.048(1-0.048)}{500}}=0.0667$

The 95% confidence interval would be given by (0.0293;0.0667)

For the normal approximation we need to check the following conditions:

$np =500* 0.048= 24>10$

$n(1-p) = 500*(1-0.48) = 260>10$

So then we satisfy both conditions and we can use the normal approximation and for this reason we can construct the confidence interval