Respuesta :
Answer:
a. 1479
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".
[tex]\hat p[/tex] estimated proportion
n represent the sample size
Me =0.02 or 2% points represent the margin of error
Solution to the problem
The population proportion have the following distribution
[tex]p \sim N(p,\sqrt{\frac{\hat p(1-\hat p)}{n}})[/tex]
We need to find a critical value in order to estimate the sample size required. 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 [tex]\alpha=1-0.95=0.05[/tex] and [tex]\alpha/2 =0.025[/tex]. And the critical value would be given by:
[tex]t_{\alpha/2}=-1.96, t_{1-\alpha/2}=1.96[/tex]
The confidence interval for the true proportion is given by the following formula:
[tex]\hat p \pm z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}[/tex]
Wher the margin of error is given by:
[tex]Me= z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}[/tex]
And we are interested in find n, solving for n we got:
[tex](\frac{Me}{z_{\alpha/2}})^2=\frac{\hat p (1-\hat p)}{n}[/tex]
[tex]n=\frac{\hat p (1-\hat p)}{(\frac{Me}{z_{\alpha/2}})^2}[/tex]
And replacing the values that we have, we got:
[tex]n=\frac{0.19 (1-0.19)}{(\frac{0.02}{1.96})^2}=1478.05[/tex]
And if we round up to th nearest integer we got that n=1479
