Question

Nielsen wants to estimate the percentage that are tuned t_Q the Tonight Show. Assume that they want 95% confidence that their sample percentage has a margin of error 2 percentage points A prior study found that 19% tune to the Tonight Show The number of households must Nielsen survey is
a. 1479,
b. 2555
c, 1042
d. 633
e. 3034

Answer 1

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".

$\hat p$ 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

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

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 $\alpha=1-0.95=0.05$ and $\alpha/2 =0.025$. And the critical value would be given by:

$t_{\alpha/2}=-1.96, t_{1-\alpha/2}=1.96$

The confidence interval for the true proportion is given by the following formula:  

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

Wher the margin of error is given by:

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

And we are interested in find n, solving for n we got:

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

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

And replacing the values that we have, we got:

$n=\frac{0.19 (1-0.19)}{(\frac{0.02}{1.96})^2}=1478.05$

And if we round up to th nearest integer we got that n=1479