Question

TV advertising agencies face growing challenges in reaching audience members because viewing TV programs via digital streaming is increasingly popular. the Harris poll reported on November 13, 2012, that 53% of 2343 American adults surveyed said they have watched digitally streamed TV programming on some type of device.a. calculate and interpret a confidence interval at the 99% confidence level for the proportion of all adult americans who have watched streamed programming.b. what sample size would be required for the width of a 99% CI to be at most .05 irrespective of the value of p?

Answer 1

Answer:

a) The 99% confidence interval would be given (0.503;0.557).

We are 99% confident that this interval contains the true population proportion.

b) $n=\frac{0.53(1-0.53)}{(\frac{0.05}{2.58})^2}=663.2$  

And rounded up we have that n=664

Step-by-step explanation:

Data given and notation  

n=2343 represent the random sample taken    

X represent the people that they have watched digitally streamed TV programming on some type of device

$\hat p=0.53$ estimated proportion of people that they have watched digitally streamed TV programming on some type of device  

$\alpha=0.01$ represent the significance level

Confidence =0.99 or 99%

z would represent the statistic for the confidence interval  

p= population proportion of people that they have watched digitally streamed TV programming on some type of device

The population proportion present the following distribution:

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

Part a) Confidence interval

The confidence interval would be given by this formula

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

For the 99% confidence interval the value of $\alpha=1-0.99=0.01$ and $\alpha/2=0.005$, with that value we can find the quantile required for the interval in the normal standard distribution.

$z_{\alpha/2}=2.58$

And replacing into the confidence interval formula we got:

$0.53 - 2.58 \sqrt{\frac{0.53(1-0.53)}{2343}}=0.503$

$0.53 + 2.58 \sqrt{\frac{0.53(1-0.53)}{2343}}=0.557$

And the 99% confidence interval would be given (0.503;0.557).

We are 99% confident that this interval contains the true population proportion.

Part b) What sample size would be required for the width of a 99% CI to be at most 0.05 irrespective of the value of p??

The margin of error for the proportion interval is given by this formula:  

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

And on this case we have that $ME =\pm 0.05$ and we are interested in order to find the value of n, if we solve n from equation (a) we got:  

$n=\frac{\hat p (1-\hat p)}{(\frac{ME}{z})^2}$   (b)  

And replacing into equation (b) the values from part a we got:

$n=\frac{0.53(1-0.53)}{(\frac{0.05}{2.58})^2}=663.2$  

And rounded up we have that n=664