Cable TV subscriptions and “cord cutters.” According to a recent Pew Research Center Survey (December 2015), 15% of U.S. adults admitted they are “cord cutters,” i.e., they canceled the cable/satellite TV service they once subscribed to. (See Exercise 2.4, p. 72) In a random sample of 500 U.S. adults, let pn represent the proportion who are “cord cutters.”

a. Find the mean of the sampling distribution of $\hat{\mathbf{p}}$.

b. Find the standard deviation of the sampling distribution of $\hat{\mathbf{p}}$.

c. What does the Central Limit Theorem say about the shape of the sampling distribution of $\hat{\mathbf{p}}$?

d. Compute the probability that $\hat{\mathbf{p}}$is less than .12.

e. Compute the probability that $\hat{\mathit{p}}$is greater than .10.

The proportion of all U.S. adults admitted that they are “cord cutters” is $p=0.15$.

A random sample of size$n=500$ is selected.

Let$\hat{p}$ represents the sample proportion of adults who admitted that they are “cord cutters.”

a

The mean of the sampling distribution of$\hat{p}$ is obtained as:

$\begin{array}{rcl}E\left(\hat{p}\right)& =& p\\ & =& 0.15\end{array}$.

Since$E\left(\hat{p}\right)=p$.

Therefore,$E\left(\hat{p}\right)=0.15$ .

b.

The standard deviation of the sampling distribution of$\hat{p}$ is obtained as:

$\begin{array}{rcl}{\sigma }_{\hat{p}}& =& \sqrt{\frac{p\left(1-p\right)}{n}}\\ & =& \sqrt{\frac{0.15\times 0.85}{500}}\\ & =& \sqrt{\frac{0.1275}{500}}\\ & =& \sqrt{0.000255}\\ & =& 0.01596\end{array}$..

Therefore,${\sigma }_{\hat{p}}=0.0160$ .

c.

Here,

$\begin{array}{rcl}n\hat{p}& =& 500\times 0.15\\ & =& 75\\ & >& 15\\ & & \end{array}$,

and

$\begin{array}{rcl}n\left(1-\hat{p}\right)& =& 500\times 0.85\\ & =& 425\\ & >& 15\\ & & \end{array}$.

The conditions to use Central Limit Theorem are satisfied. According to the Central Limit Theorem, the shape of the sampling distribution of the sample proportion is approximately symmetric (normal).

d.

The probability that $\hat{p}$is less than 0.12 is obtained as:

$\begin{array}{rcl}P\left(\hat{p}<0.12\right)& =& P\left(\frac{\hat{p}-p}{{\sigma }_{\hat{p}}}<\frac{0.12-p}{{\sigma }_{\hat{p}}}\right)\\ & =& P\left(Z<\frac{0.12-0.15}{0.0160}\right)\\ & =& P\left(Z<\frac{-0.03}{0.0160}\right)\\ & =& P\left(Z<-1.875\right)\\ & =& P\left(Z<-1.88\right)\\ & =& 0.0301\\ & & \end{array}$.

To find the probability z-table is used; the value at the intersection of -1.80 and 0.08 is the required probability.

Hence, the required probability is 0.0301.

e.

The probability that$\hat{p}$ is greater than 0.12 is obtained as:

$\begin{array}{rcl}P\left(\hat{p}>0.10\right)& =& P\left(\frac{\hat{p}-p}{{\sigma }_{\hat{p}}}>\frac{0.10-p}{{\sigma }_{\hat{p}}}\right)\\ & =& P\left(Z>\frac{0.10-0.15}{0.0160}\right)\\ & =& P\left(Z>\frac{-0.05}{0.0160}\right)\\ & =& P\left(Z>-3.125\right)\\ & =& 1-P\left(Z⩽-3.13\right)\\ & =& 1-0.0009\\ & =& 0.9991\end{array}$.

To find the probability z-table is used, the value at the intersection of -3.10 and 0.03 is the probability of the z-score less than or equal to -3.13.