Question

Do you jog? The Gallup Poll asked a random sample of 1540 adults, "Do you happen to jog?" Suppose that the true proportion of all adults who jog is $p=0.15$. $p=0.15$.

a. What is the mean of the sampling distribution of $\mathrm{p}\wedge \hat{p}$?

b. Calculate and interpret the standard deviation of the sampling distribution of ${\mathrm{p}}^{\wedge \hat{p}}$. Check that the $10\%$condition is met.

c. Is the sampling distribution of ${\mathrm{p}}^{\wedge \hat{p}}$approximately Normal? Justify your answer.

d. Find the probability that between $13\%and17\%$of people jog in a random sample of 1540 adults.

Short answer

(a)The mean of the sampling distribution of 0$\hat{p}$is equal to the population proportion:

${\mu }_{\hat{p}}=p=0.15$

(b) The $10\%$ condition is met, because the 1540 adults is less than $10\%$ of the population of all adults.

(c) since both are greater than $10$the distribution is approximately normal,

(d)the corresponding probability using table$=0.9722$

Step-by-step solution

Part (a) Step 1: Given Information

Given,$$\begin{gathered} n=1540 \\ p=15\%=0.15 \end{gathered}$$

Part (a) Step 2: Simplification

The mean of the sampling distribution is equal to the population proportion:

${\mu }_{\hat{p}}=p=0.15$

Part (b) Step 1: Given Information

Given,$$\begin{gathered} n=1540 \\ p=15\%=0.15 \end{gathered}$$

Formula used:

${\sigma }_{\hat{p}}=\sqrt{\frac{p(1-p)}{n}}$

Part (b) Step 2: Simplification

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

$\begin{array}{rcl}{\sigma }_{\hat{p}}& =& \sqrt{\frac{p(1-p)}{n}}\\ & =& \sqrt{\frac{0.15(1-0.15)}{1540}}\\ & =& 0.0091\end{array}$

The proportion of adults who jog in a random sample of 1540 adults varies on average by $0.0091$from the mean proportion of $0.15$.

The $10\%$condition is met, because the 1540 adults is less than $10\%$of the population of all adults.

Part (c) Step 1: Given Information

Given,$$\begin{gathered} n=1540 \\ p=15\%=0.15 \end{gathered}$$

10Part (c) Step 2: Simplification

The sampling distribution of $\hat{p}$is approximately normal if n p and $n(1-p)$are both at least 10 .

$$\begin{gathered} np=1540\times 0.15=231 \\ n(1-p)=1540\times (1-0.15)=1309 \end{gathered}$$

Since both are greater than the distribution is approximately normal.

Part (d) Step 1: Given Information

Given,$z=\frac{x-\mu }{\sigma }$

Part (d) Step 2: Simplification

The sampling distribution of $\hat{p}$is approximately normal with mean $0.15{\sigma }_{\hat{p}}$and standard deviation $0.0091$

The z-score is

$\begin{array}{rcl}z& =& \frac{x-\mu }{\sigma }\\ & =& \frac{0.13-0.15}{0.0091}\\ & =& -2.20\\ z& =& \frac{x-\mu }{\sigma }\\ & =& \frac{0.17-0.15}{0.0091}\\ & =& 2.20\end{array}$

Finding the corresponding probability using table

$$\begin{gathered} P(0.13\le \hat{p}\le 0.17)=P(-2.20<z<2.20)-P(z<-2.20) \\ =0.9861-0.0139 \\ =0.9722 \end{gathered}$$