Question

Do you go to church? The Gallup Poll asked a random sample of $1785$ adults whether they attended church or synagogue during the past week. Of the respondents, $44\%$ said they did attend. Suppose that $40\%$ of the adult population actually went to church or synagogue last week. Let role="math" localid="1652798797523" $\hat{p}$ be the proportion of people in the sample who attended church or synagogue.

(a) What is the mean of the sampling distribution of $\hat{p}$ ? Why?

(b) Find the standard deviation of the sampling distribution of $\hat{p}$. Check to see if the $10\%$ condition is met.

(c) Is the sampling distribution of $\hat{p}$ approximately Normal? Check to see if the Normal condition is met.

(d) Find the probability of obtaining a sample of $1785$ adults in which $44\%$ or more say they attended church or synagogue last week. Do you have any doubts about the result of this poll?

Short answer

a). The mean is $0.40$.

b). The standard deviation is $0.0115954$.

c).A normal approximation could be applied.

d). The probability is less than $0.05$. The event's occurrence is doubted.

Step-by-step solution

Part (a) Step 1: Given Information

Sample proportion $\left(\hat{p}\right)=0.44$,

Population proportion $\left(p\right)=0.40$,

Sample size $\left(n\right)=1785$.

Part (a) Step 2: Explanation

The mean of the sampling distribution can be calculated as:

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

$=0.40$

Part (b) Step 1: Given Information 

Sample proportion $\left(\hat{p}\right)=0.44$,

Population proportion $\left(p\right)=0.40$,

Sample size $\left(n\right)=1785$.

Part (b) Step 2: Explanation

The sample proportion's standard deviation is calculated as:

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

$=\sqrt{\frac{0.40(1-0.40)}{1785}}$

$=0.0115954$

Part (c) Step 1:  Given Information

Sample proportion $\left(\hat{p}\right)=0.44$,

Population proportion $\left(p\right)=0.40$,

Sample size $\left(n\right)=1785$.

Part (c) Step 2: Explanation

Here,

$np=1785(0.40)$

$=714>10$

$n(1-p)=1785(1-0.40)$

$=1071>10$

A normal approximation could be applied.

Part (d) Step 1: Given Information

Sample proportion $\left(\hat{p}\right)=0.44$,

Population proportion $\left(p\right)=0.40$,

Sample size $\left(n\right)=1785$.

Part (d) Step 2: Explanation

The probability of $44\%$of people attending church or synagogue. In addition, the poll's result is calculated as:

$P(\hat{p}\ge 0.44)=P\left(Z\ge \frac{0.44-0.40}{0.0115954}\right)$

$=P(Z\ge 3.45)$

localid="1652841525984" $=0.0003$

The probability is less than $0.05$. As a result, the event's occurrence is doubted.