Question

Do you go to church? The Gallup Poll asked a random sample of $1785$adults if they attended church during the past week. Let $\stackrel{⏜}{p}$be the proportion of people in the sample who attended church. A newspaper report claims that $40\%$of all U.S. adults went to church last week. Suppose this claim is true.

a. What is the mean of the sampling distribution of $\stackrel{⏜}{p}$?

b. Find the standard deviation of the sampling distribution of $\stackrel{⏜}{p}$. Verify that the $10\%$condition is met.

c. Verify that the sampling distribution of $\stackrel{⏜}{p}$is approximately Normal.

d. Of the poll respondents, $44\%$said they did attend church last week. Find the probability of obtaining a sample of $1785$adults in which $44\%$or more say they attended church last week, assuming the newspaper report’s claim is true.

e. Does this poll give convincing evidence against the newspaper’s claim? Explain your reasoning.

Short answer

a. The mean is $0.40$.

b. The standard deviation is $0.0115954$.

c. The sampling distribution is normal.

d. The probability is $0.03\%$.

e. Yes the pole gives convincing evidence against newspaper's claim.

Step-by-step solution

Given Information

It is given that $p=40\%=0.40$

$\hat{p}=44\%=0.44$

$n=1785$

Mean and Standard Deviation of sampling Distribution 

The mean is ${\mu }_{\hat{p}}=p=0.40$

as sample proportion is an unbiased estimator for population proportion.

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

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

$=0.0115954$

$10\%$ condition is satisfied if the sample size of$1785$is less than$10\%$of population size and this is true since there are more than$17.850$adults.

Large Count Condition

Large count condition is met as

$np=1785\times 0.4=714\ge 10$

and $n(1-p)=1785\times 0.6=1071\ge 10$

both are greater than$10$.

Explain about trueness of claim

From above parts,

${\mu }_{\hat{p}}=p=0.40$

${\sigma }_{\hat{p}}=0.0115954$

$z$score is $z=\frac{x-\mu }{\sigma }$

$=\frac{0.44-0.40}{0.0115954}$

$=3.45$

Associated probability is

$P(\hat{p}>0.60)=P(z>3.45)$

$=1-P(Z<3.45)$

$=1-0.9997$

$=0.0003$

$=0.03\%$

If this poll gives convincing evidence against newspaper's claim

From above part, associated probability $0.03\%$is small, the event of sample proportion at least $0.60$is unlikely to occur by chance.

Hence, convincing evidence is present against newspaper's claim.