Question

Do you drink the cereal milk? A USA Today poll asked a random sample of $1012$U.S. adults what they do with the milk in the bowl after they have eaten the cereal. Let $\stackrel{⏜}{p}$be the proportion of people in the sample who drink the cereal milk. A spokesman for the dairy industry claims that $70\%$of all U.S. adults drink the cereal milk. 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 $\overbrace{p}$is approximately Normal.

d. Of the poll respondents, $67\%$said that they drink the cereal milk. Find the probability of obtaining a sample of $1012$adults in which $67\%$or fewer say they drink the cereal milk, assuming the milk industry spokesman’s claim is true.

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

Short answer

a. The mean is $0.70$.

b. The standard deviation is $0.0144052$.

c. The sampling distribution of $\stackrel{⏜}{p}$is approximately normal.

d. The required probability is $0.0188$.

e. Yes.

Step-by-step solution

Given Information

It is given that $p=70\%=0.70$

$\hat{p}=67\%=0.67$

$n=1012$

Calculation of mean

The mean is:

${\mu }_{\hat{p}}=p=0.70$

It is because sample proportion is an unbiased estimator of population proportion.

Standard Deviation

Standard deviation is given by:

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

$=\sqrt{\frac{0.70(1-0.70)}{1012}}$

$=0.0144052$

The standard deviation is$0.0144052$

To explain sampling distribution of p is normal

Condition of normality is

$np=1012(0.70)=7-8.4$

and

$n(1-p)=1012(0.30)=303.6$

Both values are greater than $10$

hence, distribution is approximately normal,

To find: the probability of obtaining a sample of 1012 adults in which 67 %or fewer say they drink the cereal milk, assuming the milk industry spokesman's claim is true.

From above parts,

${\mu }_{\hat{p}}=0.70$

${\sigma }_{\hat{p}}=0.0144052$

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

$=\frac{0.67-0.70}{0.0144055}$

$=-2.08$

Probability is $P(\hat{p}\le 0.67)=P(z<-2.08)=0.0188$

The probability is less than $0.05$, it is not usual to obtain sample proportion of$67\%$. The result of poll is doubtful to be accurate.

To find that if this poll gives convincing evidence against the spokesman's claim.

From above, associated probability is

$P(\hat{p}\le 0.67)=P(z<-2.08)=0.0188$

The probability is less than $0.05$, it is not usual to obtain sample proportion of$67\%$and results are not surely accurate.