Chapter 9.1, Problem 9.4E

a)

To determine

To find the mean and standard deviation of the proportion $\widehat{p}$ of the sample who say they drink the cereal milk.

Answer to Problem 9.4E

Mean = 0.7 and standard deviation = 0.0144

Explanation of Solution

Given:

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. Suppose that 70% of all U.S. adults actually drink the cereal milk.

Formula:

Mean:

  ${\mu }_{\widehat{p}}=p$

Standard deviation:

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

Calculation:

Sample proportion = $\widehat{p}$ = 0.7

The mean = ${\mu }_{\widehat{p}}=\widehat{p}=0.7$

The standard deviation:

  ${\sigma }_{\widehat{p}}=\sqrt{\frac{0.7(1-0.7)}{1012}}\approx 0.0144$

b)

To determine

To draw a Normal curve that shows the sampling distribution of fits in this case.

Explanation of Solution

Given:

Mean = 0.7 and standard deviation = 0.0144

Formula:

2 standard deviation from the mean:

  ${\mu }_{\widehat{p}}\mp 2{\sigma }_{\widehat{p}}$

Calculation:

Therefore,

  $\begin{array}{l}{\mu }_{\widehat{p}}-2{\sigma }_{\widehat{p}}=0.7-2\times 0.0144\approx 0.671\\ {\mu }_{\widehat{p}}+2{\sigma }_{\widehat{p}}=0.7+2\times 0.0144\approx 0.729\end{array}$

So, standard normal curve is,

c)

To determine

To plot the sample result on plot.

Explanation of Solution

Given:

Mean = 0.7 and standard deviation = 0.0144

Population proportion = 0.67

Calculation:

d)

To determine

To find the probability that sample in which 67% or fewer say they drink the cereal milk.

Explanation of Solution

Given:

Mean = 0.7 and standard deviation = 0.0144

Population proportion = 0.67

Formula:

  $z=\frac{\widehat{p}-p_0}{\sqrt{\frac{p_0(1-p_0)}{n}}}$

Calculation:

First we need to find z-score,

  $z=\frac{0.67-0.7}{\sqrt{\frac{0.7(1-0.7)}{1012}}}=-2.08$

Therefore,

P(Z<-2.08) = 0.0188 …Using excel formula, =NORMSDIST(-2.08)

This probability is less than 0.05, which indicate that the true population proportion is likely to be p = 0.70