Chapter 7.2, Problem 41E

(a)

To determine

To calculate: The probability that the proportion in an SRS of 100 orders is as small as the proportion in your sample or smaller.

Answer to Problem 41E

The required probability is 0.0918.

Explanation of Solution

Given information:

Percent of orders = 90%

Number of orders = 5000

Number of orders for SRS = 100

Number of orders shipped in time = 86

Formula used:

The mean of the sampling distribution of $\widehat{p}$ is, ${\mu }_{\widehat{p}}=p$ .

The standard deviation of the sampling distribution of $\widehat{p}$ is, ${\sigma }_{\widehat{p}}=\sqrt{\frac{p\left(1-p\right)}{n}}$ .

Calculation:

Here,

  $\begin{array}{c}p=90\%\\ =0.90\end{array}$

  $x=86$

  $n=100$

Find the mean of sampling distribution of $\widehat{p}$ by dividing the number of successes by the sample size that is, $\widehat{p}=\frac{x}{n}$

Substitute 86 for x and 100 for n in the above formula $\widehat{p}=\frac{x}{n}$ .

  $\begin{array}{c}\widehat{p}=\frac{86}{100}\\ =0.86\end{array}$

For the mean of the sampling distribution of $\widehat{p}$ , we can use the formula ${\mu }_{\widehat{p}}=p$ .

Substitute 0.90 for p in the formula ${\mu }_{\widehat{p}}=p$ .

  ${\mu }_{\widehat{p}}=0.90$

For the standard deviation of the sampling distribution, use the formula ${\sigma }_{\widehat{p}}=\sqrt{\frac{p\left(1-p\right)}{n}}$ .

Substitute 0.90 for p and 100 for n in the above formula and simplify.

  $\begin{array}{c}{\sigma }_{\widehat{p}}=\sqrt{\frac{0.90\left(1-0.90\right)}{100}}\\ =\sqrt{\frac{0.90\times 0.1}{100}}\\ =\sqrt{\frac{0.09}{100}}\\ =0.03\end{array}$

Now, we find the z-score by using the formula $z=\frac{x-\mu }{\sigma }$ .

Substitute 0.86 for $x$ , 0.90 for $\mu$ and 0.03 for $\sigma$ in the formula $z=\frac{x-\mu }{\sigma }$ and simplify.

  $\begin{array}{c}z=\frac{0.86-0.90}{0.03}\\ =\frac{-0.04}{0.03}\\ \approx -1.33\end{array}$

We can find the probability by using table A as:

  $\begin{array}{c}P\left(\widehat{p}\le 0.86\%\right)=P\left(z<-1.33\right)\\ =0.0918\end{array}$

Hence, the required probability is 0.0918.

(b)

To determine

The explanation that why the result of the sample does not refute the 90% claim.

Explanation of Solution

Given information:

Percent of orders = 90%

Number of orders = 5000

Number of orders for SRS = 100

Number of orders shipped in time = 86

Formula used:

The mean of the sampling distribution of $\widehat{p}$ is, ${\mu }_{\widehat{p}}=p$ .

The standard deviation of the sampling distribution of $\widehat{p}$ is, ${\sigma }_{\widehat{p}}=\sqrt{\frac{p\left(1-p\right)}{n}}$ .

Here,

  $\begin{array}{c}p=90\%\\ =0.90\end{array}$

  $x=86$

  $n=100$

Find the mean of sampling distribution of $\widehat{p}$ by dividing the number of successes by the sample size that is, $\widehat{p}=\frac{x}{n}$

Substitute 86 for x and 100 for n in the above formula $\widehat{p}=\frac{x}{n}$ .

  $\begin{array}{c}\widehat{p}=\frac{86}{100}\\ =0.86\end{array}$

For the mean of the sampling distribution of $\widehat{p}$ , we can use the formula ${\mu }_{\widehat{p}}=p$ .

Substitute 0.90 for p in the formula ${\mu }_{\widehat{p}}=p$ .

  ${\mu }_{\widehat{p}}=0.90$

For the standard deviation of the sampling distribution, use the formula ${\sigma }_{\widehat{p}}=\sqrt{\frac{p\left(1-p\right)}{n}}$ .

Substitute 0.90 for p and 100 for n in the above formula and simplify.

  $\begin{array}{c}{\sigma }_{\widehat{p}}=\sqrt{\frac{0.90\left(1-0.90\right)}{100}}\\ =\sqrt{\frac{0.90\times 0.1}{100}}\\ =\sqrt{\frac{0.09}{100}}\\ =0.03\end{array}$

Now, we find the z-score by using the formula $z=\frac{x-\mu }{\sigma }$ .

Substitute 0.86 for $x$ , 0.90 for $\mu$ and 0.03 for $\sigma$ in the formula $z=\frac{x-\mu }{\sigma }$ and simplify.

  $\begin{array}{c}z=\frac{0.86-0.90}{0.03}\\ =\frac{-0.04}{0.03}\\ \approx -1.33\end{array}$

We can find the probability by using table A as:

  $\begin{array}{c}P\left(\widehat{p}\le 0.86\%\right)=P\left(z<-1.33\right)\\ =0.0918\end{array}$

Thus, the probability is 0.0918.

The above probability is greater than 0.05.

If the true population proportion is 0.90, then a sample can be obtained with sample proportion of 0.86 because the probability is greater than 0.05.

Hence, it is concluded that the result of the sample does not refute the 90% claim.