Chapter 9.2, Problem 40E

(a)

To determine

To Explain: the sample result gives some evidence for the alternative hypothesis.

Answer to Problem 40E

Sample proportion of 0.17 is bigger than 0.13

Explanation of Solution

Given:

  $\begin{array}{c}\alpha =0.05\\ H_0:p=0.13\\ H_1:p>0.13\\ n=100\\ x=17\end{array}$

Formula used:

  $\widehat{p}=\frac{x}{n}$

Calculation:

The sample proportion is

  $\widehat{p}=\frac{x}{n}=\frac{17}{100}=0.17$

Since the sample proportion of 0.17 is bigger than 0.13, the sample result gives some evidence for the alternative hypothesis as it agrees with the alternative hypothesis $H_1:p>0.17$ .

(b)

To determine

To Calculate: the standardized test statistic and P-value.

Answer to Problem 40E

Z=1.19

P=0.1170

Explanation of Solution

Given:

  $\begin{array}{c}\alpha =0.05\\ H_0:p=0.13\\ H_1:p>0.13\\ n=100\\ x=17\end{array}$

Formula used:

  $\begin{array}{c}\widehat{p}=\frac{x}{n}\\ z=\frac{\widehat{p}-p_0}{\sqrt{\frac{p_0\left(1-p_0\right)}{n}}}\end{array}$

Calculation:

The sample proportion is

  $\widehat{p}=\frac{x}{n}=\frac{17}{100}=0.17$

The test-statistic is

  $\begin{array}{c}z=\frac{\widehat{p}-p_0}{\sqrt{\frac{p_0\left(1-p_0\right)}{n}}}\\ =\frac{0.17-0.13}{\sqrt{\frac{0.13\left(1-0.13\right)}{100}}}\\ =1.19\end{array}$

The P-value is the probability of getting the value of the test static, or a value more extreme, when the null hypothesis is true; find the P-value using the normal probability table.

  $\begin{array}{c}P=P\left(z>1.19\right)\\ =1-P\left(Z<1.19\right)\\ =1-0.8830\\ =0.1170\end{array}$

(c)

To determine

To Explain: the conclusion would make.

Answer to Problem 40E

There is no enough convincing evidence that the true proportion of all students at researcher’s elementary school who typically walk to school is greater than 0.13.

Explanation of Solution

Given:

  $\begin{array}{c}\alpha =0.05\\ H_0:p=0.13\\ H_1:p>0.13\\ n=100\\ x=17\end{array}$

Formula used:

  $\begin{array}{c}\widehat{p}=\frac{x}{n}\\ z=\frac{\widehat{p}-p_0}{\sqrt{\frac{p_0\left(1-p_0\right)}{n}}}\end{array}$

Calculation:

The sample proportion is

  $\widehat{p}=\frac{x}{n}=\frac{17}{100}=0.17$

The test-statistic is

  $\begin{array}{c}z=\frac{\widehat{p}-p_0}{\sqrt{\frac{p_0\left(1-p_0\right)}{n}}}\\ =\frac{0.17-0.13}{\sqrt{\frac{0.13\left(1-0.13\right)}{100}}}\\ =1.19\end{array}$

The P-value is the probability of getting the value of the test static, or a value more extreme, when the null hypothesis is true, find the P-value using the normal probability table

  $P=P\left(z>1.19\right)=1-P\left(Z<1.19\right)=1-0.8830=0.1170$

If the P-value is lesser than the significance level $\alpha$ , then reject the null hypothesis:

  $P>0.05\Rightarrow \text{Fail to reject }{H}_0$

There is no enough convincing evidence that the true proportion of all students at researcher’s elementary school who typically walk to school is greater than 0.13.