Chapter 9.2, Problem 39E

(a)

To determine

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

Answer to Problem 39E

Sample proportion of 0.6833 is less than 0.80

Explanation of Solution

Given:

  $\begin{array}{l}\alpha =0.05\\ H_0:p=0.80\\ H_1:p<0.80\\ n=60\\ x=41\end{array}$

Formula used:

  $\widehat{p}=\frac{x}{n}$

Calculation:

The sample proportion is the number of successes divided by the sample size:

  $\widehat{p}=\frac{x}{n}=\frac{41}{60}=0.6833$

Since the sample proportion of 0.6833 is less than 0.80, the sample result gives some evidence for the alternative hypothesis as it agrees with the alternative hypothesis $H_1:p<0.80$ .

(b)

To determine

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

Answer to Problem 39E

Z = -2.26

P = 0.0119

Explanation of Solution

Given:

  $\begin{array}{l}\alpha =0.05\\ H_0:p=0.80\\ H_1:p<0.80\\ n=60\\ x=41\end{array}$

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

  $\widehat{p}=\frac{x}{n}=\frac{41}{60}=0.6833$

The test-statistic is

  $\begin{array}{c}z=\frac{\widehat{p}-p_0}{\sqrt{\frac{p_0\left(1-p_0\right)}{n}}}\\ =\frac{0.6833-0.80}{\sqrt{\frac{0.80\left(1-0.80\right)}{60}}}\\ =-2.26\end{array}$

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

  $P=P\left(z<-2.26\right)=0.0119$

(c)

To determine

To Explain: the conclusion would make.

Answer to Problem 39E

There is sufficient convincing proof that the proportion of all students at researcher’s high school who have a computer at home is less than 0.80.

Explanation of Solution

Given:

  $\begin{array}{l}\alpha =0.05\\ H_0:p=0.80\\ H_1:p<0.80\\ n=60\\ x=41\end{array}$

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

  $\widehat{p}=\frac{x}{n}=\frac{41}{60}=0.6833$

The test-statistic is

  $\begin{array}{c}z=\frac{\widehat{p}-p_0}{\sqrt{\frac{p_0\left(1-p_0\right)}{n}}}\\ =\frac{0.6833-0.80}{\sqrt{\frac{0.80\left(1-0.80\right)}{60}}}\\ =-2.26\end{array}$

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

  $P=P\left(z<-2.26\right)=0.0119$

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

  $P<0.05\Rightarrow \text{Reject }{H}_0$

There is sufficient convincing proof that the proportion of all students at researcher’s high school who have a computer at home is less than 0.80.