Chapter 9.2, Problem 43E

To determine

To find: that these data provide convincing evidence at the $\alpha =0.05$ significance level that the media report’s claim is correct.

Answer to Problem 43E

There is enough convincing evidence that the media report’s claim is correct.

Explanation of Solution

Given:

  $\begin{array}{c}\alpha =0.05\\ n=558\\ x=445\end{array}$

Claim is more than 75%

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 claim is either the null hypothesis or the alternative hypothesis. The null hypothesis statement is that the population proportion is equal to the value given in the claim. If the null hypothesis is the claim, then the alternative hypothesis statement is the opposite of the null hypothesis.

  $\begin{array}{c}{H}_0:p=75\%=0.75\\ H_1:p>0.75\end{array}$

Conditions

The three conditions for finding a hypothesis test for the population proportion $p$ are: Random, independent (10% condition), Normal (large counts).

Random: Satisfied, the reason is that the sample is a random sample.

Independent: Satisfied, because the sample of 558 middle school students is less than 10% of the population of the population of all middle school students .

Normal: Satisfied, because

  $\begin{array}{c}n{p}_0=558\left(0.75\right)\\ =418.5\\ \text{and }n\left(1-p_0\right)=558\left(1-0.75\right)\\ =558\left(0.25\right)\\ =139.5\end{array}$

Are both at least 10.

Since all conditions are satisfied, it is appropriate to use a hypothesis test for the population proportion $p$ .

Hypothesis test

The sample proportion is

  $\widehat{p}=\frac{x}{n}=\frac{445}{558}=0.7975$

The test-static is

  $\begin{array}{c}z=\frac{\widehat{p}-p_0}{\sqrt{\frac{p_0\left(1-p_0\right)}{n}}}\\ =\frac{0.7975-0.75}{\sqrt{\frac{0.75\left(1-0.75\right)}{558}}}\\ =2.59\end{array}$

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

  $\begin{array}{c}P=P\left(z>2.59\right)\\ =1-P\left(Z<2.59\right)\\ =1-0.9952\\ =0.0048\end{array}$

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

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

There is enough convincing evidence that the media report’s claim is correct.