Chapter 9.2, Problem 9.29E

a)

To determine

To perform hypothesis test for one sample proportion $\widehat{p}$ .

Explanation of Solution

Given:

Sample size = n = 558

Number of students are women = $x$ = 445

Formula:

Sample proportion:

  $\widehat{p}=\frac{x}{n}$

Test statistic:

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

Calculation:

The sample proportion is,

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

Population proportion is $\frac{3}{4}=0.75$

Null and alternative hypothesis:

  $\begin{array}{l}{H}_o:p=0.75\\ H_a:p>0.75\end{array}$

This corresponding to a right-tailed test, for which a z-test for one population proportion needs to be used.

Test statistic:

  $z=\frac{0.7975-0.75}{\sqrt{0.75(1-0.75)/558}}=2.59$

Test statistic z = 2.59

Using the P-value approach:

P(Z>2.59) = 1-P(Z<2.59)

P(Z>2.59) = 1-0.9952 = 0.0048

The p-value is p = 0.0048

Using excel formula, =NORMSDIST(2.59)

And a = 0.05

Since p = 0.0048 <0.051, it is concluded that the Null Hypothesis is rejected.

b)

To determine

To explain the conclusion.

Answer to Problem 9.29E

There is enough evidence to claim that more than three quarters of the students at that middle school engage in bullying behavior at 0.05 significance level.

Explanation of Solution

As p-value = 0.0048 which is < 0.05, reject H0.

Conclusion: It is concluded that the null hypothesis is rejected. Therefore, there is enough evidence to claim that more than three quarters of the students at that middle school engage in bullying behavior at 0.05 significance level. This has been done as data collected only from one school which cannot be representative of all school.