Chapter 10, Problem 10.64RE

To determine

To Explain: that there is convincing evidence of an association between gender and goals for elementary school students and carry out a test and report conclusion.

Answer to Problem 10.64RE

There is no enough evidence to help the claim that there is a connection between gender and elementary school students.

Explanation of Solution

Formula used:

  $\begin{array}{c}{\chi }^2={\displaystyle \sum \frac{{\left(O-E\right)}^2}{E}}\\ E=np\end{array}$

Calculation:

Assume

  $\alpha =0.05$

The null hypothesis statement is that the variables are not connected, where the alternative hypothesis statement that they are connected.

  $\begin{array}{c}{H}_0:\text{There is no connection between gender and goal}\text{.}\\ H_1:\text{There is no connection between gender and goal}\text{.}\end{array}$

Row and column totals are

Gender	Grades	Popular	Sports	Total
Female	130	91	30	251
Male	117	50	60	227
Total	247	141	90	478

The expected frequencies $E$

  $\begin{array}{c}{E}_{11}=\frac{r_1\times c_1}{n}=\frac{251\times 247}{478}=129.7008\\ E_{12}=\frac{r_1\times c_2}{n}=\frac{251\times 241}{478}=74.0397\\ E_{13}=\frac{r_1\times c_3}{n}=\frac{251\times 90}{478}=47.2594\\ E_{21}=\frac{r_2\times c_1}{n}=\frac{227\times 247}{478}=117.2992\\ E_{22}=\frac{r_2\times c_2}{n}=\frac{227\times 141}{478}=66.9603\\ E_{23}=\frac{r_2\times c_3}{n}=\frac{227\times 90}{478}=42.7406\end{array}$

The value of the test- statistic is

  $\begin{array}{c}{\chi }^2={\displaystyle \sum \frac{{\left(O-E\right)}^2}{E}}\\ =\frac{{\left(130-129.7008\right)}^2}{129.7008}+\frac{{\left(91-74.0397\right)}^2}{47.2594}\\ +\frac{{\left(30-47.2594\right)}^2}{47.2594}+\frac{{\left(117-117.2992\right)}^2}{117.2992}\\ +\frac{{\left(50-66.9603\right)}^2}{66.9603}+\frac{{\left(60-42.7406\right)}^2}{42.4406}\\ =21.4553\end{array}$

The degree of freedom is $df=(r-1)\left(c-1\right)=\left(2-1\right)\left(3-1\right)=2$

The P-Value is the probability of getting the value of the test statistic, or a value more extreme. And $df=2:$

  $P<0.001$

If the P-value is equal or less than to the significance level, then the alternative hypothesis is accepted and null hypothesis is rejected:

  $P<0.5\Rightarrow {\text{Reject H}}_0$

There is no enough evidence to help the claim that there is an connection between gender and elementary school students.