Chapter 10.1, Problem 10.19E

(a)

To determine

To find: the chi-square statistics and the cells contribute most of this statistic.

Answer to Problem 10.19E

  $H_0:$ There is no connection between hours gave on performance and extracurricular activities.

  $H_1:$ There is a connection between hours gave on performance and extracurricular activities.

Explanation of Solution

Given:

Extracurricular activities (hours per week)
	<2	2 to 12	>12
C or better	11	68	3
D or F	9	23	5

Formula used:

  $E=np$

Calculation:

Suppose $\alpha =0.05=5\%$

The row and column totals

Extracurricular activities (hours per week)
	<2	2 to 12	>12	Total
C or better	11	68	3	82
D or F	9	23	5	37
Total	20	91	8	119

The null hypothesis statement is that there is no connection between the variables where the alternative hypothesis statement is that there is a connection between the variables.

  $H_0:$ There is no connection between hours gave on performance and extracurricular activities.

  $H_1:$ There is a connection between hours gave on performance and extracurricular activities.

The expected frequencies $E$

  $\begin{array}{l}{E}_{11}=\frac{r_1\times c_1}{n}=\frac{82\times 20}{119}=13.7815\\ E_{12}=\frac{r_1\times c_2}{n}=\frac{82\times 91}{119}=62.7059\\ E_{13}=\frac{r_1\times c_3}{n}=\frac{82\times 8}{119}=5.5126\\ E_{21}=\frac{r_2\times c_1}{n}=\frac{37\times 20}{119}=6.2185\\ E_{22}=\frac{r_2\times c_2}{n}=\frac{37\times 91}{119}=28.2941\\ E_{23}=\frac{r_2\times c_3}{n}=\frac{37\times 8}{119}=2.4874\end{array}$

(b)

To determine

To Explain: the degrees of freedom and how significant the chi-square test is.

Answer to Problem 10.19E

  $df=2$

  $0.01<P<0.05$

reject the null hypothesis at the 5% significance level.

Explanation of Solution

Formula used:

  ${\chi }^2={\displaystyle \sum \frac{{\left(O-E\right)}^2}{E}}$

Calculation:

The value of the test-statistic is

  $\begin{array}{c}{\chi }^2={\displaystyle \sum \frac{{\left(O-E\right)}^2}{E}}\\ =\frac{{\left(11-13.7815\right)}^2}{13.7815}+\frac{{\left(68-62.7059\right)}^2}{62.7059}\\ +\frac{{\left(3-5.5126\right)}^2}{5.5126}+\frac{{\left(9-6.2185\right)}^2}{6.2185}\\ +\frac{{\left(23-28.2941\right)}^2}{291.1304}+\frac{{\left(5-17.3913\right)}^2}{17.3913}\\ =0.5614+0.4470+1.1452\\ +1.2441+0.0962+2.5381\\ =6.9264\end{array}$

It is observed that the largest chi-square subtotal is 2.5381 where corresponding cell of D or F and >12.

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

  $0.01<P<0.05$

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

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

(c)

To determine

To Explain: the brief conclusion for the study.

Explanation of Solution

At the 5% significance level, there is enough evidence to help the claim that there is an connection between the number of hours gave on the performance and extracurricular activities.