Chapter 10.1, Problem 10.14E

(a)

To determine

To Calculate: the null and alternative hypotheses for the question of interest.

Answer to Problem 10.14E

  $H_0:$ 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

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

(b)

To determine

To Calculate: the expected number of C’s or better for those who spend 2 to 12 hours per week in extracurricular activities.

Answer to Problem 10.14E

  $E_{12}=62.7059$

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:

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 expected frequencies are

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

After calculation the expected number of $C$ ’s or better who gave 2 to 12 hours per week in extracurricular activities is then $E_{12}=62.7059$ which is looking forth.