Chapter 11, Problem 1TP

To determine

To test: The research hypothesis at the .05 level of significance that boys raise their hand in class more often than girls and decide whether this is a one-or two-tailed test.

Answer to Problem 1TP

The research hypothesis is one tailed (right) test because the number of hands raised by boys is greater than girls. Boys raise their hands significantly more.

Explanation of Solution

Calculation:

The given data is represents the gender and hand up wise in which 1=”Boys” and 2=”Girls”.

Gender	Hand Up	Gender	Hand Up
1	9	1	8
2	3	2	7
2	5	1	9
2	1	2	9
1	8	1	8
1	4	2	7
2	2	2	3
2	6	2	7
1	9	2	6
1	3	1	10
2	4	1	7
1	8	1	6
2	3	1	12
1	10	2	8
2	6	2	8

The claim is,

The null hypothesis is as follows:

$H_0:$ Hands raised by boys and girls are equal.

The alternative hypothesis is as follows:

$H_1:$ Hands raise by boys are more than girls.

The t-test for difference of mean is given by:

$t=\frac{\overline{X_1}-{\overline{X}}_2}{\sqrt{\left(\frac{n_1+n_2}{n_1{n}_2}\times \frac{1}{n_1+n_2-2}\left[\left(n_1-1\right)s_1^2+\left(n_2-1\right)s_2^2\right]\right)}}\sim t_{\left(n_1+n_2-2\right)}$ (1)

$s_1^2=\frac{1}{n_1-1}{\displaystyle \sum \left(X-\overline{X}\right)}$

$s_2^2=\frac{1}{n_2-1}{\displaystyle \sum \left(Y-\overline{Y}\right)}$

$\overline{X}=\frac{1}{n_1}{\displaystyle \sum X}$ and $\overline{Y}=\frac{1}{n_2}{\displaystyle \sum Y}$

Where,

• $n_1$ is the sample size of boys.
• $n_2$ is the sample size of girls.
• $s_1^2$ is the sample variance of boys.
• $s_2^2$ is the sample variance of girls.
• $\overline{X_1}$ is the mean of boys sample.
• $\overline{X_2}$ is the mean of girls sample.

The value of ${\left(X-\overline{X}\right)}^2$ is tabulated below:

Boys	Hands Up (X)	${\left(X-\overline{X}\right)}^2$
1	9	1.147959184
1	8	0.005102041
1	4	15.43367347
1	9	1.147959184
1	3	24.29081633
1	8	0.005102041
1	10	4.290816327
1	8	0.005102041
1	9	1.147959184
1	8	0.005102041
1	10	4.290816327
1	7	0.862244898
1	6	3.719387755
1	12	16.57653061
Total	111	72.92857143

Substitute 111 for ${\displaystyle \sum X}$ and 72.928 for ${{\displaystyle \sum \left(X-\overline{X}\right)}}^2$ in above $\overline{X}$ and $s_1^2$ formula.

$\begin{array}{c}\overline{X}=\frac{111}{14}\\ =7.92\end{array}$

And,

$\begin{array}{c}{s}_1^2=\frac{72.92}{13}\\ =5.60\end{array}$

Similarly for the girls:

Girls	Hands Up (Y)	${\left(Y-\overline{Y}\right)}^2$
2	3	5.34765625
2	5	0.09765625
2	1	18.59765625
2	2	10.97265625
2	6	0.47265625
2	4	1.72265625
2	3	5.34765625
2	6	0.47265625
2	7	2.84765625
2	9	13.59765625
2	7	2.84765625
2	3	5.34765625
2	7	2.84765625
2	6	0.47265625
2	8	7.22265625
2	8	7.22265625
Total	85	85.4375

Substitute 85 for ${\displaystyle \sum Y}$ and 85.437 for ${{\displaystyle \sum \left(Y-\overline{Y}\right)}}^2$  in above $\overline{Y}$ and $s_2^2$ formula.

$\begin{array}{c}\overline{Y}=\frac{85}{16}\\ =5.31\end{array}$

And,

$\begin{array}{c}{s}_2^2=\frac{85.43}{15}\\ =5.96\end{array}$

Substitute 7.92 for $\overline{X}$, 5.60 for $s_1^2$, 5.31 for $\overline{Y}$, 5.96 for $s_2^2$, 14 for $n_1$ and 16 for $n_2$ in equation (1).

$\begin{array}{c}t=\frac{7.92-5.31}{\sqrt{\left(\frac{14+16}{14\times 16}\times \frac{1}{14+16-2}\left[\left(14-1\right)\times 5.60+\left(16-1\right)5.69\right]\right)}}\\ =\frac{2.61}{\sqrt{0.7564}}\\ =\frac{2.61}{0.8697}\\ =3.001\end{array}$

The critical value at 28 d.f and 5% level of significance is 1.701.

The calculated t value is 3.001, which is greater than tabulated value (1.701) at 5% level of significance. Hence, the null hypothesis is rejected. Therefore, hands raised by boys are more than girls. The research hypothesis is one tailed (right) test because the number of hands raised by boys is greater than girls.