Chapter 11, Problem 2TP

To determine

To test: The research hypothesis at the .01 level of significance that boys raise their hand in class more often than girls and decide Draw conclusion regarding the research hypothesis.

Answer to Problem 2TP

In comparison with one another, a one-tailed finding (see Question 1) need not be as extreme as a two-tailed finding, given the same data, to support the same conclusion that the research hypothesis is supported. The research hypothesis is two tailed test because the number of hands raised by boys might be greater or less than hands raised by girls.

Explanation of Solution

Given information:

The data is given for 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

Calculation:

The claim is;

The null hypothesis is as follows:

${\text{H}}_0:$ There is no difference between hands raised by boys and girls.

The alternative hypothesis is as follows:

${\text{H}}_{\text{1}}:$ There is a difference between hands raised by boys and girls.

Formula used:

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,

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 1% level of significance from two tailed t-table is.2.763.

Conclusion

The calculated t value is 3.001, which is greater than tabulated value (2.763) at 1% level of significance. Hence, the null hypothesis is rejected. Therefore, there is a difference between hands raised by boys and girls. The research hypothesis is two tailed test because the number of hands raised by boys is might be greater or less than hands raised by girls.