Chapter PVII, Problem 10RE

(a)

To determine

To explain: that is considerably different from the national results in the distribution of scores at this high school.

Explanation of Solution

Given:

  

Formula used:

Chi-square Statistics

  ${\chi }^2={\displaystyle \sum \frac{{\left(\text{observed}-\text{expected}\right)}^2}{\text{expected}}}$

Calculation:

  $H_0$ : The distribution of high school scores does not vary from the national results.

  $H_1$:The distribution of high school scores varies from the national results.

Score	National Distribution	Expected counts
5	11.50%	11.155
4	23.40%	22.698
3	24.90%	24.153
2	19.10%	18.527
1	21.10%	20.467

Observed count	Expected counts
26	11.155
36	22.698
19	24.153
10	18.527
6	20.467

  

  ${\chi }^2={\displaystyle \sum \frac{{\left(\text{observed}-\text{expected}\right)}^2}{\text{expected}}}=42.801$

The P-value performance is very small, close to 0.0000. Consequently, therefore dismissed the null hypothesis and concluded that the distribution of high school scores differed significantly from the national findings.

(b)

To determine

To explain: that the results of boys and girls at this school differ significantly, yes or no.

Explanation of Solution

Given:

  

Formula used:

  $SE\left(\overline{d}\right)=\frac{s_d}{\sqrt{n}}$

  $t=\frac{\overline{d}-0}{SE\left(\overline{d}\right)}$

Calculation:

  $H_0$ : No difference between performances of the girls and boys at, therefore

  $H_0:{\mu }_d=0$

  $H_1$ : difference between performances of the girls and boys at, therefore

  $H_1:{\mu }_d\ne 0$

Score	No of boys	No. of girls	Difference
5	13	13	0
4	21	15	6
3	6	13	-7
2	7	3	4
1	4	2	2
		Total	5

  

  $\begin{array}{c}SE\left(\overline{d}\right)=\frac{s_d}{\sqrt{n}}\\ =\frac{5}{\sqrt{5}}\\ =2.23607\end{array}$

  $\begin{array}{c}t=\frac{\overline{d}-0}{SE\left(\overline{d}\right)}\\ =\frac{1-0}{2.23607}\\ =0.44721\end{array}$

  $\begin{array}{c}P-value=2P\left(t>0.44721\right)\\ =2\times 0.3389\\ =0.6778\end{array}$

By seeing the P-value, which is too much large, therefore do not reject the null hypothesis. On the basis of provided sample, there is not have enough proof to conclude that there is a significance difference between performances of girls and boys at.