Question

Boys Girls Total
Speech Therapist (ST) 30 30
Neuropsychologist (NP) 45 15
Psychiatrist(P) 30 10
Total

Botlhale and Mmabatho’s final project is to determine if specialist type and gender of chil-dren are independent of each other. Use 5% level of significance.

Question 20
What is the decision with regard to the hypothesis and the conclusion about the two variables?

(1) We reject the null hypothesis and conclude that specialist type and gender of children are dependent.
(2) We reject the null hypothesis and conclude that specialist type and gender of children are independent of each other.
(3) We do reject the null hypothesis and conclude that specialist type and gender of children are independent of each other.
(4) We do reject the null hypothesis and conclude that specialist type and gender of children are dependent.
(5) We accept then null hypothesis.

Answer 1

$$\def\arraystretch{1.5} \begin{array}{c:c:c:c} & Boys & Girls & Total \\ \hline Speech\ Therapist \ (ST) & 30 & 30 & 60\\ \hline Neuropsychologist\ (NP) & 45 & 15 & 60\\\hline Psychiatrist\ (P) & 30 & 10 & 40\\ \hdashline Total & 105 & 55 & 160 \end{array}$$

$$\def\arraystretch{1.5} \begin{array}{c:c:c:c} Expected\ values& Boys & Girls & Total \\ \hline (ST) & {105\times 60\over 160}=39.375 & {55\times 60\over 160}=20.625 & 60\\ \hline (NP) & {105\times 60\over 160}=39.375 & {55\times 60\over 160}=20.625 & 60\\\hline (P) & {105\times 40\over 160 }=26.25 & {55\times 40\over 160 }=13.75 & 40\\ \hdashline Total & 105 & 55 & 160 \end{array}$$

$$\def\arraystretch{1.5} \begin{array}{c:c:c} Squared\ Distances& Boys & Girls \\ \hline (ST) & {(30-39.375)^2\over 39.375}=2.232 & {(30-20.625)^2\over 20.625}=4.261 \\ \hline (NP) & {(45-39.375)^2\over39.375}=0.804 & {(15-20.625)^2\over20.625}=1.534 \\\hline (P) & {(30-26.25)^2\over26.25}=0.536 & {(10-13.75)^2\over13.75}=1.023 \\ \end{array}$$

The following null and alternative hypotheses need to be tested:

$H_0:$ The two variables are independent

$H_1:$ The two variables are dependent

This corresponds to a Chi-Square test of independence.

Based on the information provided, the significance level is $\alpha=0.05,$ the number of degrees of freedom is $df=(r-1)(c-1)=(3-1)(2-1)=2,$ so then the rejection region for this test is

$R=\{\chi^2_{(\alpha,df)}: \chi^2_{(\alpha,df)}>5.991\}$

The Chi-Squared statistic is computed as follows:

$$\chi^2=\displaystyle\sum_{i=1}^n{(O_{ij}-E_{ij})^2\over E_{ij}}=$$

$$=2.232+0.804+0.536+4.261+1.534+1.023=10.390$$

Since it is observed that  $\chi^2=10.390>5.991=\chi^2_{(\alpha,df)},$ it is then concluded that the null hypothesis is rejected.

Therefore, there is enough evidence to claim that the two variables are dependent, at the 0.05 significance level.

The corresponding p-value for the test is $p=Pr(\chi_2^2>10.390)=0.005544.$

(1) We reject the null hypothesis and conclude that specialist type and gender of children are dependent.