Answer to Question #215969 in Statistics and Probability for Mohammad Hasan

Question

Is the probation problem the same for boys and girls at AIUB? CGPA of 100 boys and 125 girls were randomly checked. 25 boys and 18 girls were under probation. 

Answer 1

$H_0:p_1=p_2$

$H_a:p_1\ne p_2$

$\hat{p}_1=0.25$

$\hat{p}_2=0.144$

$Y_1=25$

$Y_2=18$

$n_1=100$

$n_2=125$

$Z=\dfrac{(\hat{p}_1-\hat{p}_2)-0}{\sqrt{\hat{p}(1-\hat{p})\left(\dfrac{1}{n_1}+\dfrac{1}{n_2}\right)}}$

$\hat{p}=\dfrac{Y_1+Y_2}{n_1+n_2}$

$\hat{p}=\dfrac{25+18}{100+125}$

$=\frac{43}{225}$

$Z=\dfrac{(0.25-0.144)-0}{\sqrt{\frac{43}{225}(1-\frac{43}{225})\left(\dfrac{1}{100}+\dfrac{1}{125}\right)}}$

$=2.01$

Let $\alpha=0.05$

$Cv=z_{0.025}=1.96$

Since the test statistic (2.01) is great than the critical value (1.96), we reject the null hypothesis and conclude that the probation problem is significantly different for  boys and girls at AIUB.