Answer to Question #289820 in Statistics and Probability for Enqu

Question

To compare the results of boys and girls in a class, a special test was given to 50 boys who average 67.4 with standard dedication of 5.0 and 50 girls averaged 62.8 with standard deviation of 4.6

• a. Test at 0.05 level of significance, whether the difference is significant or not
• b. Test if the difference is 3.0

Answer 1

a) We set up a null hypothesis (H₀) that there is no difference between the population means of men and women in word building. We assume the difference between the population means of two groups to be zero i.e., $H_0: D = 0.$

As we have uncorrelated means and large samples we have to apply the following formula to calculate SE_D:

$$SE_D=\sqrt{\dfrac{\sigma_1^2}{n_1}+\dfrac{\sigma_2^2}{n_2}}=\sqrt{\dfrac{(5.0)^2}{50}+\dfrac{(4.6)^2}{50}}$$

$$=0.960833$$

$$z=\dfrac{|\mu_1-\mu_2|-0}{SE_D}=\dfrac{|67.4-62.8|-0}{0.960833}$$

$$=4.7875$$

Based on the information provided, the significance level is  $α=0.05.$

The critical value for this two-tailed test is $z_c=1.96.$

Since it is observed that$|z| = 4.7875> z_c = 1.96,$ it is then concluded that the null hypothesis is rejected.

Therefore, there is enough evidence to claim that the population mean $\mu_1$ is different than $\mu_2,$ at the $\alpha = 0.05$ significance level.

b)

$$z=\dfrac{|\mu_1-\mu_2|-3}{SE_D}=\dfrac{|67.4-62.8|-3}{0.960833}$$

$$=1.6652$$

Based on the information provided, the significance level is  $α=0.05.$

The critical value for this two-tailed test is $z_c=1.96.$

Since it is observed that$|z| =1.6652< z_c = 1.96,$ it is then concluded that the null hypothesis is not rejected.

Therefore, there is not enough evidence to claim that the difference of means is different than $3,$ at the $\alpha = 0.05$ significance level.