Answer to Question #276777 in Statistics and Probability for Din

Question

7.

a. A standardized STF1093 basic statistics test was given to 50 girls and 75 boys. The girls got an average grade of 76 with a standard deviation of 6. The boys made an average grade of 82 with a standard deviation of 8. Find the 96% confidence interval for the difference in the mean score of all boys and the mean score of all girls.

(7 marks)

b. From the above scores, a further sampling of 50 girls and 75 boys revealed an average score of boys is 79, and for girls is 78. Test the hypothesis the mean differences of scores between boys and girls are not significant at 0.04 level of significance.

(8 marks)

Answer 1

a). Confidence interval

$n_1=75$

$n_2=50$

$\bar{x_1}=82$

$\bar{x_2}=76$

$s_1=8$

$s_2=6$

$\alpha=1-0.96=0.04$

Z value at 96% confidence interval = 2.05

$CI=(\bar{x_1}-\bar{x_2})\pm z_\frac{\alpha}{2}\sqrt{\frac{\sigma_1^{2}}{n_1}+\frac{\sigma_2^{2}}{n_2}}$

$CI=(82-76)\pm 2.05\sqrt{\frac{8^{2}}{75}+\frac{6^{2}}{50}}$

$CI=(6\pm2.57)$

$CI=(3.43,8.57)$

The 96% confidence interval for the difference in the mean score of all boys and the mean score of all girls is 3.43 and 8.57.

b). Hypothesis testing

Null and alternative hypothesis:

$H_0:\mu_1=\mu_2$

$H_a:\mu_1\ne\mu_2$

Test statistic:

$t=\frac{\bar{x_1}-\bar{x_2}}{\sqrt{\frac{s_1^{2}}{n_1}+\frac{s_2^{2}}{n_2}}}$

$t=\frac{79-78}{\sqrt{\frac{8^{2}}{75}+\frac{6^{2}}{50}}}$

$t=0.797$

Critical value:

Degree of freedom = 75 + 50 - 2 = 123

From t distribution table, critical value at 0.04 is 1.657.

Statistical decision:

Hence, the test statistic is less than the critical value (0.797<1.657), so we can't reject the null hypothesis.