Answer to Question #289886 in Statistics and Probability for Rob

Question

To compare the results of boys and girls in a class, a special test was given to 50 boys who averaged 67.4 with Standard deviation of 5, and 50 girls averaged 62.8 with Standard deviation of 4.6.

a) Test, at alpha=0.05, whether the difference is significant or not.

b) Test if the difference is 3.0.

Answer 1

Boys

$n_1=50\\\bar x=67.4\\s_1=5$

Girls

$n_2=50\\\bar x_2=62.8\\s_2=4.6$

$a)$

We test the following hypotheses,

$H_0:\mu_1-\mu_2=0\\vs\\H_1:\mu_1-\mu_2\not=0$

We apply t distribution to perform this test as follows,.

The test statistic is given as,

$t_c={(\bar x_1-\bar x_2)-(\mu_1-\mu_2)\over \sqrt{{s_1^2\over n_1}+{s_2^2\over n_2}}}={(67.4-62.8)-0\over\sqrt{{25\over50}+{21.16\over 50}}}={4.6\over\sqrt{0.9232}}={4.6\over0.96083297}=4.7875$

$t_c$ is compared with the table value at $\alpha=0.05$ with $n_1+n_2-2=50+50-2=98$ degrees of freedom.

The table value is,

$t_{{0.05\over2},98}=t_{0.025,98}=1.984467$

The null hypothesis is rejected if $|t_c|\gt t_{0.025,98}.$

Now,

$|t_c|=4.7875\gt t_{0.025,98}=1.984467,$ and we reject the null hypothesis and conclude that there is sufficient evidence to show that the difference in means between boys and girls is significant at 5% level of significance.

$b)$

The hypothesis tested are,

$H_0:\mu_1-\mu_2=3\\vs\\H_1:\mu_1-\mu_2\not=3$

$t_c={(\bar x_1-\bar x_2)-(\mu_1-\mu_2)\over \sqrt{{s_1^2\over n_1}+{s_2^2\over n_2}}}=={(67.4-62.8)-3\over\sqrt{{25\over50}+{21.16\over 50}}}={(4.6-3)\over\sqrt{0.9232}}={1.6\over 0.96083297}=1.6652$

$t_c$ is compared with the table value at $\alpha=0.05$ with $n_1+n_2-2=50+50-2=98$ degrees of freedom.

The table value is,

$t_{{0.05\over2},98}=t_{0.025,98}=1.984467$

The null hypothesis is rejected if $|t_c|\gt t_{0.025,98}.$

Since $|t_c|=1.6652\lt t_{0.025,98}=1.984467,$ we fail to reject the null hypothesis and conclude that there is sufficient evidence to show that the difference in means between boys and girls is 3 at 5% significance level.