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## Here's the Solution to this Question

We have that,

Boys

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

Girls

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

Before we go to test the means first, we have to test their variability using F-test. This is to know whether the population variances are either equal or unequal in order to apply the appropriate test for the difference in means.

Therefore, we first test,

$H_0:\sigma_1^2=\sigma^2_2\\vs\\H_1:\sigma_1^2\not=\sigma^2_2$

The test statistic is,

$F_{cal}={s_1^2\over s_2^2}={5\over4.6}=1.086957$

The critical value is given as,

$F_{\alpha,n_1-1,n_2-1}=F_{0.05,49,49}=1.607289$ and reject the null hypothesis if $F_{cal}\gt F_{0.05,49,49}$

Since

$F_{cal}=1.086957\lt F_{0.05,49,49}=1.607289$, we accept the null hypothesis that the population variances for boys and girls are equal.

$a)$

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,

$t_c={(\bar x_1-\bar x_2)-(\mu_1-\mu_2)\over \sqrt{sp^2({1\over n_1}+{1\over n_2})}}$

where $sp^2$ is the pooled sample variance given as,

$sp^2={(n_1-1)s_1^2+(n_2-1)s_2^2\over n_1+n_2-2}={(49\times25)+(49\times21.16)\over98}={2261.84\over98}=23.08$

Therefore,

$t_c={(67.4-62.8)-0\over \sqrt{23.08({1\over 50}+{1\over 50})}}={4.6\over0.96083297}=4.7875$

$t_c$ 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,

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

$t_c={(\bar x_1-\bar x_2)-(\mu_1-\mu_2)\over \sqrt{sp^2({1\over n_1}+{1\over n_2})}}={(67.4-62.8)-3\over\sqrt{23.08({1\over50}+{1\over 50})}}={(4.6-3)\over\sqrt{0.9232}}={1.6\over 0.96083297}=1.6652$

$t_c$ 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.