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## Here's the Solution to this Question

Diet A

$n_1=10\\\bar x_1={\sum x\over n_1}={120\over10}=12$

$s_1^2={\sum x^2-{(\sum x)^2\over n_1}\over n_1-1}={1560-1440\over9}=13.33333$

Diet B

$n_2=12\\\bar x_2={\sum x\over n_2}={179\over12}=14.92$

$s_2^2={\sum x^2-{(\sum x)^2\over n_2}\over n_2-1}={2989-2670.083\over11}=28.9924245$

Before we test for the difference in means we have to test their variability using F-test.

We test,

$H_0:\sigma_1^2=\sigma_2^2\\vs\\H_1:\sigma_1^2\not=\sigma_2^2$

The test statistic is,

$F_c={s_2^2\over s_1^2}={28.9924245\over13.3333333}=2.1744$

The table value is,

$F_{\alpha,n_1-1,n_2-1}=F_{0.05,11,9}= 3.102485$ and we reject the null hypothesis if $F_c\gt F_{\alpha,n_1-1,n_2-1}$

Since $F_c=2.1744\lt F_{0.05,11,9}= 3.102485$, we fail to reject the null hypothesis and conclude that the means for both diets are equal.

Now,

The hypothesis tested are,

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

The test statistic is,

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

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

$sp^2={(n_1-1)s_1^2+(n_2-1)s_2^2\over n_1+n_2-2}={(9\times13.3333333)+(11\times28.9924245)\over20}={438.91667\over20}=21.9458335$

Therefore,

$t_c={(12-14.91667)\over \sqrt{21.95({1\over 10}+{1\over 12})}}={2.91667\over2.0058}=1.4541(4dp)$

$t_c$ is compared with the table value at $\alpha=0.05$ with $n_1+n_2-2=10+12-2=20$ degrees of freedom.

The table value is,

$t_{{0.05\over2},20}=t_{0.025,20}= 2.085963$

The null hypothesis is rejected if $t_c\gt t_{0.025,20}.$

Since, $t_c=1.4541\lt t_{0.025,20}=2.085963$, we fail to reject the null hypothesis and conclude that there is no sufficient evidence to show that the two sample means for diet A and diet B differ significantly regarding their effect on weight increase at 5% level of significance.