Answer to Question #289796 in Statistics and Probability for Beki

Question

The net weight of sample of bottles filled by two different machines produced by two different 

manufactures, are (in grams) Machine 1-5, 8, 7, 6, 9, 7 Machines 2-8, 10, 11, 9, 12, 14, 9 

At the 0.05 level is the mean might of the bottled filed by machine 2 are greater than the mean weight of 

the bottles filled by machine 1?

Answer 1

$n_1=6\\ n_2=7$

The sample means for the two machines are given as,

$\bar{x_1}={\sum x\over n_1}={42\over6}=7\\\bar{x_2}={\sum x\over n_2}={73\over 7}=10.4286$

The sample variances for both machines are derived using the following formula,

${\sum(x_i^2)-{(\sum x_i)^2\over n_i}}\over n_i-1$

Therefore,

$s^2_1={304-294\over 5}=2\\s^2_2={787-761.29\over 6}=4.2857$

The hypotheses tested are,

$H_0:\mu_1=\mu_2\\vs\\H_1:\mu_1\lt\mu_2$

To perform this test, we shall apply the student's t distribution since the population variance is unknown. We shall proceed as follows,

The test statistic is given as,

$t_c={(\bar x_1-\bar x_2)\over \sqrt{{s_1^2\over n_1}+{s_2^2\over n_2}}}={(7-10.4286)\over\sqrt{{2\over6}+{4.2857\over 7}}}={-3.4285714\over\sqrt{ 0.94557823}}=-3.5259$

$t_c$ is compared with the table value at $\alpha =0.05$ with $v$ degrees of freedom. The number of degrees of freedom, $v$ is given as,

$v={({s_1^2\over n_1}+{s_2^2\over n_2})^2\over {({s_1^2\over n_1})^2\over n_1-1}+{({s_2^2\over n_2})^2\over n_2-1}}$

$v={0.94557823^2\over 0.02222222+0.06247397}=10.55\approx 11$

The table value is given by, $t_{\alpha, v}=t_{0.05, 11}=-1.796$.

We reject the null hypothesis if, $t_c\lt t_{0.05,11}$.

Since $t_c=-3.5259\lt t_{0.05,11}=-1.796,$ we reject the null hypothesis and conclude that there is sufficient evidence to show that the mean weight of bottles filled by machine 2 is greater than the mean weight of bottles filled by machine 1 at 5% level of significance.