Question

The net weights (in grams) of a sample of bottles filled by a machine manufactured by Edne, and the net weights of a sample filled by a similar machine manufactured by Orno, Inc., are; Edne 8 7 6 9 7 5 Orno 10 7 11 9 12 14 9 8 Testing the claim at the 0.05 level that the mean weight of the bottles filled by the Orno machine is greater than the mean weight of the bottles filled by the Edne machine, what is the critical value for this test?

Answer 1

Answer:

$t=\frac{(10 -7)-(0)}{\sqrt{\frac{2.268^2}{8}+\frac{1.414^2}{6}}}=3.036$  

$p_v =P(t_{12}>3.036) =0.0051$  

So with the p value obtained and using the significance level given $\alpha=0.05$ we have $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the mean of the group 1 (Orno) is significantly higher than the mean for the group 2 (Edne).  

Step-by-step explanation:

The system of hypothesis on this case are:  

Null hypothesis: $\mu_1 \leq \mu_2$  

Alternative hypothesis: $\mu_1 > \mu_2$  

Or equivalently:  

Null hypothesis: $\mu_1 - \mu_2 \leq 0$  

Alternative hypothesis: $\mu_1 -\mu_2 > 0$  

Our notation on this case :  

$n_1 =8$ represent the sample size for group 1 (Orno)  

$n_2 =6$ represent the sample size for group 2  (Edne)

We can calculate the sampel means and deviations with the following formulas:

$\bar X =\frac{\sum_{i=1}^n X_i}{n}$

$s=\sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}}$

$\bar X_1 =10$ represent the sample mean for the group 1  

$\bar X_2 =7$ represent the sample mean for the group 2  

$s_1=2.268$ represent the sample standard deviation for group 1  

$s_2=1.414$ represent the sample standard deviation for group 2  

If we see the alternative hypothesis we see that we are conducting a right tailed test.

On this case since the significance assumed is 0.05 and we are conducting a bilateral test we have one critica value, and we need on the right tail of the distribution $\alpha/2 = 0.05$ of the area.  

The distribution on this case since we don't know the population deviation for both samples is the t distribution with $df=8+6 -2=12$ degrees of freedom.

We can use the following excel code in order to find the critical value:

"=T.INV(1-0.05,12)"

And the rejection zone is: (1.78,infinity)

The statistic is given by this formula:  

$t=\frac{(\bar X_1 -\bar X_2)-(\mu_{1}-\mu_2)}{\sqrt{\frac{s^2_1}{n_1}+\frac{S^2_2}{n_2}}}$  

And now we can calculate the statistic:  

$t=\frac{(10 -7)-(0)}{\sqrt{\frac{2.268^2}{8}+\frac{1.414^2}{6}}}=3.036$  

The degrees of freedom are given by:  

$df=8+6-2=12$

And now we can calculate the p value using the altenative hypothesis:  

$p_v =P(t_{12}>3.036) =0.0051$  

So with the p value obtained and using the significance level given $\alpha=0.05$ we have $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the mean of the group 1 (Orno) is significantly higher than the mean for the group 2 (Edne).