Respuesta :
Answer:
At 0.05 level of significance, the abrasive wear of material 1 exceeds that of material 2 by more than 2 units
Step-by-step explanation:
We hypothesize that mean difference between abrasive wear of material 1 and material 2 is greater than 2.
So we write the null hypothesis [tex]H_0 : \mu_1 - \mu_2 >2[/tex],
and the alternative hypothesis [tex]H_1: \mu_1 - \mu_2 \leq 2[/tex].
We will find the T-score as well as the p-value. If the p-value is less than the level of significance, we will reject the null hypothesis, i.e. we will conclude that the abrasive wear of material 1 is less than that of material 2. Otherwise, we will accept the null hypothesis.
Since the variance is unknown and assumed to be equal, we will use the pooled variance
[tex]s_p^2 = \frac{(n_1-1)s_1^2 + (n_2-1)s_2^2}{n_1+n_2-2} = 20.05[/tex],
where [tex]n_1 = 12, n_2 =10, s_1 =4, s_2 = 5[/tex].
The mean of material 1 and material 2 are [tex]\mu_1 =85, \mu_2=81[/tex] respectively and mean difference [tex]d[/tex] is equal to 4. The hypothesize difference [tex]d_0[/tex] is equal to 2.
To find the T-score, we use the following formula
[tex]T = \frac{d - d_0}{\sqrt{\frac{s_p^2}{n_1} + \frac{s_p^2}{n_2} }}[/tex]
Substituting all the values into the T-score formula gives us [tex]T = 1.04[/tex], and the respective p-value is equal to 0.31. This means we have enough statistical evidence not to reject the null hypothesis, and at 5% significance level, the abrasive wear of material 1 exceeds that of material 2 by more than 2 units.
