Respuesta :
Answer:
At a significance level of 0.1 (90% reliability), there is not enough evidence to support the claim that there is significant difference in mean mineral deposits in the two holes.
Step-by-step explanation:
This is a matched-pair t-test for the difference.
We have to calculate the difference d for each pair, and then treat his as the sample for our test.
For example, the difference for the first pair is:
[tex]d_1=X_{1h,1}-X_{2h,1}=5.5-5.7=-0.2[/tex]
Then, the sample for the differences between each pair is:
[tex]d=[-0.2 , -0.2 , -0.1 , 2.6 , 0.7 , 0.9 , 1.7 , -1.6 , 1 , 1.1 , -1.7 , -0.3 , 0.1 , -1.5 , -1.2][/tex]
The sample mean and standard deviation are:
[tex]M=\dfrac{1}{n}\sum_{i=1}^n\,x_i\\\\\\M=\dfrac{1}{15}((-0.2)+(-0.2)+(-0.1)+. . .+(-1.2))\\\\\\M=\dfrac{1.3}{15}\\\\\\M=0.09\\\\\\s=\sqrt{\dfrac{1}{n-1}\sum_{i=1}^n\,(x_i-M)^2}\\\\\\s=\sqrt{\dfrac{1}{14}((-0.2-0.09)^2+(-0.2-0.09)^2+(-0.1-0.09)^2+. . . +(-1.2-0.09)^2)}\\\\\\s=\sqrt{\dfrac{22.38}{14}}\\\\\\s=\sqrt{1.6}=1.26\\\\\\[/tex]
The claim is that there is significant difference in mean mineral deposits in the two holes.
Then, the null and alternative hypothesis are:
[tex]H_0: \mu=0\\\\H_a:\mu> 0[/tex]
The significance level is 0.1.
The sample has a size n=15.
The sample mean is M=0.09.
As the standard deviation of the population is not known, we estimate it with the sample standard deviation, that has a value of s=1.26.
The estimated standard error of the mean is computed using the formula:
[tex]s_M=\dfrac{s}{\sqrt{n}}=\dfrac{1.26}{\sqrt{15}}=0.325[/tex]
Then, we can calculate the t-statistic as:
[tex]t=\dfrac{M-\mu}{s/\sqrt{n}}=\dfrac{0.09-0}{0.325}=\dfrac{0.09}{0.325}=0.277[/tex]
The degrees of freedom for this sample size are:
[tex]df=n-1=15-1=14[/tex]
This test is a right-tailed test, with 14 degrees of freedom and t=0.277, so the P-value for this test is calculated as (using a t-table):
[tex]\text{P-value}=P(t>0.277)=0.393[/tex]
As the P-value (0.393) is bigger than the significance level (0.1), the effect is not significant.
The null hypothesis failed to be rejected.
At a significance level of 0.1, there is not enough evidence to support the claim that there is significant difference in mean mineral deposits in the two holes.
