Question

Mineral deposits in diamond mines are measured in percentages. Scientists study these deposits by drilling two identical holes simultaneously at different locations. An important question in this study is whether there is a difference in the true means of deposits in the two holes. A sample of data from 15 pairs of holes are collected and presented below. Use an appropriate hypothesis test with 90% reliability to determine whether there is a difference in mean mineral deposits in the two holes.

Location 1st Hole 2nd hole
1 5.5 5.7
2 11.0 11.2
3 5.9 6.0
4 8.2 5.6
5 10.0 9.3
6 7.9 7.0
7 10.1 8.4
8. 7.4 9.0
9 7.0 6.0
10 9.2 8.1
11 8.3 10.0
12 8.3 8.6
13 10.5 10.4
14 5.5 7.0
15 10.0 11.2

Answer 1

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.

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:

`d_1=X_(1h,1)-X_(2h,1)=5.5-5.7=-0.2`

Then, the sample for the differences between each pair is:

`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]`

The sample mean and standard deviation are:

`M=(1)/(n)\sum_(i=1)^n\,x_i\\\\\\M=(1)/(15)((-0.2)+(-0.2)+(-0.1)+. . .+(-1.2))\\\\\\M=(1.3)/(15)\\\\\\M=0.09\\\\\\s=\sqrt{(1)/(n-1)\sum_(i=1)^n\,(x_i-M)^2}\\\\\\s=\sqrt{(1)/(14)((-0.2-0.09)^2+(-0.2-0.09)^2+(-0.1-0.09)^2+. . . +(-1.2-0.09)^2)}\\\\\\s=\sqrt{(22.38)/(14)}\\\\\\s=√(1.6)=1.26\\\\\\`

The claim is that there is significant difference in mean mineral deposits in the two holes.

Then, the null and alternative hypothesis are:

`H_0: \mu=0\\\\H_a:\mu> 0`

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:

`s_M=(s)/(√(n))=(1.26)/(√(15))=0.325`

Then, we can calculate the t-statistic as:

`t=(M-\mu)/(s/√(n))=(0.09-0)/(0.325)=(0.09)/(0.325)=0.277`

The degrees of freedom for this sample size are:

`df=n-1=15-1=14`

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):

`\text{P-value}=P(t>0.277)=0.393`

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.