Question

A traditional method of verifying mineralization grades in mining is to drill twinned​ holes, i.e., the drilling of a new​ hole, or​ "twin," next to an earlier drillhole. Geologists use data collected at both holes to estimate the total amount of a particular chemical compound present at the drilling site. The data in the associated table represent the​ chemicals' percentages for a sample of 15 twinned holes drilled at a diamond mine. The geologists want to know if there is any evidence of a difference in the true chemical compound means of all original holes and their twin holes drilled at the mine. Complete parts a through e below.

Location   1st Hole   2nd Hole
1   5.3   5.4
2   11.4   11.5
3   5.9   6
4   8.1   5.4
5   9.6   8.9
6   8   7.3
7   10.6   8.6
8   7   8.5
9   7.2   6.2
10   9.3   8.1
11   8.2   9.8
12   8.4   7.8
13   10.4   10.2
14   5.4   6.9
15   10.1   11.4

a. Explain why the data should be analyzed as paired differences. Choose the correct answer below.

A. Since geologists use the twinned holes as an estimate and make their conclusions based solely on the sum of material in both​ holes, the pair of holes twinned should be used.

B. Since geologists use the combination of the first and second holes to make an​ estimation, the observations are not independent long dash—each set of twinned holes should be treated as a pair.

C. Since geologists use only the first hole to make as an observation and the second hole as a​ check, each set of twinned holes should be treated as a pair.

D. Since geologists intend to find evidence of a difference from each​ hole, the sum of the material found in all second holes is subtracted from the sum of the material found in all first​ holes, which exactly explains the use of paired differences.

b. Compute the difference between the​ "1st hole" and​ "2nd hole" measurements for each drilling location. Complete the table below.

Location	1 st Hole	2 nd Hole	Difference		Location	1st Hole	2nd Hole	Difference
1	5.3	5.4			9	7.2	6.2
2	11.4	11.5			10	9.3	8.1
3	5.9	6			11	8.2	9.8
4	8.1	5.4			12	8.4	7.8
5	9.6	8.9			13	10.4	10.2
6	8	7.3			14	5.4	6.9
7	10.6	8.6			15	10.1	11.4
8	7	8.5			​(Type integers or decimals. Do not​ round.)

c. Find the mean and standard deviation of the differences found in part b.

x overbar d=

s Subscript sd=

​(Round to three decimal places as​ needed.)

d. Use the summary statistics found in part b to find a 99​% confidence interval for the true mean difference​ ("1st hole" minus​ "2nd hole") in chemical compound measurements.

​(__,__​)

​(Round to two decimal places as​ needed.)

e. Interpret the interval from part d. Can the geologists conclude that there is no evidence of a difference in the true means of all original holes and their twin holes drilled at the​ mine? Choose the correct interpretation below.

A. With 99​% confidence, the true mean difference in chemical compound percentages falls within the interval.

B.There is a 99​% probability that the true mean difference in chemical compound percentages falls within the interval.

C.With 99​% ​confidence, the sample mean difference in chemical compound percentages falls within the interval.

D.In 99 of 100​ locations, the actual difference in chemical compound percentages in each hole will fall within the interval.

The geologists (can, can not) conclude that there is no evidence of a difference in the true means of all original holes and their twin holes drilled at the mine because (1, 0, -1) (is, is not)

contained within the interval.

Answer 1

a)

Since geologists use the combination of first and second holes to make an estimation, the observation are not independent, each set of twinned holes should be treated as pair

b)

Sample #1	Sample #2	difference , Di =sample1-sample2
5.3	5.4	-0.1
11.4	11.5	-0.1
5.9	6	-0.1
8.1	5.4	2.7
9.6	8.9	0.7
8	7.3	0.7
10.6	8.6	2
7	8.5	-1.5
7.2	6.2	1
9.3	8.1	1.2
8.2	9.8	-1.6
8.4	7.8	0.6
10.4	10.2	0.2
5.4	6.9	-1.5
10.1	11.4	-1.3

c)

mean of difference ,    D̅ =ΣDi / n =   0.193
std dev of difference , Sd =    √ [ (Di-Dbar)²/(n-1) =    1.291

d)

sample size ,    n =    15          
Degree of freedom, DF=   n - 1 =    14   and α =    0.01  
t-critical value =    t α/2,df =    2.9768   [excel function: =t.inv.2t(α/2,df) ]      
                  
std dev of difference , Sd =    √ [ (Di-Dbar)²/(n-1) =    1.2909          
                  
std error , SE = Sd / √n =    1.2909   / √   15   =   0.3333
margin of error, E = t*SE =    2.9768   *   0.3333   =   0.9922
                  
mean of difference ,    D̅ =   0.193          
confidence interval is                   
Interval Lower Limit= D̅ - E =   0.193   -   0.9922   =   -0.7989
Interval Upper Limit= D̅ + E =   0.193   +   0.9922   =   1.1855
                  
so, confidence interval is (   -0.80 < Dbar <   1.19 )  

e)

With 99​% confidence, the true mean difference in chemical compound percentages falls within the interval

The geologists ( can not) conclude that there is no evidence of a difference in the true means of all original holes and their twin holes drilled at the mine because ( 0,) (is, )

contained within the interval.