Problem 3 (35 points, 10 points for Part (1), (2), or (4), 5 points for part (3)

Question

Problem 3 (35 points, 10 points for Part (1), (2), or (4), 5 points for part (3)). Penner and Watts
(1991) investigated whether the time to drill holes in rock differs using “dry” or “wet” drilling. In
dry drilling, compressed air is used to flush the cuttings, and in wet drilling water is used. Each method is used on 12 rocks. The drilling times (in 1/1000 minutes) for the two methods are:

Dry Drilling 727 965 904 987 847 918 814 750 804 989 902 939

Sample Mean: 878.83

Sample Variance: 8004.979

Wet Drilling 607 549 762 665 588 798 704 772 780 599 603 699

Sample Mean: 677.17

Sample Variance: 7601.97

(1) Use the stem-and-leaf display to assess the normal and equal variance assumptions for two sample t-test. [Hint: For this stem-and-leaf display, you should use first digit as the stem and the last two digits as leaf.]

(2) Perform the appropriate hypothesis test about whether the time to drill holes in rock differs using “dry” or “wet” drilling. Use the p-value approach and a significance level of 0.01. (3) Construct a 99% two-sided confidence interval for the true difference in the mean drilling time.

(4) Perform the appropriate hypothesis test about equal variance assumption for two sample t-test using the significance level of 0.05.

Explanation / Answer

(1) Use the stem-and-leaf display to assess the normal and equal variance assumptions for two sample t-test. [Hint: For this stem-and-leaf display, you should use first digit as the stem and the last two digits as leaf.]

We use the R software to create back-to-back stem-and-leaf plots:

> dry <- c(727,965,904,987,847,918,814,750,804,989,902,939)
> wet <- c(607,549,762,665,588,798,704,772,780,599,603,699)
> library(aplpack)
Loading required package: tcltk
Warning message:
package ‘aplpack’ was built under R version 3.2.3
> stem.leaf.backback(dry,wet)
____________________________
1 | 2: represents 120, leaf unit: 10
       dry       wet  
____________________________
          | 5* |4      1  
          | 5. |89     3  
          | 6* |00     5  
          | 6. |69    (2)
1      2| 7* |0      5  
2      5| 7. |6789   4  
5    410| 8* |          
          | 8. |          
7   3100| 9* |          
3    886| 9. |          
          | 10* |          
____________________________
n:      12       12   
____________________________

(2) Perform the appropriate hypothesis test about whether the time to drill holes in rock differs using “dry” or “wet” drilling. Use the p-value approach and a significance level of 0.01.

The t.test is applied to carry out the appropriate hypothesis test about whether the time to drill holes in rock differs using "dry" or "wet" drilling.

> t.test(dry,wet,var.equal=TRUE)

        Two Sample t-test

data: dry and wet
t = 5.592, df = 22, p-value = 1.273e-05
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
126.8757 276.4576
sample estimates:
mean of x mean of y
878.8333 677.1667

Since the p-value is much lesser that the significance level of 0.01, we reject the null hypothesis and conclude that the time to drill is different for the two methods.

(3) Construct a 99% two-sided confidence interval for the true difference in the mean drilling time.

> t.test(dry,wet,var.equal=TRUE,conf.level=0.99)

        Two Sample t-test

data: dry and wet
t = 5.592, df = 22, p-value = 1.273e-05
alternative hypothesis: true difference in means is not equal to 0
99 percent confidence interval:
100.0127 303.3207
sample estimates:
mean of x mean of y
878.8333 677.1667

Thus, the 99% confidence intereval is (100.0127, 303.3207).

(4) Perform the appropriate hypothesis test about equal variance assumption for two sample t-test using the significance level of 0.05.

We perform the var.test to carry out the F-test for equality of variances for the two methods:

> var.test(wet,dry)

        F test to compare two variances

data: wet and dry
F = 0.94967, num df = 11, denom df = 11, p-value = 0.9333
alternative hypothesis: true ratio of variances is not equal to 1
95 percent confidence interval:
0.2733878 3.2988576
sample estimates:
ratio of variances
         0.9496671

Since the p-value is very large, we can't conclude that the variances of the two methods are different.

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