Question

Using a completely randomized design, three drill bits were tested to assess their rate of rock penetration (measured in feet per hour) in oil fields. Twelve drilling locations in Texas were used for tests—with four drilling sites randomly assigned to each of the three bits. The resultant data are:

Drill Bit 1	35.2	30.1	37.6	34.3
Drill Bit 2	25.8	29.7	26.6	30.1
Drill Bit 3	14.7	28.9	23.3	16.2

Output from a MINITAB ANOVA include:

Source	D of F	Sum of Squares	Mean Square	F-Ratio	p-Value
Drill Bit	2	366.6	183.3	9.50	0.006
Error	9	173.7	19.3
Total	11	540.2

Drill Bit	N	Mean	Std Dev
1	4	34.300	3.127
2	4	28.050	2.167
3	4	20.775	6.589

1. Calculate the pooled estimate of the standard deviation, s.

2. Use the Tukey’s Multiple Comparison Procedure to compare the mean rates of penetration for the three drill bits. Identify the means that differ. Use α = 0.01.

Answer 1

(a)

First Calculate the sample sample variance for three groups:

For Drill Bit 1:

For Drill Bit 2:

For Drill Bit 3:

Now we have sample variance for three groups:

= 9.78 , = 4.697 and = 43.409

n1 = 4, n2 = 4 and n3 = 4

Now calculate the pooled estimate of the standard deviation, s

Use this formula when groups sample sizes are same:

Pooled Estimate Standard Deviation, s =

Where k is the number of groups.

s =

s =

s =

Pooled Estimate Standard Deviation, s = 4.40

(b).

Use the Tukey’s Multiple Comparison Procedure:

Using Minitab:

First Enter the data:

Steps:

Stats-->Anova-->One Way(Unstacked)

Select all data in Response box and click on Comparisons.

Tick the Tukey's Family error rate and write 1 in box for 99% Tukey's Multiple Comparisons then click OK and Continue.

Output:

Anova Table:

Tukey's Multiple Comparisons:

Here the Drill Bit 1 and Drill Bit 3 are significant difference because the line(interval) does not contain the zero.

Then we can say there is a significant difference.