Question

Part B only:

Two types of drill bits are being tested for wear. The two types of drill bits are each tested on 10 different machines, each used for a variety of parts. The data on wear in mm, over a 2-week testing cycle is shown below.

	Drill Bit
Machine	1	2
1	3.2	2.1
2	3.0	2.4
3	2.4	2.2
4	1.4	1.3
5	3.4	2.8
6	2.0	1.9
7	2.8	2.2
8	3.5	2.5
9	3.1	2.2
10	3.6	2.3

1. Is there evidence to suggest that the two types of drill bits wear differently? Use a hypothesis test with α=0.05 to make your conclusions.
2. Is it believable that Drill Bit 2 has significantly less wear than Drill Bit 1? Use a 99% CI to make this decision and state your conclusions.

Answer 1

Solution :

Here two samples each has size 10 ,so here we have two use the two sample t test .

a)

Hypothesis :

Level of significance   = 0.05

Where &   are the population means of drill bit 1 and drill bit 2 respectively.

Test statistic :

Under the Null (H₀)the test statistic is;

Where ,

And are the sample mean of sample1 and sample 2

and are the sample standard deviations of sample 1 and sample 2

n1 and n2 are sample sizes of each sample

here samples are drill bit

Test statistic t is ;

  

t = 2.53

Also T - critical Value is;

t_c = t_{(0.05,18) (from statistical t table when two tail = 0.05 ,df = 18) df = n1+n2 - 2}

t_c = 2.101

Conclusion ;

If t > t_c then reject the null hypothesis at level of significance.

Here also t = 2.53 > 2.101 , so we may reject the null hypothesis at 5% level of significance.

It is concluded that the null hypothesis Ho is rejected. Therefore, there is enough evidence to claim that population mean is different than ​, at the 0.05 significance level.

There is enough evedence to two types of drill bits wear differently.

b)

Hypothesis :

Hypothesis :

99% Confidence interval(CI)

ie Level of significance   = 0.01

Where &   are the population means of drill bit 1 and drill bit 2 respectively.

Test Statistic :

t = 2.53

t_c = t_{(0.01,18) (from statistical table when one tail = 0.01 & df = 18)}

t_c = 2.552

Conclusion ;

If t > t_c then reject the null hypothesis at level of significance.

Here also t = 2.53 < 2.552 , so we may not reject the null hypothesis at 1% level of significance.

It is concluded that the null hypothesis Ho is not rejected. Therefore, there is not enough evidence to claim that the population mean is greater than ​, at the 0.01 significance level.

or There is not enough evidence to claim that Drill bit 2 has significantly less thanDrill bit 1

Minitab output :

a)

Two-sample T for X vs Y

N Mean StDev SE Mean
X 10 2.840 0.709 0.22
Y 10 2.190 0.396 0.13

Difference = μ (X) - μ (Y)
Estimate for difference: 0.650
95% CI for difference: (0.111, 1.189)
T-Test of difference = 0 (vs ≠): T-Value = 2.53 P-Value = 0.021 DF = 18
Both use Pooled StDev = 0.5741

b)

Two-Sample T-Test and CI: X, Y

Two-sample T for X vs Y

N Mean StDev SE Mean
X 10 2.840 0.709 0.22
Y 10 2.190 0.396 0.13

Difference = μ (X) - μ (Y)
Estimate for difference: 0.650
99% lower bound for difference: -0.005
T-Test of difference = 0 (vs >): T-Value = 2.53 P-Value = 0.010 DF = 18
Both use Pooled StDev = 0.5741

For Critical Value T table is given below;