Question

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 2.1 1 3.2 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 3.5 2.5 9 3.1 2.2 10 3.6 2.3 (a) Is there evidence to suggest that the two types of drill bits wear differently? Use a hypothesis test with a 0.05 to make your conclusions. (b) 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

Answer)

First we need to find the mean and s.d for 1 and 2

1

u1 = 2.84

S1 = 0.709

N1 = 10

2

u2 = 2.19

S2 = 0.3957

N2 = 10

As the population standard deviation is unknown, we will use t distribution to conduct the test.

Null hypothesis Ho : u1 = u2

Alternate hypothesis Ha : u1 is not equal to u2

Test statistics t = (u1-u2)/standard error

Standard error = √{(s1^2/n1)+(s2^2/n2)} =0.25676048956

t = 2.532

Degrees of freedom is is given by

Smaller of n1-1, n2-1

So, df = 9

For df 9 and test statistics of 2.532,p-value from t distribution is = 0.03213

As the obtained p-value is less than the given significance level of 0.05

We reject the null hypothesis

And we have enough evidence to support the claim that the means are different.

2)

For degrees of freedom 9 and 99% confidence level, critical value t from t table is = 3.25

MOE (margin of error) = 3.25*standard error (as calculated in first)

MOE = 0.83447159107

Confidence interval is given by

(U1-u2)-MOE to (u1-u2)+MOE

−0.18447159107 to 1.48447159107

As the given confidence interval contains the value 0

That is no difference

We fail to reject Ho

And we do not have enough evidence to support the claim