Two companies manufacture a rubber material intended for use in an automotive application. The part will be subjected to abrasive wear in the field application, so we decide to compare the material produced by each company in a test. Twenty-five samples of material from each company are tested in an abrasion test, and the amount of wear after 1000 cycles is observed. For company 1, the sample mean and standard deviation of wear are

xÌ… = 20 milligrams / 1000cycless
s1= 2 milligrams/1000cycles and for company 2, you obtain

xÌ… 2 = 15 milligrams /1000cycles

and
s2= 8 milligrams / 1000cycles

Required:
a. Do the data support the claim that the two companies produce material with different mean wear? Use α = 0.05, and assume that each population is normally distributed but that their variances are not equal. What is the P-value for this test?
b. Do the data support a claim that the material from company 1 has higher mean wear than the material from company 2? Use the same assumptions as in part (a).
c. Construct confidence intervals that will address the questions in parts (a) and (b) above.

Step-by-step explanation:

Due to the time it took to solve this question, I had little time to type the answers fully

The answers are contained fully in the attachment I have uploaded.

A. H0: u1 = u2

H1: u1 not equal to u2

The t statistics was solved to be 3.03

The degree of freedom was solved to be 26

Given this information, this is a 3 tailed test, critical value at 5% = +-2.056

T stat > critical value

We therefore reject the null hypothesis at 5% and conclude they have different mean wear

The p value is calculated using excel and the result is 0.00544

We reject null at 5% since p value is less than 0.05

B. H0: u1 = u2

H1: u1>u2

Alpha = 0.05

Following a,

Test stat = 3.03

Df = 26

Critical value for 1 tailed test = 1.706

Test stat > 1.706

We reject H0 and conclude company 1 has higher mean wear.

We find p value for a right tailed distribution using excel

Tdist(3.03,26,1)

= 0.002722

We reject h0

C. Confidence interval

1.610 < u < 8.390

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