Engineers at a tire manufacturing corporation wish to test a

Engineers at a tire manufacturing corporation wish to test a new tire material for increased durability. To test the tires under realistic road conditions, new front tires are mounted on each of 11 company cars, one tire made with a production material and the other with the experimental material. After a fixed period the 11 pairs were measured for wear. The amount of wear for each tire (in mm) is shown in the table:

Construct the 99% confidence interval for the difference based on these data.

Test, at the 1% level of significance, the hypothesis that the mean wear with the experimental material is less than that for the production material.

State null and alternative hypotheses:

Determine distribution of test statistic and compute its value:

Construct the rejection region:

Make your decision:

State your conclusion:

Solution

Q1.
CI = x1 - x2 ± t a/2 * Sqrt ( sd1 ^2 / n1 + sd2 ^2 /n2 )
Where,
x1 = Mean of Sample 1, x2 = Mean of sample2
sd1 = SD of Sample 1, sd2 = SD of sample2
a = 1 - (Confidence Level/100)
ta/2 = t-table value
CI = Confidence Interval
Mean(x1)=4.873
Standard deviation( sd1 )=1.276
Sample Size(n1)=11
Mean(x2)=4.218
Standard deviation( sd2 )=1.183
Sample Size(n1)=11
CI = [ ( 4.873-4.218) ±t a/2 * Sqrt( 1.628176/11+1.399489/11)]
= [ (0.66) ± t a/2 * Sqrt( 0.28) ]
= [ (0.66) ± 3.169 * Sqrt( 0.28) ]
= [-1.01 , 2.32]

Q2.
Set Up Hypothesis
Null Hypothesis , There Is No-Significance between them Ho: u1 = u2
Alternate, the mean wear with the experimental material is
less than that for the production material- H1: u1 > u2
Test Statistic
X(Mean)=4.873
Standard Deviation(s.d1)=1.276 ; Number(n1)=11
Y(Mean)=4.218
Standard Deviation(s.d2)=1.183; Number(n2)=11
we use Test Statistic (t) = (X-Y)/Sqrt(s.d1^2/n1)+(s.d2^2/n2)
to =4.873-4.218/Sqrt((1.62818/11)+(1.39949/11))
to =1.25
| to | =1.25
Critical Value
The Value of |t | with Min (n1-1, n2-1) i.e 10 d.f is 2.764, Reject Ho, if to>2.764
We got |to| = 1.24849 & | t | = 2.764
Make Decision
Hence Value of |to | < | t | and Here we Do not Reject Ho
P-Value:Right Tail -Ha : ( P > 1.2485 ) = 0.12015
Hence Value of P0.01 < 0.12015,Here We Do not Reject Ho