# Answer in Statistics and Probability for blossomqt #248890

In one company, two machines producing tire material intended for commercial airplane tires are subjected to resistance abrasion test in the field application. The QC engineer conducted their regular inspection of the two machines’ outputs. Upon test, the following data was recovered from the 29 selected samples. For Machine 1, the sample average and standard deviation was recorded to be 20mg per 1000 cycles and 2mg per 1000 cycles. For Machine 2, the sample average and standard deviation is 15mg and 8mg per 1000 cycles. Do the data support the claim that the two machines produce material with the same mean wear? Use α =0.10, and assume that each population is normally distributed but that their variances are equal.

Let us first make a summary of the provided data for each machine.

Machine 1

“n_1=29”

“bar{x}_1=20mg”

“S_1=2mg”

Machine 2

“n_2=29”

“bar{x}_2=15mg”

“S_2=8mg”

The hypothesis tested is,

“H_0:mu_1=mu_2”

“Against”

“H_1:mu_1not=mu_2”

To perform this test, we shall use the students’ t-distribution as described below.

The t-test statistic is given as,

“T=(bar{x}_1-bar{x}_2)/sqrt{(Sp^2(1/n_1+1/n_2))}” where “Sp^2” is the pooled sample variance which we consider because the populations are normally distributed with unknown but equal variances.

Now,

“Sp^2=((n_1-1)S_1^2+(n_2-1)S_2^2)/(n_1+n_2-2)”

“Sp^2=((29-1)4+(29-1)64)/(29+29-2)=1904/56=34”

Therefore,

“T=(20-15)/sqrt{34(1/29+1/29)}=5/sqrt{68/29}=5/ 1.531283=3.265236”

“T” is compared with the t-distribution table value at “alpha=0.1” with “(n_1+n_2-2)” degrees of freedom. The table value is given as,

“t_{alpha/2,n_1+n_2-2}=t_{0.05,56}=1.672522” and the null hypothesis is rejected if, “Tgt t_{0.05,56}”

Since “T=3.265236gt t_{0.05,56}=1.672522”, we reject the null hypothesis and conclude that there is no sufficient evidence to show that the mean wear of tires produced by the two machines are same.

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