The Melting points of two alloys The melting points of two a

The Melting points of two alloys...

The melting points of two alloys used in formulating solder were investigated by melting 21 samples from each alloy. The sample mean and sample standard deviation for Alloy 1 was x = 420 degree F and s1 = 4 degree F, and for Alloy 2, they were X_2 = 426 degree F and s2 = 3 degree F. Assume samples are from Normal populations with the same variance (i.e.sigma 1^2 - sigma2 ^2 =sigma^2). Let mU_1 be the true mean melting point for Alloy 1, and let mu 2 be the mean for Alloy 2. (Formula 5B) Calculate a pooled estimate of the common variance a2. What are the associated degrees of freedom? Do the data support the claim that Alloy 1 has a lower mean melting point than Alloy 2? Test H0: mU_1 - mU_2 =0 versus Ha: mU_1 - mU_2

Solution

Set Up Hypothesis
Null Hypothesis, There Is NoSignificance between them Ho: u1 > u2
Alternative, Alloy 1 has lower mean than Alloy 2, H1: u1 < u2
Test Statistic
X (Mean)=420; Standard Deviation (s.d1)=4
Number(n1)=21
Y(Mean)=426; Standard Deviation(s.d2)=3
Number(n2)=21
Value Pooled variance S^2= (n1-1*s1^2 + n2-1*s2^2 )/(n1+n2-2)
S^2 = (20*16 + 20*9) / (42- 2 )
S^2 = 12.5
we use Test Statistic (t) = (X-Y)/Sqrt(S^2(1/n1+1/n2))
to=420-426/Sqrt((12.5( 1 /21+ 1/21 ))
to=-6/1.0911
to=-5.4991
| to | =5.4991
Critical Value
The Value of |t | with (n1+n2-2) i.e 40 d.f is 1.684
We got |to| = 5.4991 & | t | = 1.684
Make Decision
Hence Value of | to | > | t | and Here we Reject Ho
P-Value: Left Tail - Ha : ( P < -5.4991 ) = 0
Hence Value of P0.05 > 0,Here we Reject Ho

We claim that Alloy 1 has lower mean than Alloy