5. The melting points of two alloys used in formulating solder were investigated

Question

5. 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 1- 420°F and s 1-4°F, and for Alloy 2, they were X2 = 426"F and s2-3°F. Assume samples are melting point for Alloy 1, and let Ha be the mean for Alloy 2, (Formula 5B) (a) Calculate a pooled estimate of the common variance . 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 Ho: 1 -Ha = 0 versus Hai 1 -Ha

Explanation / Answer

Given that,
mean(x)=420
standard deviation , s.d1=4
number(n1)=21
y(mean)=426
standard deviation, s.d2 =3
number(n2)=21
a.
calculate 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

b.
null, Ho: u1 = u2
alternate, alloy 1 has lower mean melting point than alloy 2, H1: u1 < u2
level of significance, = 0.05
from standard normal table,left tailed t /2 =1.684
since our test is left-tailed
reject Ho, if to < -1.684
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=-6/1.091089
to=-5.499091
| to | =5.499091
critical value
the value of |t | with (n1+n2-2) i.e 40 d.f is 1.684
we got |to| = 5.499091 & | 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
ANSWERS
---------------
null, Ho: u1 = u2
alternate, H1: u1 < u2
test statistic: -5.499091
critical value: -1.684
decision: reject Ho
p-value: 0

we support that alloy 1 has lower mean melting point than alloy 2

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