Dehween Salt concentratioa ae roadw 2. The melting points of two alloys used in
ID: 3324576 • Letter: D
Question
Dehween Salt concentratioa ae roadw 2. The melting points of two alloys used in formulating solder were investigated by melting 21 samples of each alloy y. Alloy 1, the sample mean and standard deviation were found to be X1 -425 Degrees Fahrenheit and S1-2.5 Degrees Fahrenheit. And for Alloy 2, the sample mean and standard deviation were found to be X2-42048 Degrees Fahrenhs and S2-2.34 Degrees Fahrenheit. Assume that both Populations are Normally distributed. A) Do the sample data support the clain, that both alloys have the same YARIANCE or meting pointe Peim , Hypothesis test, list all the reguired Ssteps in order to answer the question.!L hers.(1s Points) 3) Repeat only the three required steps to test the same Hypothesis as in part A but use t eiection rule.(9 Points)Explanation / Answer
2.a.
Given that,
mean(x)=425
standard deviation , s.d1=2.5
number(n1)=21
y(mean)=420.48
standard deviation, s.d2 =2.34
number(n2)=21
null, Ho: u1 = u2
alternate, H1: u1 != u2
level of significance, = 0.05
from standard normal table, two tailed t /2 =2.021
since our test is two-tailed
reject Ho, if to < -2.021 OR if to > 2.021
calculate pooled variance s^2= (n1-1*s1^2 + n2-1*s2^2 )/(n1+n2-2)
s^2 = (20*6.25 + 20*5.476) / (42- 2 )
s^2 = 5.863
we use test statistic (t) = (x-y)/sqrt(s^2(1/n1+1/n2))
to=425-420.48/sqrt((5.863( 1 /21+ 1/21 ))
to=4.52/0.747
to=6.049
| to | =6.049
critical value
the value of |t | with (n1+n2-2) i.e 40 d.f is 2.021
we got |to| = 6.049 & | t | = 2.021
make decision
hence value of | to | > | t | and here we reject Ho
p-value: two tailed ( double the one tail ) - ha : ( p != 6.0489 ) = 0
hence value of p0.05 > 0,here we reject Ho
ANSWERS
---------------
null, Ho: u1 = u2
alternate, H1: u1 != u2
test statistic: 6.049
critical value: -2.021 , 2.021
decision: reject Ho
p-value: 0
we have enough evidence to support the claim
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