Let X_1 and X_2 denote the resistance of a type of a capacitor from Company A an
ID: 3159686 • Letter: L
Question
Let X_1 and X_2 denote the resistance of a type of a capacitor from Company A and B respectively. Assume that X_1 is N(mu_1, sigma^2) and X_2 is N(mu_2, sigma^2). (Notice that the variances are the same for the two populations). A simple random sample of n = 10 capacitors from company A has a sample mean resistance of x_1 = 800 megohms and a sample variance of s^2_1 = 40000 megohms. A simple random sample of m = 8 capacitors from company B has a sample mean resistance of x_2 = 810 megohms and a sample variance of s^2_2 = 42000 megohms. Test if the mean resistance of capacitors from company A is different from the mean resistance of capacitors from company B at the 5 percentage significance level. (Set up the hypothesis, include the test statistics, and state your conclusion).Explanation / Answer
Set Up Hypothesis
Null , There Is No-Significance between them Ho: u1 = u2
Alternate, A is diffrent from B - H1: u1 != u2
Test Statistic
X(Mean)=800
Standard Deviation(s.d1)=200 ; Number(n1)=10
Y(Mean)=810
Standard Deviation(s.d2)=204.939015319192; Number(n2)=8
we use Test Statistic (t) = (X-Y)/Sqrt(s.d1^2/n1)+(s.d2^2/n2)
to =800-810/Sqrt((40000/10)+(42000/8))
to =-0.104
| to | =0.104
Critical Value
The Value of |t | with Min (n1-1, n2-1) i.e 7 d.f is 2.365
We got |to| = 0.10398 & | t | = 2.365
Make Decision
Hence Value of |to | < | t | and Here we Do not Reject Ho
P-Value: Two Tailed ( double the one tail ) - Ha : ( P != -0.104 ) = 0.92
Hence Value of P0.05 < 0.92,Here We Do not Reject Ho
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