The strength of the wire produce by A has a mean of 4,500kg and a standard deviation of 200kg . Company B has a mean of 4000kg and a standard deviation of 300kg. If 50wires of company A and 100wires of company B are selected at random and tested for strength, what is the probability that the sample mean strength of A will be at least 600kg more than that of B?
Answers
Answer:
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Step-by-step explanation:
Given Data :
For Sample 1
x̄1 = 4500
n1= 50
s1= 200
For Sample 2
x̄2= 4000
n2 = 100
s2= 300
α =1- 0.99 = 0.01
Degree of freedom
2 2 (sis + 01 02 n1 . df = = 2 2 2 s1 h1 s m2. + h1-1 02-1
= 136.0445
= 136
99% confidence interval for population mean μ :
Point estimate:
x̄1 - x̄2 = 4500 - 4000
= 500.0
Critical value
critical value at α= 0.01 and df= 136 is
tα/2,df= 2.612
.....from student's t distribution table
Margin of error:
ME=tα/2,df × V(s1^2/n1+s2^2/n2)
= 2.612 ×V( 200.0 ^2/ 50.0 + 300.0 ^2/ 100.0 )
= 107.6955
Margin of error is 107.6955
Now
99% confidence interval is
CI=(x̄1 - x̄2) ± ME
=500.0 ± 107.6955
= (392.3045, 607.6955)
Therefore 99 per cent confidence limits on the difference in the average strength of the populations of wires produced by the two companies is (392.3, 607.7)
