Q: In a random sample of 85 automobile engine crankshaft bearing, 10 have a surf
ID: 3222370 • Letter: Q
Question
Q: In a random sample of 85 automobile engine crankshaft bearing, 10 have a surface finish that is rougher than the specificals allow.
(Please answer all parts A, B, and C in this picture, also please include how you found the answers, dont just post your final calculations.)
1. In a random sample of 85 automobile engine crankshaft bearings, 10 have a surface finish that is rougher than the specifications allow. a. Check the required conditions, construct a 98% confidence interval and explain what your confidence interval says in the context of the problem. b. Construct a 93% lower confidence bound and explain what your lower confidence bound says in the context of the problem. c. If it is desirable to estimate the 9% of crankshaft bearings that is is rougher than the specifications allow with t 0.05 precison and 95% confidence, what sample size would you recommend?Explanation / Answer
a.
Confidence Interval For Proportion
CI = p ± Z a/2 Sqrt(p*(1-p)/n)))
x = Mean
n = Sample Size
a = 1 - (Confidence Level/100)
Za/2 = Z-table value
CI = Confidence Interval
No. of success(x)=10
Sample Size(n)=85
Sample proportion = x/n =0.118
Confidence Interval = [ 0.118 ±Z a/2 ( Sqrt ( 0.118*0.882) /85)]
= [ 0.118 - 2.326* Sqrt(0.001) , 0.118 + 2.33* Sqrt(0.001) ]
= [ 0.037,0.199]
b.
Confidence Interval For Proportion
CI = p ± Z a/2 Sqrt(p*(1-p)/n)))
x = Mean
n = Sample Size
a = 1 - (Confidence Level/100)
Za/2 = Z-table value
CI = Confidence Interval
No. of success(x)=10
Sample Size(n)=85
Sample proportion = x/n =0.118
Confidence Interval = [ 0.118 ±Z a/2 ( Sqrt ( 0.118*0.882) /85)]
= [ 0.118 - 1.812* Sqrt(0.001) , 0.118 + 1.81* Sqrt(0.001) ]
= [ 0.055,0.181]
Interpretations:
1) We are 93% sure that the interval [0.055 , 0.181 ] contains the true population proportion
2) If a large number of samples are collected, and a confidence interval is created
for each sample, 93% of these intervals will contains the true population proportion
c.
Compute Sample Size ( n ) = n=(Z/E)^2*p*(1-p)
Z a/2 at 0.05 is = 1.96
Sample Proportion = 0.5
ME = 0.05
n = ( 1.96 / 0.05 )^2 * 0.5*0.5
= 384.16 ~ 385
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