a study is undertaken to assess he effect on assembly of the manufacturin tolera

Question

a study is undertaken to assess he effect on assembly of the manufacturin tolerance of the outside diameter of a mating hole in a wheel. a random sample of ten rods and ten wheels are selected for assembly. differences (in inches) between the measured hole diameter and shaft diameter are as follows

0.01, 0.09, 0.15, -0.01, 0.11, 0.06, -0.03, 0.13, 0.8, -0.4

estimate the proportion of assemblies with interfernce (can't assemble) using a point interval estimate (90% confidence) if the distribution differnce is

i) Not known

ii) Normal

b) the number of defects per inspected PC-X based on a random sample of 15 from a days' production is

1,3,1,0,2,0,0,1,1,1,0,1,2,1,1

i) Analyze these data and present your results

ii) Estimate the probability that a randomly selected PC will have at least 3 defects

Explanation / Answer

i) Not known:-

Out of 10 differences, 3 are negative and 7 are positive, meaning only 3 assemblies have interference. So, point estimate of the proportion with interference = 3/10 = 0.3

Since true proportion isn’t known and any proportion in range (0,1) is equally likely, and given number of interferences in the sample of ten are binomial distributed, proportion of interferences follow beta distribution with parameter a=3+1 and b=7+1.

90% confidence interval lower limit = BETAINV(0.05, 4, 8) = 0.135075
90%confidence interval upper limit = BETAINV(0.95, 4, 8) = 0.564374

So, 90% confidence interval is (0.135075, 0.564374)

ii) Normal:-

Point estimate for proportion is again 3/10 = 0.3
Normally approximating, estimated standard deviation = sqrt(0.3*(1-0.3)/10) = 0.144914
For 90% confidence limit, z = 1.64485
90% confidence interval lower limit = 0.3 - 1.64485*0.144914 = 0.061638
90%confidence interval upper limit = 0.3 + 1.64485*0.144914 = 0.538362

So, 90% confidence interval is (0.061638, 0.538362)

b) the number of defects per inspected PC-X based on a random sample of 15 from a days' production is

1,3,1,0,2,0,0,1,1,1,0,1,2,1,1

i) Analyze these data and present your results:-

Mean number of defects = (1+3+1+0+2+0+0+1+1+1+0+1+2+1+1)/15 = 1

So, average number of defects per PC is 1.

ii) Estimate the probability that a randomly selected PC will have at least 3 defects:-

Number of defects per PC can be approximated by Poisson distribution with mean = 1.

P(x>=3)

= 1 - p(x=0) - p(x=1) - p(x=2)

= 1 - 1^0*e^(-1)/0! - 1^1*e^(-1)/1! - 1^2*e^(-1)/2!

= 0.080301

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