2. A manufacturer of coil springs is interested in implementing a quality contro
ID: 3054525 • Letter: 2
Question
2. A manufacturer of coil springs is interested in implementing a quality control system to monitor his production process. As part of this quality system, it is decided to record the number of nonconforming coil springs in each production batch of size 50. During 40 days of production, 40 batches. Here are numerical summaries of the data. mean sd 0% 25% 50% 75% 100% n 9.325 4.485804 3 68 12 19 40 Let X be a random variable representing the number of nonconforming coil springs in a production batch of size 50. Its mean is denoted ? and its standard deviation is ? (a) Give a point estimate for ? and ? b) Compute the (estimated) standard error for the estimate of the mean. (c) Construct a 95% confidence interval for ,. (d) Without computing the 90% confidence interval for ?, would you expect the length of the interval to be smaller or larger than the 95% confidence interval? Explain your answer in a few sentences.Explanation / Answer
a)
mean = 9.325
std.dev = 4.485804
b)
Standard error = s/sqrt(n)
= 4.485804/sqrt(40) = 0.7093
c)
CI for = 95%
n = 40
mean = 9.325
z-value of 95% CI = 1.9600
std. dev. = 4.485804
SE = std.dev./sqrt(n)
= 4.485804/sqrt(40)
= 0.70927
ME = z*SE
= 1.96 * 0.70927
= 1.39014
Lower Limit = Mean - ME
= 9.325 - 1.39014
= 7.93486
Upper Limit = Mean + ME
= 9.325 + 1.39014
= 10.71514
95% CI (7.9349 , 10.7151 )
d)
Expect length of the interval be smaller. As the confidence level decreases confidence interval also decreases
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