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. 8 mean 9.325 4.485804 Let X be a random variable representing the number of nonconforming coil springs in a production batch of size 50. Its mean is denoted y and its standard deviation is (a) (1.5] Give a point estimate for and o. (b) (1.5) Compute the (estimated) standard error for the estimate of the mean. (e) (1.5) Construct a 95% confidence interval for . (d) (1.5) Without computing the 90% confidence interval for y, 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,

Answer 1

Solution :- Given That :- *= 9.325 9.325 40 4.485804 CI= 95% o=

a ) Point Estimate = 9.325

b ) Standard Error = σx = σ / √ n

Standard Error = σx = 4.485804 / √ 40

   Standard Error = σx = 0.70927

c ) 95 % CI

Solution :- Given That :- CI= 95% = n = o= 9.325 40 4.485804 At 95% confidence Level' z' is, a = 1 - 95% a = 1 - 0.95 a = 0.05 a/2 = 0.025 Za/2 = (1.96) At 95% Confidence Interval for Mean is,

Cl for Mean Cl for mean Z-value of 95% CI std. dev. SE = std.dev./sqrt(n) ME = 2*SE Lower Limit = Mean - ME Upper Limit = Mean + ME 95% CI 95% 40 9.325 1.9600 4.485804 0.70927 1.3901 7.93486 10.71514 (7.9349, 10.7151 )

At 95 % CI for mean is ( 7.9349 , 10.7151 )

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d ) A 95% confidence interval is larger than a 90%

A 90% Confidence Interval would be narrower than a 95% Confidence Interval. This occurs because the as the precision of the confidence interval increases (ie CI width decreasing), the reliability of an interval containing the actual mean decreases (less of a range to possibly cover the mean).