1.) A sample of 25 circuit boards yields 2 defective boards after testing. What

Question

1.) A sample of 25 circuit boards yields 2 defective boards after testing. What would be confidence interval for the true proportion of circuit boards in the lot from which the sample was drawn? Use the Clopper-Pearson Interval and also a normal approximation interval. In your own words which do feel is better to use and why?

2.) At a later date another twenty five sample circuit boards were obtained from a different lot and were tested for defectives. This time none of the boards tested were defective. What would a confidence interval for the proportion of defective circuit boards be based on this new data?

3.) Using the information in part a) of this problem, if the lot from which the sample was collected consisted of 500 circuit boards what would a confidence interval be for “the number” of defective circuit boards in the entire lot?

Explanation / Answer

1)

normal approximation confidence interval for proportion

=(p^ - z *sqrt(p^*q^/n) , p^ +z *sqrt(p^*q^/n) )

p^ = X/n = 2/25 = 0.08

for 95% , z = 1.96

hence

(0.08 - 1.96 * sqrt(0.08*0.92/25, 0.08 +1.96 * sqrt(0.08*0.92/25))

=(-0.026346 , 0.186345)

= (0.000000, 0.186345)                {proportion can't be negative.}

Clopper-Pearson Interval :-

Sample X   N Sample p         95% CI
1       2      25 0.080000    (0.009840, 0.260306)

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