The article \"Repeatability and Reproducibility for Pass/Fail Data\"† reported that in n = 48 trials in...

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The article \"Repeatability and Reproducibility for Pass/FailData\"† reported that in n = 48 trials in a particularlaboratory, 16 resulted in ignition of a particular type ofsubstrate by a lighted cigarette. Let p denote thelong-run proportion of all such trials that would result inignition. A point estimate for p is p̂ =_______ (roundedto three decimal places). A confidence interval for p witha confidence level of approximately 95% is
0.333 + (1.96)2/96
1 + (1.96)2/48
± (1.96)
(0.333)(0.667)/48 + (1.96)2/9216
1 + (1.96)2/48

= 0.345 ± 0.129 = (0.216, 0.474)

This interval is quite wide because a sample size of 48 is not atall large when estimating a proportion.

The traditional interval is0.333 ± 1.96
(0.333)(0.667)
48
= 0.333 ±  (rounded to three decimal places) =
0.200,  
(rounded to three decimal places)These two intervals would be inmuch closer agreement were the sample size substantiallylarger.

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Let n 48 trials in a particular laboratory 16resulted in ignition of a particular type of substrate by a lightedcigarette Let    See Answer
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