Question

In: Statistics and Probability

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

The article "Repeatability and Reproducibility for Pass/Fail Data"† reported that in n = 48 trials in a particular laboratory, 16 resulted in ignition of a particular type of substrate by a lighted cigarette. Let p denote the long-run proportion of all such trials that would result in ignition. A point estimate for p is p̂ =_______ (rounded to three decimal places). A confidence interval for p with a 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 at all 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 in much closer agreement were the sample size substantially larger.

Solutions

Expert Solution

Let n = 48 trials in a particular laboratory, 16 resulted in ignition of a particular type of substrate by a lighted cigarette. Let p denote the long-run proportion of all such trials that would result in ignition.

A point estimate for p is p̂ =16/48=0.333 (rounded to three decimal places). A confidence interval for p with a confidence level of approximately 95% is

The value of 1.96 is based on the fact that 95%

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

The traditional interval is

= 0.333 ± 1.33 =(0.200,0.466)

These two intervals would be in much closer agreement were the sample size substantially larger.


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