Please find the answers for Q5, Q6. (0.0308140 is not the answer for Q5.) Proble

Question

Please find the answers for Q5, Q6.

(0.0308140 is not the answer for Q5.)

Problem 2 A simple random sample of 60 widgets is drawn from a manufacturing lot of 700 widgets. The widgets in the sample are run until they fail. 56 of the widgets in the sample last more than 3,000 hours. An unbiased estimate of the fraction of widgets in the manufacturing lot that will last more than 3,000 hours is (Q3) 0.933333333 The finite population correction for the standard error of the sample percentage is (Q4) 0.956866607 The largest the standard error of the sample percentage could be is (Q5) The bootstrap estimate of the standard error of the sample percentage is (Q6) The bootstrap estimate of the standard error of the sample percentage is (Q7) B: biasedv

Explanation / Answer

Q.5.The largest the standard error of the sample percentage could be

SE = square root of ( p(1 ? p)/ n) =0.03220

Q.6 The bootstrap estimate of the standard error of the sample percentage:

By central Limit theorem , approximate distribution of sample proportion (p) is normal with

Standard Error=square root of ( p(1 ? p)/ n) =0.03220

The variability of the bootstrap statistics is similar to the variability of the sample statistics

  

Related Questions

Navigate

• Browse (All)
• Browse P
• Subjects

---