21. Researchers want to estimate the mean leakage, measured in nanometers (nm),

Question

21. Researchers want to estimate the mean leakage, measured in nanometers (nm), of a newflling material for cavities using a simple random sample of 10 trials. Assuming that the population distribution is approximately normal and the population standard deviation is 14nm·find the margin of error (E) for a 90% confidence interval for the population mean a. 11.09 b. 10.08 c. 7.28 d. 7.47 22. Using the data in #21 above, suppose the data stated that the mean ol the sample was 18.8nm. What is the LCL (lower confidence limit)? a. 9.36 b. 11.52 d. 12.48

Explanation / Answer

21. n = 10 = 14 z = 1.645

Margin of error = z / n

= 1.645 * 14 / 10

= 7.28.

The correct answer is c.

22.x' = 18.8 n = 10 = 14 z = 1.645

The lower confidence limit for 90% is:

LCL = 18.8 - 1.645 * 14 / 10

= 11.52.

The correct answer is b.

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