Question

A company produces tens of thousands of pills per day that are expected to weigh 50 mg each with 95% confidence that the weight will be between 48 mg and 52 mg.

a) What standard deviation does this imply?

b) A sample of 20 pills is randomly selected from each day’s batch and is found to have a mean weight 51 mg and a standard deviation of 0.75 mg. Compute the standard deviation of the mean of the sample.

c) Compute the confidence interval for the mean based on the information from part b.

d) Does the result from part b and c suggest that the process might be out of control i.e. does the confidence interval include the population mean? Explain your logic for answering this question in 1-2 sentences.

Answer 1

A) Margin of error = (52 - 48)/2 = 2

At 95% confidence interval the critical value is z_0.025 = 1.96

z_0.025 * = 2

Or, 1.96 * = 2

Or, = 2/1.96

Or, = 1.02

B) s/ = 0.75/ = 0.1677

C) At 95% confidence interval the critical value is t^* = 2.093

The 95% confidence interval for population mean is

+/- t^* * s/

= 51 +/- 2.093 * 0.1677

= 51 +/- 0.351

= 50.649, 51.351

d) No, the confidence interval doesn't include the population mean 50 mg. So the process might be out of control.