Question

1Define - Confidence Interval and Confidence Level. (4 pts)

2. A sample of 100 of a certain machine part found an average width of 0.755 in with a standard deviation of 0.012 in. Find the 99% confidence interval of the true mean of the width of the parts. (What distrubution should be used?

Answer 1

z @ 99.0% = 2.5758

Here, we will use the normal approximation,
Since we know that

Required confidence interval = (0.755-2.5758(0.0012), 0.755+2.5758(0.0012))
Required confidence interval = (0.755 - 0.0031, 0.755 + 0.0031)
Required confidence interval = (0.7519, 0.7581)

Interpretion: We are 99.0% confident that the true mean of the population lie between the interval 0.7519 and 0.7581.

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