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# Point Estimates & Confidence Interval

A simple random sample of 8 employees is selected from a large firm. For the 8 employees, the number of days each was absent during the past month was found to be
1, 3, 3, 3, 1, 5, 4, and 3
(a) What is the point estimate for , the mean number of days absent for the firm's employees? Round your answer to nearest hundredth.
=
(b) What is the point estimate for , the variance of the number of days absent? Round your answer to nearest hundredth.
=

Determine the limits of the 90% confidence interval for , given that
n = 86; = 60; and = 10.
95% CI = (n1,n2) where n1=lower limit and n2=upper limit

There are 300 students enrolled in Business Statistics. Historically, exam scores are normally distributed with a standard deviation of 21.25. Your instructor randomly selected a sample of 30 examinations and finds a mean of 67.2. Determine a 90% confidence interval for the mean score for all students taking the course.
90% CI = to

#### Solution Preview

A simple random sample of 8 employees is selected from a large firm. For the 8 employees, the number of days each was absent during the past month was found to be
1, 3, 3, 3, 1, 5, 4, and 3
X X - 2.88 (X - 2.875)^2
1 -1.88 3.5344
3 0.12 0.0144
3 0.12 0.0144
3 0.12 0.0144
1 -1.88 3.5344
5 2.12 4.4944
4 1.12 1.2544
3 0.12 0.0144
23 12.8752
(a) What is the point estimate for , the mean number of days absent for the firm's employees? Round your answer to nearest hundredth.
=
Point estimate for µ = Sample mean ( )
=
...

#### Solution Summary

The solution provides step by step method for the calculation of confidence interval for population mean and point estimates of population mean and variance. Formula for the calculation and Interpretations of the results are also included.

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