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A sample of 35 different payroll departments found that employees worked an average of 240.6 days a year. If the population standard deviation is 18.8 days, find the 90% confidence interval for the average number of days μ worked by all employees who are paid through payroll departments.
- 232.4 < μ < 248.8
- 230.9 < μ < 250.3
- 236.8 < μ < 244.4
- 235.4 < μ < 245.8
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Answer
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Step 1:: First, we need to find the margin of error for the 90% confidence interval.
where $z$ is the z-score corresponding to the desired confidence level, $\sigma$ is the population standard deviation, and $n$ is the sample size.
The formula for the margin of error for a population mean is:
Step 2:: For a 90% confidence interval, the z-score is 1.645 (you can find this value in a standard normal distribution table).
E = 1.645 \times \frac{18.8}{\sqrt{35}} \approx 5.41
Substituting the given values into the formula, we get:
Step 3:: Now, we can find the lower and upper bounds of the confidence interval by adding and subtracting the margin of error from the sample mean:
\mu_{upper} = \bar{x} + E = 240.6 + 5.41 = 246.01
Step 4:: Since we are asked to find the 90% confidence interval for the average number of days worked by all employees who are paid through payroll departments, and the interval we found is (235.19, 246.01), we need to round the interval endpoints to two decimal places to match the given answer choices.
Final Answer
Therefore, the correct answer choice is: - 235.4 < μ < 245.8
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