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:

Final Answer

Therefore, the correct answer choice is: - 235.4 < μ < 245.8