In a time-use study, 20 randomly selected managers were found to spend a mean of 2.40 hours each day on paperwork. The standard deviation of the 20 scores is 1.30 hours. Also, the sample data appear to have a bell-shaped distribution. Construct the 95% confidence interval for the mean time spent on paperwork by all managers.
To construct a confidence interval for the mean time spent on paperwork by all managers, we can use the formula:
Confidence interval = sample mean +/- margin of error
First, let's calculate the margin of error using the following formula:
Margin of error = (critical value) * (standard deviation / √(sample size))
Since the sample data appear to have a bell-shaped distribution and we want to construct a 95% confidence interval, we need to find the critical value associated with a 95% confidence level. This corresponds to a z-score of 1.96 (from the standard normal distribution).
Now, let's substitute the given values into the formula:
Sample mean = 2.40 hours
Standard deviation = 1.30 hours
Sample size = 20
Critical value (z-score) = 1.96
Margin of error = (1.96) * (1.30 / √(20))
To calculate the square root of the sample size, √(20), we find that it is approximately 4.472.
Margin of error = (1.96) * (1.30 / 4.472) ≈ 0.450
Now, we can calculate the confidence interval:
Confidence interval = 2.40 +/- 0.450
Lower bound = 2.40 - 0.450 ≈ 1.950 hours
Upper bound = 2.40 + 0.450 ≈ 2.850 hours
Therefore, the 95% confidence interval for the mean time spent on paperwork by all managers is approximately 1.950 to 2.850 hours.