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 the 20 scores is 1.30. 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.
Formula:
CI95 = mean ± 1.96(sd/√n)
Your data:
mean = 2.40
sd = 1.30
n = 20
Plug the values into the formula and calculate the interval.
I hope this will help get you started.
To construct a confidence interval for the mean time spent on paperwork by all managers, we can use the formula:
Confidence interval = (sample mean) ± (critical value * standard error)
First, let's calculate the critical value. Since the sample size is small (n = 20) and the sample data appear to have a bell-shaped distribution, we can use a t-distribution instead of a normal distribution to find the critical value.
For a 95% confidence interval with 19 degrees of freedom (n-1), the two-tailed t-critical value is obtained by looking up the t-value in the t-distribution table. In this case, the critical value is approximately 2.093.
Next, let's calculate the standard error:
Standard error = (sample standard deviation) / √(sample size)
Given that the sample standard deviation is 1.30 and the sample size is 20, the standard error is:
Standard error = 1.30 / √(20) = 0.290
Now we can plug the values into the formula to calculate the confidence interval:
Confidence interval = (sample mean) ± (critical value * standard error)
Confidence interval = 2.40 ± (2.093 * 0.290)
Confidence interval = 2.40 ± 0.606
Therefore, the 95% confidence interval for the mean time spent on paperwork by all managers is (1.794, 2.994). This means that we are 95% confident that the true mean time spent on paperwork falls within this interval.