Question

A hospital administrator is examining the amount of time its care teams spend with patients. He finds that the care teams are spending an average of 22 minutes per patient with a standard deviation of 17 minutes, distributed normally. What amount of time separates the lowest 60% of the means of time spent with patient from the highest 40% in a sample size of 40 care teams? Use the z-table below:

Answer 1

SEM = 17 / sqrt(40) = 2.68

Next, we can calculate the z-scores using the formula:

z = (x - μ) / SEM

where x is the sample mean, μ is the population mean (which we don't know), and SEM is the standard error of the mean.

For the lowest 60% of means, we want to find the z-score that corresponds to the 20th percentile (since 60% is the area below the 80th percentile). Using the z-table, we find that the z-score for the 20th percentile is -0.84.

z = -0.84

-0.84 = (x - μ) / 2.68

-2.25 = x - μ

For the highest 40% of means, we want to find the z-score that corresponds to the 60th percentile (since 40% is the area above the 60th percentile). Using the z-table, we find that the z-score for the 60th percentile is 0.25.

z = 0.25

0.25 = (x - μ) / 2.68

0.67 = x - μ

Now we have two equations with two unknowns (x and μ):

-2.25 = x - μ

0.67 = x - μ

Solving for x, we get:

x = -0.79 (for the lowest 60% of means)

x = 0.93 (for the highest 40% of means)

Therefore, the amount of time that separates the lowest 60% of means from the highest 40% is:

0.93 - (-0.79) = 1.72 minutes.