Question

Suppose you computed the profit per surgery for each of 300 patients who underwent hip replacement surgery at your hospital. Profit per surgery is approximately normally distributed and the mean and standard error of profit per surgery in this sample are $97 and $49, respectively. Recall that the 95th percentile of the standard normal distribution is 1.65 and the 97.5th percentile of the standard normal distribution is 1.96. The lower bound of the 95% confidence interval of profit per surgery is (in dollars) . (Truncate to whole dollars)."

Answer 1

Solution:

We are given that: Profit per surgery is approximately normally distributed and the mean and standard error of profit per surgery in this sample are $97 and $49, respectively.

That is: Mean =

Standard Error =

We have to find lower bound of the 95% confidence interval of profit per surgery.

Formula:

For 95% confidence interval, we use 97.5th percentile, since 95% area is middle area and reamining 5% area is distributed in two tails equally, that is 0.05 /2 = 0.025 in left tail and 0.025 in right tail.

Thus total area = Area in left tail + Area of confidence interval = 0.025 + 0.95 = 0.9750

Thus we use z value of area 0.9750

that is z value of 97.5th percentile = 1.96

Thus lower bound is:

( in dollars)