Answer to Question #293819 in Statistics and Probability for Emu

Question

In the length of of hospitalization study conducted by several cooperating hospitals,a random sample of 64 peptic ulcer patients was drawn from the list of all peptic ulcer ever admitted to the participating hospital and the length of hospitalization per admission was determined for each. The mean length of hospitalization was found to be 8.25 days.the population standard deviation is known to be 3 days.calculate the 90,95 and 99% confidence interval and compare the intervals.

Answer 1

$n=64\\ \overline{x} =8.25\\ \sigma=3$

Solution : 95% confidence interval:

$z_{0.05/2}=1.96 \\ CI = \bar{x}±Z_{0.05/2} \times \frac{s}{\sqrt{n}} \\ = 8.25 ±1.96 \times \frac{3}{\sqrt{64}} \\ \\ =(7.515, 8.985)$

90%confidence interval:

$z_{0.10/2}=1.645 \\ CI = \bar{x}±Z_{0.10/2} \times \frac{s}{\sqrt{n}} \\ = 8.25 ±1.645 \times \frac{3}{\sqrt{64}} \\ \\ =(7.633, 8.867)$

99% confidence interval:

$z_{0.01/2}=2.576 \\ CI = \bar{x}±Z_{0.01/2} \times \frac{s}{\sqrt{n}} \\ = 8.25 ±2.576 \times \frac{3}{\sqrt{64}} \\ \\ =(7.284, 9.216)$

From above intervals, we see that the width of a confidence interval increases as the confidence level increases.