Question

1. A hospital administrator wants to estimate the mean length of stay for all inpatients in the hospital. Based on a random sample of 100 patient records during the previous year, she reports that “The sample mean was 5.3 days. In repeated random samples of this size, the sample mean should be expected to fall within 1.0 day of the population mean about 95% of the time.”

1. 1. If the administrator has done her statistics correctly, what is the 95% confidence interval for the population mean of length of stay? Show your calculation process and Interpret the confidence interval.
2. 2. What are the critical values of t that the administrator used to determine this confidence interval?
3. 3. What is the estimated standard error of the sample mean?
4. 4. What is the standard deviation of the sample?

Answer 1

(a)

Margin of Error = MOE = 1

Confidence Interval:

5.3 1

= (4.3,6.3)

So,

Confidence Interval is given by:

4.3 < < 6.3

Interpretation:

The 95% Confidence Interval (4.3, 6.3) is a range of values that is likely to contain unknown population mean length of stay for all inpatients in the hospital. If we draw a random samples many times, 95% of the confidence intervals will contain the population mean.

(b)

ndf = 100 - 1= 99

= 0.05

From Table, critical values of t = 1.9842

(c)

MOE = t SE

Substituting values, e get:

1 = 1.9842 X SE

So,

SE = 1/1.9842 = 0.5040

(d)
SE = s/

Substituting the values, we get:

0.5040 = s/

So,

s = 0.5040 X 10

= 5.04