In: Statistics and Probability
The average length of stay in U.S. hospitals 4.5 days. The length of stay in days of a random sample of patients at a local hospital was recorded.
3,6,3,7,8,9,4,6,5,5,7,3,3,5,8,5
A). Use this data to construct a 99% confidence interval. What are the upper and lower bounds?
B). Based on your confidence interval, is the length of stay at this local hospital significantly different from the national average at the 1% significance level?
The average length of stay in U.S. hospitals 4.5 days. The length of stay in days of a random sample of patients at a local hospital was recorded.
3,6,3,7,8,9,4,6,5,5,7,3,3,5,8,5
A). Use this data to construct a 99% confidence interval. What are the upper and lower bounds?
Confidence Interval Estimate for the Mean |
|
Data |
|
Sample Standard Deviation |
1.9653 |
Sample Mean |
5.4375 |
Sample Size |
16 |
Confidence Level |
99% |
Intermediate Calculations |
|
Standard Error of the Mean |
0.4913 |
Degrees of Freedom |
15 |
t Value |
2.9467 |
Interval Half Width |
1.4478 |
Confidence Interval |
|
Interval Lower Limit |
3.9897 |
Interval Upper Limit |
6.8853 |
99% CI = ( 3.9897, 6.8853)
B). Based on your confidence interval, is the length of stay at this local hospital significantly different from the national average at the 1% significance level?
The 99% CI ( 3.9897, 6.8853) contains the value 4.5. Therefore, the length of stay at this local hospital not significantly different from the national average at the 1% significance level.