In: Math
Construct 90%, 95%, and 99% confidence intervals for the population mean, and state the practical and probabilistic interpretations of each. Indicate which interpretation you think would be more appropriate to use. Explain why the three intervals that you construct are not of equal width. Indicate which of the three intervals you would prefer to use as an estimate of the population mean, and state the reason for your choice.
In a length of hospitalization study conducted by several cooperating hospitals, a random sample of 64 peptic ulcer patients was drawn from a list of all peptic ulcer patients ever admitted to the participating hospitals 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.
For each confidence interval indicate
a) Z-critical values ,
b) Error,
c) confidence interval
For 90% confidence interval
a)
Level of Significance , α =
0.1
' ' '
z value= z α/2= 1.645 [Excel formula
=NORMSINV(α/2) ]
b)
Standard Error , SE = σ/√n = 3.000 /
√ 64 = 0.3750
margin of error, E=Z*SE = 1.6449
* 0.375 = 0.617
c)
confidence interval is
Interval Lower Limit = x̅ - E = 8.25
- 0.617 = 7.633
Interval Upper Limit = x̅ + E = 8.25
- 0.617 = 8.867
90% confidence interval is (
7.63 < µ < 8.87 )
we are 90% confident that true mean length of
hospitalization lies within confidence interval
--------------------
For 95% confidence interval
a)
Level of Significance , α =
0.05
' ' '
z value= z α/2= 1.9600 [Excel
formula =NORMSINV(α/2) ]
b)
Standard Error , SE = σ/√n = 3.000 /
√ 64 = 0.3750
margin of error, E=Z*SE = 1.9600
* 0.375 = 0.735
c)
confidence interval is
Interval Lower Limit = x̅ - E = 8.25
- 0.735 = 7.515
Interval Upper Limit = x̅ + E = 8.25
- 0.735 = 8.985
95% confidence interval is (
7.52 < µ < 8.98 )
we are 95% confident that true mean length of
hospitalization lies within confidence interval
-----------------------
for 99% confidence interval
a)
Level of Significance , α = 0.01
' ' '
z value= z α/2= 2.5758
b)
Standard Error , SE = σ/√n = 3.000 /
√ 64 = 0.3750
margin of error, E=Z*SE = 2.5758
* 0.375 = 0.966
c)
confidence interval is
Interval Lower Limit = x̅ - E = 8.25
- 0.966 = 7.284
Interval Upper Limit = x̅ + E = 8.25
- 0.966 = 9.216
99% confidence interval is (
7.28 < µ < 9.22 )
we are 99% confident that true mean length of
hospitalization lies within confidence interval
---------------
three intervals that constructed are not of equal width because z critical values is different at each confidence level.