Download A hospital administrator estimates the mean length of stay for all

A hospital administrator estimates the
mean length of stay for all inpatients
is at least 5 days. We sampled 100
patients and we know that the
standard deviation is 1.1 days. If we
are testing the hypothesis  H o :   5
H A :   5
Find the power of the test if
the length of stay is really 4.6 days.
Mean =
x
n
Median
*Measures the center of a
sample and is very sensitive to
outliers.
*the middle valued in an ordered
list of observations
*Not sensitive to outliers
*Average of the squared
differences from the mean
Variance
Standard
Deviation
s2



xx
n 1
Range
*Average deviation from the
mean

2
S xx
s 

n 1
2
x
2
x



n 1
Highest - Lowest
n
2
s s
2
# of deviations above or below the mean
– used from comparisons.
The average on a chemistry test was 81 and the
standard deviation was 6. Jill made a 78. The
average on the Calculus test was 85 and the
standard deviation was 3. Jill made an 81. Which
test was actually better for Jill?
Chemistry:
xx
z
s
78  81
z
6
z  0.5
Calculus:
xx
s
81  85
z
3
z  1.3
z
Jill did better on the chemistry test because it’s
only a half deviation below the mean.

*only use if normally
distributed
1 standard deviation
68%
 2 standard deviations
95%
 3 standard deviations
99.7%
List 1
X
L2
List 2
Mean = 13.8
Mean = 23.8
6
16
St. Dev = 6.1
St. Dev = 6.1
10
20
Median =15
Median =25
15
25
Range = 16
16
26
22
32
L2 – Add 10 to each x
Range = 16
*Mean & Median increase by same
amount x
*St. Dev. & Range don’t change
List 1
List 2
X
L3
Mean = 13.8
Mean = 276
6
120
St. Dev = 6.1
St. Dev = 121.98
10
200
Median = 15
Median = 300
15
300
Range = 16
Range = 320
16
320
22
440
Multiply List 1 by 20
All were multiplied by 20!