Example
A hospital administrator claims that the average length of stay of
patients admitted to the hospital is 5 days. An insurance company is
investigating the possibility that the average length of stay is actually
greater than 5 days. They select a random sample of 100 patients
from last year and they find that the sample average length of stay of
patients is 6 days. We assume that the population is approximately
Normal with
σ
= 3 days. Does the data support the insurance
company’s claim at level
α
=
.
01?
0
Identify Parameter of Interest
True mean time a person is hospitalized 
μ
TH
1
Identify test:
n
= 100
≥
30
⇒
ztest for means
2
Formulate
H
0
and
H
A
: (Hint: start with
H
A
)
H
0
:
μ
TH
= 5 (innocent until proven guilty)
H
A
:
μ
TH
>
5 (company’s suspicion)
Example
3
Compute test statistic (zscore)
z
obs
=
¯
x

μ
0
(
s
or
σ
)
/
√
n
=
6

5
3
/
√
100
≈
3
.
33
4
Using
Table A2
compute pvalue:
H
A
:
μ
TH
>
5
pvalue=
P
(
Z
≥
z
obs
) =
P
(
Z
≥
3
.
33)
z
obs
= 3
.
33
Table A2
→
.
9996 (Area to the left of 3.33)
pvalue=
P
(
Z
≥
3
.
33) = 1.9996 = .0004
Interpretation:
The probability of getting a sample mean
of 6 days (
¯
X
) or greater (see
H
A
), if the true mean
hospitalization time was 5 days (
H
0
), is .0004 or .04%.
5
Decision and Conclusion:
level
α
=
.
01
since pvalue = .0004
<
.01 =
α
, we reject the
H
0
We have
very strong evidence
at
α
level .01 that the
mean hospitalization time at the specific hospital is greater
than 5 days.
Example
From extensive records it is known that the duration of treating a
disease by standard therapy has a mean of 15 days. Some researchers
think that a new therapy can reduce the treatment time. We
consider a sample of 25 patients receiving this new therapy and their
time to recovery are recorded. From this sample we observe an
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 Spring '08
 Papachristau
 Statistics, Statistical hypothesis testing, Tails

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