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AMS312_Spring2007
Mar 29th
Chapters 6&7 (continued):
1、 Inference on one population proportion p for large sample test: summary:
Same: H 0 : p  p0
H a : p p0
Same test statistic Z 0 
H 0 : p  p0
H 0 : p  p0
H a : p  p0
H a : p  p0
pˆ  p0
 N (0,1) when Ho is true
p0 (1  p0 ) / n
At the significance level  , we reject Ho in favor of Ha if:
Z0  Z
| Z0 | Z
Z 0   Z
P-value= P( Z0  Z | H 0 ) P-value= P(Z0  Z | H 0 )
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P-value= P(| Z0 | Z | H 0 )
2
Definition: P-value (observed significance level) is the probability to observe the test statistic
value or more extreme values than the significance test statistic value—Zo, given that Ho is true
(= a value that is at least as extreme as the test statistic value observed)
* Make conclusion using the P-value.
At the significance level  , we reject Ho in favor of Ha if the P-value is less than  .
* P-value is more informative.
2、Inference on one population mean  for large sample test: Summary:
Same:
H 0 :   0
H 0 :   0
H 0 :   0
H a :   0
H a :   0
H a :   0
Same Test statistic:
Pivotal quantity: Z 
X 
s
n
N (0,1) when  is unknown.
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AMS312_Spring2007
Z
X 

N (0,1) When  is known.
n
(*Z follows exactly N(0,1) distribution when the population is normal)
Test statistic: Z 0 
X  0
s
n
N (0,1) or Z 0 
X  0

N (0,1)
n
At the significance level  , we reject Ho if:
Z0  Z
| Z0 | Z
Z 0   Z
2
P-values are the same as above (proportion)
Derivation:
  P (Type 1 error)=P (reject Ho|Ho)
 P(| Z0 | c | H 0 )  P(Z0  c | H 0 )  P(Z 0  c | H 0 )  2P(Z 0  c | H 0 )
Hence P( Z 0  c | H 0 )  
2
Therefore, we reject Ho at  if: | Z 0 | Z
2
Example 3a (Problem 6.2.10 from Page439)
At a class research project, Rosaura wants to see whether the stress of final exams elevates the
blood pressures of freshmen women. When they are not under any untoward duress, healthy
18-year-old women have systolic blood pressures that average 120mmHg with a standard
deviation of 12mm Hg. If Rosaura finds that the average blood pressure for the 50 women in
statistics 101 on the day of the final exam is 125.2, what should she conclude? Set up and test an
appropriate hypothesis.
Solution: This is test on one population mean for large sample.
  12   0.05
N=50, X  125.2 , 0  120
Z0 
125.2  120
122
3
50
H 0 :   120 v.s. H a :   120
Since Zo=3>1.645, we reject Ho at   0.05
P-value=0.013, since P-value<0.05, we reject Ho
If   0.01 : since now P-value is still smaller than 0.01, we reject Ho.
Example 3b
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AMS312_Spring2007
At a class research project, Rosaura wants to see whether the stress of final exams elevates the
blood pressures of freshmen women. When they are not under any untoward duress, healthy
18-year-old women have systolic blood pressures that average 120mmHg. If Rosaura finds that
the average blood pressure for the 50 women in statistics 101 on the day of the final exam is 125.2
and with a standard deviation of 13 mm Hg, what should she conclude? Set up and test an
appropriate hypothesis.
Solution: Z 0 
125.2  120
132
 2.8 Now p-value is 0.0026<0.05, hence we reject Ho.
50
3、Inference on one population mean  , small sample test, normal population:
Same:
H 0 :   0
H 0 :   0
H 0 :   0
H a :   0
H a :   0
H a :   0
Now the pivotal quantity is:
T
X 
s
n
tn 1 Exactly
William S. Gosset (1876 -1937), inventor of the Student’s t-distribution.
For complete biography of William S. Gosset, please refer to:
http://en.wikipedia.org/wiki/William_Sealey_Gosset
and http://www.swlearning.com/quant/kohler/stat/biographical_sketches/bio12.1.html
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AMS312_Spring2007
Comparison of T distribution and the Standard normal distribution
T distribution is heavy-tailed than the standard normal distribution.
When n increase, t-distribution would get closer, and eventually converge to N (0,1)
Definition (t-distribution)
Let Z
 2 k (chi-square distribution with k degrees of freedom, a special
N (0,1) and W
member of the Gamma distribution). Furthermore, Z and W are independent. Then
T
Z
W
tk
k
i .i .d .
Theorem (Sampling from the normal population): If X 1 , X 2 ..... X n
(a) X
(b)
N ( ,
(n  1)

2
S2
(c) X and S
2
n
)
 2 n 1
are independent
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N (  ,  2 ) , then: