Download March 20th There are 3 cases. Test on one population mean, normal

March 20th
There are 3 cases.
1. Test on one population mean, normal population,  2 known : Pivotal Quantity method
2. Test on one population mean, any population, large sample : Pivotal Quantity method
3. Test on one population mean, normal population,  2 unknown : Pivotal Quantity method
i .i .d .
3rd case : X 1 , X 2 , , X n ~ N (  ,  2 ) , 2 is unknown
1. Point Estimator for  : X ~ N (  ,
2
n
)
X is NOT a pivotal quantity since  2 is unknown.
2.
Z
X 
~ N (0,1)
 n
This is also NOT a pivotal quantity since  is unknown.
3. Theorem. Sample from normal population Z ~ N (0,1)
W
(n  1) S 2
Definition. T 

2
~  n21
Z
W
~ tn1  T 
(n  1)
X 
( Z and W are independent.)
S n
 T is a pivotal quantity for 
4.
a.
 H 0 :   0

 H a :  0
Test Statistic : T0 
X  0 H 0
~ tn1
S n
  P(Reject H 0 | H 0 )  P(T0
(  is usually 0.05.)
c | H 0 :   0 )
c  tn 1,
At the significance level  , we reject H 0 in favor of H a if T0
b.
 H 0 :   0

 H a :  0
Test Statistic : T0 
X  0 H 0
~ tn1
S n
  P(Reject H 0 | H 0 )  P(T0
c | H 0 :   0 )
Reject H 0 in favor of H a if T0
tn 1,
c  tn 1,
tn 1,
c.
 H 0 :   0

 H a :   0
Test Statistic : T0 
X  0 H 0
~ tn1
S n
  P(Reject H 0 | H 0 )
 P( T0
 2  P(T0


2
 P(T0
c | H0 )
c | H0 )
c | H0 )
At the significance level  , we reject H 0 in favor of H a if T0
tn1, 2
c  tn 1, 2
Example. Prospective salespeople for an encyclopedia company are now being affected a sales training
program. Previous data indicate that the average number of sales per month for those who do not
participate is 33. To determine whether the training program is effective, a random sample of 35 new
employees is given the sales training and then sent.
One month later, the mean and standard deviation for the number of encyclopedia sold are 35 and 8.4,
respectively. Do the data present sufficient evidence to indicate that the training program enhances sales?
Use   0.05
Solution.
This is the 2nd case : large sample ( n
Pivotal Quantity : Z 
30 )
X 
~ N (0,1)
S n
 H 0 :   33

 H a :  33
Test Statistic : Z 0 
X  0
35  33

 1.41
S n 8.4 35
At   0.05 , we would reject H 0 in favor of H a if Z 0
Z0.05  1.645
 We do not reject H 0
If you are told the population is normal, then you should use the t-test – because the t-test is exact while
z-test is approximate in this situation.
Example. Suppose that the population is normal in the above example.
Solution.
 H 0 :   33

 H a :  33
Test Statistic : T0 
X  0
35  33

 1.41
S n 8.4 35
At   0.05 , we would reject H 0 in favor of H a if T0
tn 1,  t34,.05  1.69  exact test
Because T0  1.41 is not greater than 1.69, we cannot reject H 0
d.