ECO 252
Fall 2000
Formulas
Second Exam
Dr. Andrews
x

n
x
s
n
p 
 (1   )
n
Introductory information: (this is not a question)
A company is trying to decide whether to change advertising companies. They compare a trial company’s
results to their current company, and decide whether or not to switch. The following problems are each
separate ways to test the same idea.
Part I. 5 points each:
1.
2.
3.
4.
Set up the null and alternative hypothesis in words.
What is the type I error in this case?
What is the type II error in this case?
What is an appropriate level of α? Explain your answer.
Part II. 10 points each
Irrespective of your answer in 4, use α= .05 for the remainder of the exam.
For each problem state the null and alternative hypotheses in symbols, show all your work, and interpret
your results. A p-value calculation is only required for the second test. (The conclusion you reach may
differ for different approaches, i.e., you may reject in one test and fail to reject in another.)
Test One: Over the past several years our current advertisement is seen by an average customer .3 times
each week with a standard deviation of 2 times. In a test of the new advertiser, 36 individuals saw the
advertisement an average of .4 times per week with a standard deviation of 5 times. Explain why it is
absolutely necessary to sample at least 30 customers for this test.
Test Two: In a sample of 64 individuals 24 had seen our current ad at least once in the last week while 35
had seen the new advertiser’s ad at least once in the last week. For this question, also calculate and interpret
the p-value.
Test three: To test recall, 5 individuals were asked 8 questions about the advertisements the day after they
had been exposed to the ad. The number of correct responses is recorded below. Extra columns are added
for your convenience, use whichever ones you want.
Person
Old ad
New ad
1
8
7
2
7
7
3
6
8
4
2
4
5
5
8
Hypothesis testing two samples
Two Samples
Means
Proportions
H o :  1   2   0
p1  p2
p1 (1  p1 ) p2 (1  p2 )

n1
n2
TS 
CV  Z
Independent
H o : 1  2 ? 0
Dependent
H o : d ? 0
d
sd
n
CV  t ,n 1
TS 
Ho :1   2
TS 
2
s LG
2
s sm
CV  F.10,( nto p 1),( nb o t 1)
REJ
Unequal 
TS 
x1  x 2
s12
s 22

n1
n2
CV  Z
FTR
equal 
TS 
s 2p 
x1  x 2
1
1
s 2p ( 
)
n1 n2
(n
1
 1) s12  ( n2  1) s 22 
n1  n2  2 
CV  t ,n  n  2