Statr session 17 and 18

Learning Objectives
• Understand how to test a hypothesis about a single
population parameter:
– Proportion (using z-statistic)
– Variance (using c2-statistic)

• Calculate the probability of Type II error when failing
to reject the null hypothesis
• Test hypotheses and construct confidence intervals
about the difference in two population means using
the Z statistic.
about the difference in two population means using
the t statistic.

Learning Objectives (continued)
about the difference in two related populations.
about the differences in two population proportions.

Hypothesis Test of p
Suppose a company held 26% of market share for
several years. Due to a massive marketing effort and
improved product quality, company officials believe
that the market share has increased, and they want to
prove it statistically. In a random sample of 140 users,
48 used their product. Does this present evidence
that their market share has increased? Test it at the
5% level of significance.

1 – Establish the null and alternative hypotheses
H0: p = 0.26 vs. Ha: p > 0.26
2 – Determine the appropriate statistical test:
• Z-test for proportions:
•

ˆ
z  p p
pq
n

Appropriate if the following two conditions are met:
 The sample was randomly selected from the
population
 np>= 5 and nq >=5. For our data, 140(0.26) = 36.4 > 5
and 140 (0.74) = 103.6 > 5, so this condition is met.
3 – Set a, the Type I error rate / significance level
Choose the common value of a = 0.05


ˆ
p  p  0.3430.26  0.083  2.24
z  pq
(0.26)(10.26) 0.037
n
140

A small business has 37 employees. Because of the uncertain
demand for its product, the company usually pays overtime on
any given week. The company assumed that about 50 total
hours of overtime per week is required and that the variance on
this figure is about 25. Company officials want to know whether
the variance of overtime hours has changed. The data below are
a random sample of 16 weeks of overtime in hours per week.
Assume hours of overtime are normally distributed. Use these
data to test the null hypothesis that the variance of overtime
data is 25.
57 56 52 44
46 53 44 44
48 51 55 48
63 53 51 50

Step 1: Hypothesize

H0: 2 = 25
Ha: 2  25
Step 2: Variances follow a chi-squared distribution with n -1
degrees of freedom, underlying population has a normal
distribution. The test statistic is:

c2

n 1

(n 1)s2


2

• Step 3: Choose a = 0.10 (so a/2 = 0.05)
• Step 4: The degrees of freedom are 16 – 1 = 15.
The lower and upper critical chi-square values are

c2(1 – 0.05), 15 = c2 0.95, 15 = 7.3 and c2 0.05, 15 = 25.0
• Step 5: The data are listed in the text.
• Step 6: The sample variance is s2 = 28.1.
The observed chi-square value is calculated as
c2 = (n-1)s2 / 2 = (15) 28.1 / 25 = 16.9

• Step 7: The observed chi-square value is in the
nonrejection region because
c2 0.95, 15 = 7.3 < c2observed = 16.9 < c2 0.05, 15 = 25.0

• Step 8: This result indicates to the company
managers that the variance of weekly overtime
hours is about what they expected.

df = 7

.05

.95

.05
0
7.3

25.0

Solving for Type II Errors
• When the null hypothesis is not rejected, then
either a correct decision is made or an incorrect
decision is made.
• If an incorrect decision is made, that is, if the null
hypothesis is not rejected when it is false, then a
Type II error has occurred.

Solving for Type II Errors (Soft Drink)
• Suppose a test is conducted on the following
hypotheses about the amount of liquid in a 12
ounce soft drink can: H0: = 12 ounces vs. Ha: < 12
ounces, the sample size is 60, and the sample mean
is 11.985, and population standard deviation () is
assumed to be 0.10.
• The first step in determining the probability of a
Type II error is to calculate a critical value for the
sample mean (or proportion or variance, etc.).

xc  
zc 
/ n
xc 12
1.645 
0.10 / 60
xc 11.979

In testing the null hypothesis
by the critical value method,
this value is used as the cutoff
for the nonrejection region.
For any sample mean obtained
that is less than 11.979, the
null hypothesis is rejected.
Any sample mean greater than
11.979, the null hypothesis is
not rejected.

The Type II Error rate (b) varies with different values of the
true parameter. For example, if the true mean as 11.99, the
corresponding z-value for b is

xc  
1
z1 
/ n
11.979 11.99
z1 
0.10 / 60
z1  0.85

• Recall that a Type II error is made when you fail to reject
when you should. Thus, you want to calculate
Px  xc |   1   Px  11 .979 |   11 .99   P( z  0.85 )  .802

• Thus, there is an 80.2% chance of committing a Type II error
if the alternative mean is 11.99. That is quite a high chance
of being wrong (but then again 11.99 is so close to 12, so
you would need a lot of data to show that those are
statistically different).
• Note, you only need to be concerned about type II errors
since you would have failed to reject the null hypothesis [t =
-1.16 > tc = -1.645 (for a=0.05)]

Operating Characteristic
and Power Curve
• Because the probability of committing a Type II error
changes for each different value of the alternative
parameter, it is common to examine a series of possible
alternative values.
• The power of a test is the probability of rejecting the null
hypothesis when it is false.
• Power = 1 - b.

• Calculating two sample means and using the
difference in the two sample means is used
to test the difference in the population
• The Central Limit theorem states that the
difference in two sample means is normally
distributed for large sample sizes (both n1
and n2 > 30) regardless of the shape of the
population

Hypothesis Testing for Differences
Between Means: The Wage Example
As an example, we want to conduct a hypothesis test
to determine whether the average annual wage for an
advertising manager is different from the average
annual wage of an auditing manager. Because we are
testing to determine whether the means are different,
it might seem logical that the null and alternative
hypotheses would be
Ho: μ1 = μ2
Ha: μ1 ≠ μ2
where advertising managers are Population 1 and
auditing managers are Population 2.


H 0 : 1  2
H a : 1  2
a =0.05, a/2 = 0.025, z0.025 = 1.96
The two hypotheses can also be expressed as:

H0 : 1  2  0
H a : 1  2  0
Analysis is testing whether there is a difference in the
average wage. This is a two tailed test.

Rejection
Region

Rejection
Region

Rejection
Region
a
2

Rejection
Region
a

 .025

2

 .025

Non Rejection Region

Z c  1.96

0

Z c  1.96

Critical Values
If z < - 1.96 or z > 1.96, reject Ho.
If - 1.96  z  1.96, do not reject Ho.

n  32
x  70.700
1

1




n
x

 16.253

1

 264.164

2
1

2




 34

2
2
2
2

 62.187
 12.900
 166.411


z  (70.700 62.187)  (0)  2.35
264.160  166.411
32
34
Since the observed value of 2.35 is greater than 1.96,
reject the null hypothesis. That is, there is a significant
difference between the average annual wage of advertising
managers and the average annual wage of auditing
managers.


 0
H :    0
Ho :

1

2

1

Rejection
Region

Rejection
Region

2

a

a
 .025
2

a
 .025
2


X

1



X2

Critical Values

X X
1

2

If z  1.96 or z  1.96, reject H 0 .
Rejection
Region

Rejection
Region

a
2

z
a

 .025

2

 .025


Z

c

 2.33

If  1.96  z  1.96, do not reject H 0 .

0
Critical Values

Z

c

 2.33



( x1  x2 )  ( 1   2)
2
 12  2

n1 n2

(70.700- 62.187)- (0)
 2.35
264.164 166.411

32
34

Since z  2.35  1.96 , reject H 0 .

Demonstration Problem 1
A sample of 87 professional working women showed that the
average amount paid annually into a private pension fund per
person was $3352. The population standard deviation is $1100.
A sample of 76 professional working men showed that the
average amount paid annually into a private pension fund per
person was $5727, with a population standard deviation of
$1700. A women’s activist group wants to “prove” that women
do not pay as much per year as men into private pension funds.
If they use α = .001 and these sample data, will they be able to
reject a null hypothesis that women annually pay the same as
or more than men into private pension funds? Use the eightstep hypothesis-testing process.

Demonstration Problem 1 (Step 1)

Rejection
Region

Ho: 1   2  0
Ha: 1   2  0

a .001

Z

c

 3.08

0
Critical Value

Demonstration Problem 1 (Steps 2 -7)
Women
x1  $3,352

 1  $1,100

Rejection
Region

 3.08

x  x      
z
 
n n
1

c

 2  $1,700
n2  76

n1  87

a .001

Z

Men
x2  $5,727

0
Critical Value

If z < - 3.08, reject Ho.
If z   3.08, do not reject Ho.

2

1

2

2

1

2

1



2

2

3352  5727  0  10.42
2

2

1100  1700
87
76

Since z = - 10.42 < - 3.08, reject Ho.

(Step 8 – Business implications)
• The evidence is substantial that women, on
average, pay less than men into private pension
funds annually.

• The probability of obtaining an observed z value of
-10.42 is virtually zero.

Confidence Interval
• Sometimes the solution(s) is/are to take a
random sample from each of the two
populations and study the difference in the
two samples.
• Formula for confidence interval to estimate
(µ1 - µ2).
• Designating a group as group one, and
another as group two is an arbitrary
decision.

A consumer test group wants to determine the difference in gasoline
mileage of cars using regular unleaded gas and cars using premium
unleaded gas. Researchers for the group divided a fleet of 100 cars of
the same make in half and tested each car on one tank of gas. Fifty of
the cars were filled with regular unleaded gas and 50 were filled with
premium unleaded gas. The sample average for the regular gasoline
group was 21.45 miles per gallon (mpg), and the sample average for
the premium gasoline group was 24.6 mpg. Assume that the
population standard deviation of the regular unleaded gas population
is 3.46 mpg, and that the population standard deviation of the
premium unleaded gas population is 2.99 mpg. Construct a 95%
confidence interval to estimate the difference in the mean gas
mileage between the cars using regular gasoline and the cars using
premium gasoline.

Regular

Premium

n
n  50
x  21.45 x
  3.46 

 50

1

2

1

1

2

95% Confidence z = 1.96

 24.6
2
 2.99

x  x  z   
2

1

1

2
2

2

 
n n

2

1

2

n n
3.46  2.99      21.45  24.6  1.96
50
50
1

21.45  24.6  1.96

     x1  x 2   z

2

2

1

2

2

1

2

1

2

 4.42      1.88
1

2

2

3.46 2.992

50
50
2

Hypothesis Test for Two Populations
with population variances unknown
• Hypothesis test - compares the means of two
samples to see if there is a difference in the two
population means from which the sample comes.
This is used when σ2 is unknown and samples are
independent.
• Assumes that the measurement is normally
distributed.

Hypothesis Test for Two Populations
with population variances unknown
If σ is unknown, it can be estimated by pooling the
two sample variances and computing a pooled sample
standard deviation

t Statistic to test the
Difference in Means: 12 = 22

t

( x1  x2 )  ( 1   2 )
2
2
s1 (n1  1)  s2 (n2  1)
n1  n2  2

1
1

n1 n2

Hernandez Manufacturing
Company
At the Hernandez Manufacturing Company, an
application of this test arises. New employees are
expected to attend a three-day seminar to learn about
the company. At the end of the seminar, they are
tested to measure their knowledge about the
company. The traditional training method has been
lecture and a question-and-answer session.
Management decided to experiment with a different
training procedure, which processes new employees
in two days by using DVDs and having no questionand-answer session.

Hernandez Manufacturing Company
If this procedure works, it could save the company thousands of
dollars over a period of several years. However, there is some
concern about the effectiveness of the two-day method, and
company managers would like to know whether there is any
difference in the effectiveness of the two training methods. The
managers randomly select a group of 15 from the old method
(Method A) and a group of 12 from the proposed method
(Method B) and test all on a set of questions. The following are
the scores of the groups.
Training Method A
56 50 52 44
47 47 53 45
42 51 42 43

Training Method B
52 59 54 55 65
48 52 57 64 53
44 53 56 53 57

(Steps 1– 4)

 
H :  
Ho:
a

1

0
2

1

2

0

a .05

 .025
2 2
df  n1  n2  2  15  12  2  25
t0.25, 25  2.060
If t < - 2.060 or t > 2.060, reject Ho.
If - 2.060  t  2.060, do not reject Ho.

Rejection
Rejection
Region
Region

Rejection
Rejection
Region
Region

a
.025
2

a
.025
2

t

.025, 25

 2.060

0
0
Critical Values
Critical Values

t

. 025, 25

 2.060

(Step 5)
Training Method A

Training Method B

56

51

45

47

52

43

42

53

52

50

42

48

47

44

44

59

57

53

52

56

65

53

55

53

54

64

57

n1  15

n2  12

x1  47.73

x2  56.5

s  19.495

s  18.273

2
1

2
2

(Step 6-7)
t

( x1  x2 )  (1   2 )
2
1

2
2



s
s

n1
n2


( x1  x2 )  (1   2 )
2
2
s1 (n1  1)  s2 (n2  1)
n1  n2  2

 47.73  56.50   0
19.495 14   18.27311
15  12  2

1
1

n1 n2

1
1

15 12

 5.20
2

2
2
 s1
s2 
n  n 
 1
2 
df 
 25
2
2
2
2
 s1 
 s2 




 n1    n2 
n1  1
n2  1

If t < - 2.060 or t > 2.060, reject Ho.
Since t = -5.20 < -2.060,reject Ho.

Business Implications (Step 8)
• The conclusion is that there is a significant difference in the
effectiveness of the training methods.
• Given that training method B scores are significantly higher
and the fact that the seminar is a day shorter than method A
(saving time and money), it makes business sense to adopt
method B as the standard training method.

Statistical Inferences for Two
Related Populations

• Dependent samples
 Used in before and after studies
 After measurement is not independent of
the before measurement

Hypothesis Testing
• Researcher must determine if the two samples are
related to each other
• The technique for related samples is different from
the technique used to analyze independent
samples
• Matched pairs test requires the two samples to be
of the same size

Dependent Samples
•
•
•

Before and after measurements on the same individual
Studies of twins
Studies of spouses
Individual

Before

After

1

32

39

2

11

15

3

21

35

4

17

13

5

30

41

6

38

39

7

14

22

Hypothesis Testing
• The matched pair t test for dependent measures
uses the sample difference, d, between individual
matched samples as the basic measurement of
analysis
• An analysis of d converts the problem from a two
sample problem to a single sample of differences
• The null hypothesis states that the mean
population difference is zero
• An assumption for the test is that the differences of
two populations are normally distributed

Hypothesis Testing:
Formulas for Dependent Samples
d D
t
sd
n
df  n  1
n  number of pairs
d = sample differencein pairs
D = mean population difference
st = standard deviation of sample difference
d = mean sample difference

d

d

n

sd 

( d  d ) 2
n 1
( d ) 2
d 
n
n 1
2



Hypothesis Testing:
Degree of Freedom
• Analysis of data by this method involves calculating
a t value with a critical value obtained from the
table
• n in the degrees of freedom (n – 1) is the number
of matched pairs or scores

P/E Ratios for Nine Randomly
Selected Companies
Suppose a stock market investor is interested in
determining whether there is a significant difference
in the P/E (price to earnings) ratio for companies from
one year to the next. In an effort to study this
question, the investor randomly samples nine
companies from the Handbook of Common Stocks
and records the P/E ratios for each of these companies
at the end of year 1 and at the end of year 2.

P/E Ratios for Nine Randomly
Selected Companies
Company

Year 1
P/E Ratio

Year 2
P/E Ratio

1

8.9

12.7

2

38.1

45.4

3

43.0

10.0

4

34.0

27.2

5

34.5

22.8

6

15.2

24.1

7

20.3

32.3

8

19.9

40.1

9

61.9

106.5

Hypothesis Testing for Dependent Samples:
P/E Ratios for Nine Companies

Ho : D  0
Ha : D  0
a  .01
df  n  1  9  1  8
t.005,8  3.355
If t < - 3.355 or t > 3.355, reject Ho.

Rejection
Region

Rejection
Region

a
.005
2

a
.005
2

t

 3.355

0
Critical Value

t

 3.355

P/E Ratios for Nine Companies

Ho : D  0
Ha : D  0

d  5.033
sd  21.599
t

 5.033  0
 0.70
21.599
9

Since -3.355  t = -0.70  3.355, do not reject Ho

P/E Ratios for Nine Companies – Software output

Confidence Intervals for
Mean Population Difference
• Researcher can be interested in estimating the
mean difference in two populations for related
samples
• This requires a confidence interval of D (the mean
population difference of two related samples) to be
constructed

Confidence Interval for
Mean Population Difference of Related Samples

d t

s

d

 D  d t

n
df  n  1

s

d

n

Difference in Number of New-House Sales
Realtor

d

8

11

-3

2

19

30

-11

3

5

6

-1

4

sd  3.27

May 2011

1

d  3.39

May 2010

9

13

-4

5

3

5

-2

6

0

4

-4

7

13

15

-2

8

11

17

-6

9

9

12

-3

10

5

12

-7

11

8

6

2

12

2

5

-3

13

11

10

1

14

14

22

-8

15

7

8

-1

16

12

15

-3

17

6

12

-6

18

10

10

0

Confidence Interval for Mean Difference
in Number of New-House Sales
df  n  1  18  1  17
t .005,17  2.898
d t

s

d

n

 D  d t

s

d

n

3.27
3.27
 D  3.39  2.898
18
18
 3.39  2.23  D  3.39  2.23
 3.39  2.898

 5.62  D  1.16

The analyst estimates with a 99% level of confidence that the
average difference in new-house sales for a real estate
company in Indianapolis between 2005 and 2006 is between
-5.62 and -1.16 houses.

ˆ
ˆ ˆ
( p1  p2 )  ( p1  p2 ) p  proportion from sample 1
z
ˆ
p  proportion from sample 2
p1  q1 p2  q2

n  size of sample 1
n1
n2
n  size of sample 2
1

2

1

2

p  proportion from population 1
p  proportion from population 2
q  1- p
q  1- p
1

2

1

1

2

2

Statistical Inference about Two
Population Proportions: Hypothesis Testing
• Because population proportions are
unknown, an estimate of the Standard
Deviation of the difference in two sample
proportions is made by using sample
proportions as point of estimates of the
population proportion

Z Formula to Test the Difference

in Population Proportions

p
ˆ
Z 

1

P 



2

  p

1



p

 1
1 

 p  q 



n2 
 n1
x1  x2

n n
ˆ
ˆ
n p n p
n n
1



ˆ
p

1

2

2

1

q  1 p

1

2

2

2



Statr session 17 and 18

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Statr session 17 and 18