1. St. Edward’s University
SLIDES BY
John Loucks
St. Edward’s University .

2. Chapter 11 Comparisons Involving Proportions and a Test of Independence
Inferences About the Difference Between
Two Population Proportions
Hypothesis Test for Proportions
of a Multinomial Population
Test of Independence

3. Inferences About the Difference Between Two Population Proportions
Interval Estimation of p1 - p2
Hypothesis Tests About p1 - p2

4. Inferences About the Difference Between Two Population Proportions
Let:
p1 denote the proportion for population 1
p2 denote the population for population 2
To make an inference about p1 - p2 we will select two independent random samples consisting of n1 units from population 1 and n2 units from population 2.
Let:
denote the sample proportion for population 1
denote the sample proportion for population 2

5. Sampling Distribution of
Expected Value
Standard Deviation (Standard Error)
where: n1 = size of sample taken from population 1
n2 = size of sample taken from population 2

6. Sampling Distribution of
If the sample sizes are large, the sampling distribution
of can be approximated by a normal probability
distribution.
The sample sizes are sufficiently large if all of these
conditions are met:
n1p1 > 5
n1(1 - p1) > 5
n2p2 > 5
n2(1 - p2) > 5

7. Sampling Distribution of
p1 – p2

8. Interval Estimation of p1 - p2
Interval Estimate
where:
Point Estimate is
Margin of Error is

9. Interval Estimation of p1 - p2
Example: Market Research Associates
Market Research Associates is conducting
research to evaluate the effectiveness of a client’s
new advertising campaign. Before the new
campaign began, a telephone survey of 150
households in the test market area showed 60
households “aware” of the client’s product.
The new campaign has been initiated with TV
and newspaper advertisements running for three
weeks.

10. Interval Estimation of p1 - p2
Example: Market Research Associates
A survey conducted immediately after the new
campaign showed 120 of 250 households “aware”
of the client’s product.
Does the data support the position that the
advertising campaign has provided an increased
awareness of the client’s product?

11. Point Estimator of the Difference Between Two Population Proportions
p1 = proportion of the population of households
“aware” of the product after the new campaign
p2 = proportion of the population of households
“aware” of the product before the new campaign
= sample proportion of households “aware” of the
product after the new campaign
product before the new campaign

12. Interval Estimation of p1 - p2
For = .05, z.025 = 1.96:
(.0510)
Hence, the 95% confidence interval for the difference
in before and after awareness of the product is
-.02 to +.18.

13. Interval Estimation of p1 - p2
Excel Formula Worksheet A B C D E 1
Sur2
Sur1
Survey 2 (from Popul.1)
Survey 1 (from Popul.2) 2 No
Yes
Sample Size
250
150 3
No. of "Yes"
=COUNTIF(A2:A251,"Yes")
=COUNTIF(B2:B151,"Yes") 4
Samp. Propor.
=D3/D2
=E3/E2 5 6
Confid. Coeff.
0.95 7
Lev. Of Signif.
=1-D6 8 z
Value
=NORM.S.INV(1-D7/2,TRUE) 9 10
Std. Error
=SQRT(D4*(1-D4)/D2+E4*(1-E4)/E2) 11
Marg. of Error
=D8*D10 12 13
Pt. Est. of Diff.
=D4-E4 14
Lower Limit
=D13-D11 15
Upper Limit
=D13+D11
Note: Rows are not shown.

14. Interval Estimation of p1 - p2
Excel Value Worksheet A B C D E 1
Sur2
Sur1
Survey 2 (from Popul.1)
Survey 1 (from Popul.2) 2 No
Yes
Sample Size
250
150 3
No. of "Yes"
120 60 4
Samp. Propor.
0.48
0.40 5 6
Confid. Coeff.
0.95 7
Lev. Of Signif.
0.05 8 z
Value
1.960 9 10
Std. Error
0.0510 11
Marg. of Error
0.0999 12 13
Pt. Est. of Diff.
0.080 14
Lower Limit
-0.020 15
Upper Limit
0.180
Note: Rows are not shown.

15. Hypothesis Tests about p1 - p2
Hypotheses
We focus on tests involving no difference between
the two population proportions (i.e. p1 = p2)
H0: p1 - p2 < 0
Ha: p1 - p2 > 0
Left-tailed
Right-tailed
Two-tailed

16. Hypothesis Tests about p1 - p2
Pooled Estimate of Standard Error of
where:

17. Hypothesis Tests about p1 - p2
Test Statistic

18. Hypothesis Tests about p1 - p2
Example: Market Research Associates
Can we conclude, using a .05 level of significance,
that the proportion of households aware of the
client’s product increased after the new advertising
campaign?

19. Hypothesis Tests about p1 - p2
p -Value and Critical Value Approaches
1. Develop the hypotheses.
H0: p1 - p2 < 0
Ha: p1 - p2 > 0
p1 = proportion of the population of households
“aware” of the product after the new campaign
p2 = proportion of the population of households
“aware” of the product before the new campaign

20. Hypothesis Tests about p1 - p2
p -Value and Critical Value Approaches
2. Specify the level of significance.
a = .05
3. Compute the value of the test statistic.

21. Hypothesis Tests about p1 - p2
p –Value Approach
4. Compute the p –value.
For z = 1.56, the p–value = .0594
5. Determine whether to reject H0.
Because p–value > a = .05, we cannot reject H0.
We cannot conclude that the proportion of households
aware of the client’s product increased after the new
campaign.

22. Hypothesis Tests about p1 - p2
Critical Value Approach
4. Determine the critical value and rejection rule.
For a = .05, z.05 = 1.645
Reject H0 if z > 1.645
5. Determine whether to reject H0.
Because 1.56 < 1.645, we cannot reject H0.
We cannot conclude that the proportion of households
aware of the client’s product increased after the new
campaign.

23. Hypothesis Tests about p1 - p2
Excel Formula Worksheet A B C D E 1
Sur2
Sur1
Survey 2 (from Popul.1)
Survey 1 (from Popul.2) 2 No
Yes
Sample Size
=COUNTA(A2:A251)
=COUNTA(B2:B151) 3
Resp. of Interest 4
Count for Resp.
=COUNTIF(A2:A251,D3)
=COUNTIF(B2:B151,E3) 5
Sample Propor.
=D4/D2
=E4/E2 6 7
Hypoth. Value 8
Point Est. of Diff.
=D5-E5 9 10
Pooled Est. of p
=(D2*D5+E2*E5)/(D2+E2) 11
Standard Error 12
Test Statistic
=(D8-D7)/D11 13 14
-Value (lower tail)
=NORM.S.DIST(D12,TRUE) 15
-Value (upper tail)
=1-NORM.S.DIST(D12,TRUE) 16
-Value (two tail)
=2*MIN(D14,D15)
=SQRT(D10*(1-D10)*(1/D2+1/E2))
Note:
Rows
are not shown.

24. Hypothesis Tests about p1 - p2
Excel Value Worksheet A B C D E 1
Sur2
Sur1
Survey 2 (from Popul.1)
Survey 1 (from Popul.2) 2 No
Yes
Sample Size
250
150 3
Resp. of Interest 4
Count for Resp.
120 60 5
Sample Propor.
0.48
0.40 6 7
Hypoth. Value 8
Point Est. of Diff.
0.08 9 10
Pooled Est. of p
0.450 11
Standard Error 12
Test Statistic
1.557 13 14
-Value (lower tail)
0.940 15
-Value (upper tail)
0.060 16
-Value (two tail)
0.120
0.0514
Note:
Rows
are not shown.

25. Hypothesis Test for Proportions of a Multinomial Population
In this case, each element of a population is assigned to one and only one of several classes or categories.
Such a population is a multinomial population.
The multinomial distribution can be thought of as an extension of the binomial distribution.
On each trial of a multinomial experiment:
One and only one of the outcomes occurs
Each trial is assumed to be independent
The probabilities of the outcomes remain the same for each trial

26. 1. State the null and alternative hypotheses.
Hypothesis (Goodness of Fit) Test for Proportions of a Multinomial Population
1. State the null and alternative hypotheses.
H0: The population follows a multinomial
distribution with specified probabilities
for each of the k categories
Ha: The population does not follow a
multinomial distribution with specified
probabilities for each of the k categories

27. Hypothesis (Goodness of Fit) Test for Proportions of a Multinomial Population
2. Select a random sample and record the observed
frequency, fi , for each of the k categories.
3. Assuming H0 is true, compute the expected
frequency, ei , in each category by multiplying the
category probability by the sample size.

28. Hypothesis (Goodness of Fit) Test for Proportions of a Multinomial Population
4. Compute the value of the test statistic.
where:
fi = observed frequency for category i
ei = expected frequency for category i
k = number of categories
Note: The test statistic has a chi-square distribution
with k – 1 df provided that the expected frequencies
are 5 or more for all categories.

29. Hypothesis (Goodness of Fit) Test for Proportions of a Multinomial Population
5. Rejection rule:
p-value approach:
Reject H0 if p-value < a
Critical value approach:
Reject H0 if
where  is the significance level and
there are k - 1 degrees of freedom

30. Multinomial Distribution Goodness of Fit Test
Example: Finger Lakes Homes (A)
Finger Lakes Homes manufactures four models
of prefabricated homes, a two-story colonial, a log
cabin, a split-level, and an A-frame. To help in
production planning, management would like to
determine if previous customer purchases indicate
that there is a preference in the style selected.

31. Multinomial Distribution Goodness of Fit Test
Example: Finger Lakes Homes (A)
The number of homes sold of each model for 100
sales over the past two years is shown below.
Split A-
Model Colonial Log Level Frame
# Sold

32. Multinomial Distribution Goodness of Fit Test
Hypotheses
H0: pC = pL = pS = pA = .25
Ha: The population proportions are not
pC = .25, pL = .25, pS = .25, and pA = .25
where:
pC = population proportion that purchase a colonial
pL = population proportion that purchase a log cabin
pS = population proportion that purchase a split-level
pA = population proportion that purchase an A-frame

33. Multinomial Distribution Goodness of Fit Test
Rejection Rule
Reject H0 if p-value < .05 or c2 >
With  = .05 and
k - 1 = = 3
degrees of freedom
Do Not Reject H0
Reject H0 2
7.815

34. Multinomial Distribution Goodness of Fit Test
Expected Frequencies
Test Statistic
e1 = .25(100) = e2 = .25(100) = 25
e3 = .25(100) = e4 = .25(100) = 25 =
= 10

35. Multinomial Distribution Goodness of Fit Test
Conclusion Using the p-Value Approach
Area in Upper Tail
c2 Value (df = 3)
Because c2 = 10 is between and , the
area in the upper tail of the distribution is between
.025 and .01.
The p-value < a . We can reject the null hypothesis.

36. Multinomial Distribution Goodness of Fit Test
Conclusion Using the Critical Value Approach
c2 = 10 > 7.815
We reject, at the .05 level of significance,
the assumption that there is no home style
preference.

37. Multinomial Distribution Goodness of Fit Test
Excel Worksheet (showing data)
Note: Rows are not shown.

38. Multinomial Distribution Goodness of Fit Test
Excel Formula Worksheet C D E F G H I 1
Hyp.
Observed
Expect.
Sq'd.
Sq.Diff./ 2
Categ.
Prop.
Frequency
Freq.
Diff.
Exp.Freq. 3
Col.
0.25
=COUNTIF(B2:B101,"Col")
=D3*$E$7
=E3-F3
=G3^2
=H3/F3 4
Log
=COUNTIF(B2:B101,"Log")
=D4*$E$7
=E4-F4
=G4^2
=H4/F4 5
Split-L
=COUNTIF(B2:B101,"Spl")
=D5*$E$7
=E5-F5
=G5^2
=H5/F5 6
A-Fr.
=COUNTIF(B2:B101,"Afr")
=D6*$E$7
=E6-F6
=G6^2
=H6/F6 7
Total
=SUM(E3:E6)
=SUM(I3:I6) 8 9 10
=I7 11
=E9-1 12
=CHISQ.DIST.RT(E10,E11)
Categories
Degr. of Free. p
-Value
Test Statistic
Note: Columns A-B and rows are not shown.

39. Multinomial Distribution Goodness of Fit Test
Excel Value Worksheet C D E F G H I 1
Hyp.
Observed
Expect.
Sq'd.
Sq.Diff./ 2
Categ.
Prop.
Frequency
Freq.
Diff.
Exp.Freq. 3
Col.
0.25 30 25 5 4
Log 20 -5
Split-L 35 10
100 6
A-Fr. 15
-10 7
Total 8 9 11 12
0.0186
Categories
Degr. of Free. p
-Value
Test Statistic
Note: Columns A-B and rows are not shown.

40. Test of Independence
Another important application of the chi-square distribution involves using sample data to test for the independence of two variables.
To test whether two variables are independent, one sample is selected and crosstabulation is used to summarize the data for the two variables simultaneously.

41. Test of Independence 1. Set up the null and alternative hypotheses.
H0: The column variable is independent of
the row variable
Ha: The column variable is not independent
of the row variable
2. Select a random sample and record the observed
frequency, fij , for each cell of the contingency table.
3. Compute the expected frequency, eij , for each cell.

42. Test of Independence 4. Compute the test statistic.
5. Determine the rejection rule.
Reject H0 if p -value < a or
where  is the significance level and,
with n rows and m columns, there are
(n - 1)(m - 1) degrees of freedom.

43. Test of Independence Example: Finger Lakes Homes (B)
Each home sold by Finger Lakes Homes can be
classified according to price and to style. Finger
Lakes’ manager would like to determine if the
price of the home and the style of the home are
independent variables.

44. Test of Independence Example: Finger Lakes Homes (B)
The number of homes sold for each model and
price for the past two years is shown below. For
convenience, the price of the home is listed as either
$200,000 or less or more than $200,000.
Price Colonial Log Split-Level A-Frame
< $200,
> $200,

45. Test of Independence Hypotheses
H0: Price of the home is independent of the
style of the home that is purchased
Ha: Price of the home is not independent of the
style of the home that is purchased

46. Test of Independence Expected Frequencies
Price Colonial Log Split-Level A-Frame Total
< $200K
> $200K
Total

47. Test of Independence Rejection Rule
With  = .05 and (2 - 1)(4 - 1) = 3 d.f.,
Reject H0 if p-value < .05 or 2 > 7.815
Test Statistic
= =

48. Test of Independence Conclusion Using the p-Value Approach
Area in Upper Tail
c2 Value (df = 3)
Because c2 = is between and 9.348, the
area in the upper tail of the distribution is between
.05 and .025.
The p-value < a . We can reject the null hypothesis.

49. Test of Independence Conclusion Using the Critical Value Approach
We reject, at the .05 level of significance,
the assumption that the price of the home is
independent of the style of home that is
purchased.

50. Test of Independence Excel Worksheet (showing data) A B C D E 1 Home
Price ($)
Style 2
>200K
Colonial 3
<=200K
Log 4 5
A-Frame 6 7
Split-Level 8 9 10
Note: Rows are not shown.

51. Test of Independence Excel Worksheet (showing Pivot Table)J 1 2
Price ($)
Colonial
Log
Split-Lev.
A-Frame 3
<=200K 18 6 19 12 55 4
>200K 14 16 45 5
Grand Total 30 20 35 15
100
Count of Home
Preference
Grand Tot.
Note: Columns A-D are not shown.

52. Test of Independence Excel Formula WorksheetG H I J 1 2
Price ($)
Colonial
Log
Split-Lev.
A-Frame 3
<=200K 18 6 19 12 55 4
>200K 14 16 45 5
Grand Total 30 20 35 15
100
Count of Home
Preference 7
Expected Frequencies 8 9 10 11
=I5*J3/J5
=I5*J4/J5
=H5*J3/J5
=H5*J4/J5
=F5*J3/J5
=G5*J3/J5
=F5*J4/J5
=G5*J4/J5 p
-Value
=CHISQ.TEST(F3:I4,F9:I10)
Grand Tot.
Note: Columns A-D are not shown.

53. Test of Independence Excel Value WorksheetG H I J 1 2
Price ($)
Colonial
Log
Split-Lev.
A-Frame 3
<=200K 18 6 19 12 55 4
>200K 14 16 45 5
Grand Total 30 20 35 15
100
Count of Home
Preference 7
Expected Frequencies 8 9 10 11
8.25
6.75
19.25
15.75
16.50
11.00
13.50
9.00 p
-Value
0.0274
Grand Tot.
Note: Columns A-D are not shown.

54. End of Chapter 11