LESSON 2.2
INFERENCES ABOUT
TWO PROPORTIONS
MGT204 - Unit 2
Hypothesis Testing with Two Samples
In this Unit
In MGT201, we introduced methods of inferential statistics:
confidence intervals and testing claims made about population
parameters.
● The objective of this unit is to extend the methods for
estimating values of population parameters and the methods
for testing hypotheses to situation involving two sets of sample
data instead of just one.
● See Lesson 2.1 to review these concepts.
●
In this Lesson
We present methods for:
Testing a claim made about the two population proportions
and
2. Constructing a confidence interval estimate of the difference
between the two population proportions.
1.
Notation for Two Proportions
For population 1 we let
p1 = population proportion
n1 = size of the sample
x1 = number of successes in the sample
(sample proportion)
(complement of )
The corresponding notations
apply to population 2.
Pooled Sample Proportion
The Pooled sample proportion is denoted by
given by:
and is
Requirements
1. The sample proportions are from two simple random sample that
are independent (Samples are independent if the sample values
selected from one population are not related to or somehow
naturally paired or matched with sample values selected from the
other population.)
2. For each of the two samples, the number of successes is at least 5
and the number of failures is at least 5 (That is
for each of the two samples.)
Test Statistic for Two Proportions (with H0: p1 = p2)
where p1 - p2 = 0 (Assumed in the
null hypothesis)
Critical values
Use Appendix B.1 - Click to access
Based on the significance level α, find the area under the curve
up to the critical value by finding 1 - α/2
Finally, express the critical value as a z-score, find the z-score in
Appendix B.1 equal to the area of the curve.
Confidence Interval Estimate of p1 - p2
The confidence interval estimate of the difference p1 - p2 is:
where the margin of error E is given by:
Rounding: Round the confidence interval limits to three significant
digits.
Example 1 - Do Airbags Save Lives?
The table below lists results from a simple random sample of front-seat
occupants involved in car crashes (based on data from “Who Wants Airbags?”
by Meyer and Finney, Chance, Vol. 18, No 2). Use a 0.05 significance level to
test the claim that the fatality rate of occupants is lower for those in cars
equipped with airbags.
Airbag Available
Occupant Fatalities
Total number of Occupants
No Airbag Available
41
52
11,541
9,853
Example 1 - Do Airbags Save Lives?
Step 0: Given
Occupant Fatalities
Total number of Occupants
Airbag Available
No Airbag Available
41
52
11,541
9,853
α = 0.05
Claim: Fatality rate of occupants is lower for those in cars equipped with airbags.
Symbolically: The claim that the fatality rate is lower for those with airbags can be expressed
as p1 < p2
p1 represents the fatality rate in Airbags available
p2 represents the fatality rate in no airbags available.
Example 1 - Do Airbags Save Lives?
Step 1: State Null and Alternative Hypotheses
Since the claim of p1 < p2 does not contain equality, it becomes the
alternative hypothesis. The null hypothesis is the statement of equality.
For tests of hypotheses made about two population proportions, we
consider only tests having a null hypothesis of p1 = p2. So we have:
H0: p1 = p2
H1: p1 < p2 (Original Claim)
Since H1 contains the less than symbol <, this will be a left-tailed test
Example 1 - Do Airbags Save Lives?
Step 2: Determine which Test Statistic to use given the information you have:
Since it is a proportion, we will use the normal distribution (z-test)
Example 1 - Do Airbags Save Lives?
Step 3: The Decision Rule
Given the level of significance, type of test, and which test statistic being used,
we find the critical value(s) and find the area of rejection.
We refer to Appendix B.1 and find that an area of α = 0.05 in the left tail
corresponds to the critical value of z = -1.645
Reject H0
z = -1.645
Example 1 - Do Airbags Save Lives?
Step 4: Calculate the Test Statistic
Example 1 - Do Airbags Save Lives?
Step 5: Interpret Results
The test statistic of z = -1.91 does fall in the critical region bounded by
the critical value of z = -1.645.
Therefore we reject the null hypothesis and accept the alternative
hypothesis (the original claim).
We conclude that there is sufficient evidence to support the claim that
the proportion of accident fatalities for occupants in cars with airbags is
less than the proportion of fatalities for occupants in cars without
airbags.
Confidence Intervals
Using the format given earlier, we can construct a confidence interval
estimate of the difference between population proportions (p1 - p2). If a
confidence interval estimate of p1 - p2 does not include 0, we have
evidence suggesting that p1 and p2 have different values.
Example 2 - Airbags
Use the sample data given in Example 1 to construct a 90% confidence
interval estimate of the difference between the two population
proportions. What does the result suggest about the effectiveness of
airbags in an accident?
Example 2 - Airbags
Step 0: Given
Airbag Available
Occupant Fatalities
Total number of Occupants
No Airbag Available
41
52
11,541
9,853
With a 90% confidence level, zα/2 = 1.645 (from Appendix B.1).
From Example 1:
Example 2 - Airbags
Step 1: Calculate the value of the margin of error E as shown.
Example 2 - Airbags
Step 2: Construct the confidence interval
Example 2 - Airbags
Step 3: Interpret results.
The confidence interval limits do not contain 0, implying that there is a significant
difference between the two proportions.
p1
The confidence interval suggests that the fatality rate is lower for occupants in cars
with airbags than for occupants in cars without airbags. (The interval contains only
p2
negative numbers)
p1 - p2 is always negative
The confidence interval also provides an estimate of the amount of the difference
between the two fatality rates.
Practice
2.2 Handout