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Section 11.1
Hypothesis Testing – Two
Means (Large, Independent
Samples)
HAWKES LEARNING SYSTEMS
Hypothesis Testing (Two or More Populations)
math courseware specialists
11.1 Hypothesis Testing – Two Means
(Large, Independent Samples)
Definitions:
•
Hypothesis – a theory or premise, usually the claim that
someone is investigating.
•
Null Hypothesis, H0 – describes the currently accepted
value for the population parameter.
•
Alternative Hypothesis, Ha – describes the claim that is
being tested; the mathematical opposite of the null
hypothesis.
•
Hypothesis Test – compares the merit of the two
competing hypotheses by examining the data that is
collected.
HAWKES LEARNING SYSTEMS
Hypothesis Testing (Two or More Populations)
math courseware specialists
11.1 Hypothesis Testing – Two Means
(Large, Independent Samples)
Determine the null and alternative hypotheses (Left-Tailed):
Write the null and alternative hypotheses for the claim
that population 1’s mean is less than population 2’s
mean.
Solution:
Claim: Population 1’s mean is less than population 2’s
mean.
m1 < m2 or m1 – m2 < 0
Mathematical opposite: m1 ≥ m2 or m1 – m2 ≥ 0
H0: m1 – m2 ≥ 0
Ha: m1 – m2 < 0
Current accepted belief
Testing hypothesis
HAWKES LEARNING SYSTEMS
Hypothesis Testing (Two or More Populations)
math courseware specialists
11.1 Hypothesis Testing – Two Means
(Large, Independent Samples)
Determine the null and alternative hypotheses (Right-Tailed):
Write the null and alternative hypotheses for the claim
that population 1’s mean is more than 20 units above
population 2’s mean.
Solution:
Claim: Population 1’s mean is more than 20 units above
population 2’s mean.
m1 > m2 + 20 or m1 – m2 > 20
Mathematical opposite: m1 ≤ m2 + 20 or m1 – m2 ≤ 20
H0: m1 – m2 ≤ 20 Current accepted belief
Ha: m1 – m2 > 20 Testing hypothesis
HAWKES LEARNING SYSTEMS
Hypothesis Testing (Two or More Populations)
math courseware specialists
11.1 Hypothesis Testing – Two Means
(Large, Independent Samples)
Determine the null and alternative hypotheses (Two-Tailed):
Write the null and alternative hypotheses for the claim that
population 1’s mean is not equal to population 2’s mean.
Solution:
Claim: Population 1’s mean is not equal to population 2’s
mean.
m1 ≠ m2 or m1 – m2 ≠ 0
Mathematical opposite: m1 = m2 or m1 – m2 = 0
H0: m1 – m2 = 0
Ha: m1 – m2 ≠ 0
Current accepted belief
Testing hypothesis
HAWKES LEARNING SYSTEMS
Hypothesis Testing (Two or More Populations)
math courseware specialists
11.1 Hypothesis Testing – Two Means
(Large, Independent Samples)
Test Statistic for Large Samples, n ≥ 30:
• A p-value is the probability of obtaining a sample
more extreme than the one observed on your
data, when H0 is assumed to be true.
• To find the p-value, first calculate the z-score
from the sample data and then find the
corresponding probability for that z-score.
HAWKES LEARNING SYSTEMS
Hypothesis Testing (Two or More Populations)
math courseware specialists
11.1 Hypothesis Testing – Two Means
(Large, Independent Samples)
Conclusions for a Hypothesis Testing Using p-Values:
1. If p ≤ a, then reject the null hypothesis.
2. If p > a, then fail to reject the null hypothesis.
HAWKES LEARNING SYSTEMS
Hypothesis Testing (Two or More Populations)
math courseware specialists
11.1 Hypothesis Testing – Two Means
(Large, Independent Samples)
Steps for Hypothesis Testing:
1. State the null and alternative hypotheses.
2. Set up the hypothesis test by choosing the
test statistic and stating the level of
significance.
3. Gather data and calculate the necessary
sample statistics.
4. Draw a conclusion by comparing the p-value
to the level of significance.
HAWKES LEARNING SYSTEMS
Hypothesis Testing (Two or More Populations)
math courseware specialists
11.1 Hypothesis Testing – Two Means
(Large, Independent Samples)
Draw a conclusion:
Two universities in the same state are bitter rivals. Each university believes
that their students are more physically fit than the other. To test the claim that
there is a difference in the fitness of the students at each university, 36
students at the first university were surveyed and found to exercise on
average 2.9 hours a week with a standard deviation of 1.1 hours. Thirty-eight
students at the second university were also surveyed and found to have an
average of 2.7 exercise hours a week with a standard deviation of 1.0 hour.
Use a 0.05 level of significance to perform a hypothesis test with the given
data.
Solution:
First state the hypotheses:
H0: m1 – m2 = 0
Ha: m1 – m2 ≠ 0
Next, set up the hypothesis test and state the level of
significance:
a = 0.05
Reject if p < a, or if p < 0.05.
HAWKES LEARNING SYSTEMS
Hypothesis Testing (Two or More Populations)
math courseware specialists
11.1 Hypothesis Testing – Two Means
(Large, Independent Samples)
Solution (continued):
Gather the data and calculate the necessary sample statistics:
n1 = 36, 1 = 2.9, s1 = 1.1, n2 = 38, 2 = 2.7, s2 = 1.0
0.82
Since this is a two-tailed test, p = 0.2061 x (2) = 0.4122.
Finally, draw a conclusion:
Since p is greater than a (0.05), we will fail to reject the null
hypothesis. There is not sufficient evidence at the 0.05 level
of significance to say that there is a difference in the fitness of
the students at each university.
HAWKES LEARNING SYSTEMS
Hypothesis Testing (Two or More Populations)
math courseware specialists
11.1 Hypothesis Testing – Two Means
(Large, Independent Samples)
Draw a conclusion:
A drug manufacturer claims that its new cholesterol drug, when used together
with a healthy diet and exercise plan, lowers cholesterol by over 20 points
more than simply changing a patient’s diet and exercise regimen. To test the
claim, a group of 55 patients with high cholesterol is chosen to take the drug
along with a change in diet and exercise. Over the course of 3 months, this
group lowers their cholesterol by an average of 44.7 points with a standard
deviation of 6.8 points. Another 55 patients change their diet and exercise
regimen, but do not take the drug. This group lowers their cholesterol by an
average of 23.1 points, with a standard deviation of 5.3 points. Test the claim
using a 0.01 level of significance.
Solution:
First state the hypotheses:
H0: m1 – m2 ≤ 20
Ha: m1 – m2 > 20
Next, set up the hypothesis test and state the level of
significance:
a = 0.01
Reject if p < a, or if p < 0.01.
HAWKES LEARNING SYSTEMS
Hypothesis Testing (Two or More Populations)
math courseware specialists
11.1 Hypothesis Testing – Two Means
(Large, Independent Samples)
Solution (continued):
Gather the data and calculate the necessary sample statistics:
n1 = 55, 1 = 44.7, s1 = 6.8, n2 = 55, 2 = 23.1, s2 = 5.3
1.38
Since this is a right-tailed test, p = 0.0838.
Finally, draw a conclusion:
Since p is greater than a (0.01), we will fail to reject the null
hypothesis. There is not sufficient evidence at the 0.01 level
of significance to support the drug manufacturer’s claim that
the new drug, when used together with a healthy diet and
exercise plan, lowers cholesterol by more than 20 points.
μ1 - μ2 = 0
μ1 - μ2 ≠ 0
Population Means are the same
Population Means are the different
=
(8.83 – 9.04) - 0
SQRT( 0.3932 + 0.6722 )
34
36
. = -1.6065
μ1 - μ2 = 0
μ1 - μ2 ≠ 0
Population Means are the same
Population Means are the different
Reject the Null, if p-value < a
μ1 - μ2 = 0
μ1 - μ2 ≠ 0
Population Means are the same
Population Means are the different
Reject if z < -za or |z| > | za | or p-value < a
Reject if |z| > |1.96| or z < -1.96
Computed-z is -1.6065 and it is Not < -1.96 or |1.6065| is Not > |1.96|
p-value is 0.108 and it is greater than 0.05
Don’t reject the Null Hypothesis. Not enough evidence that they are different.
 s12 s 22 
Std.Error =   
 n1 n2 
μ1 - μ2 = 0
μ1 - μ2 ≠ 0
Population Means are the same
Population Means are the different
Critical-Z
c
One-Tailed Test
Two-Tailed Test
0.90
1.28
±1.645
0.95
1.645
±1.96
0.98
2.05
±2.33
0.99
2.33
±2.575
Confidence Interval = (μ1 – μ2) ± (Critical-Z x Std Error)
Confidence Interval = (μ1 – μ2) ± (Margin of Error)
Std Error = SQRT(
3.952
251
+
3.032
270
) = 0.3101
 s12 s 22 
Std.Error =   
 n1 n2 
Lower = (6.93 - 8.21) - (1.96 x 0.3101) = -1.8878
Upper = (6.93 - 8.21) + (1.96 x 0.3101) = -0.6722
 s12 s 22 
Std.Error =   
 n1 n2 