Transcript Section 11.2

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Section 11.2
Hypothesis Testing – Two
Means(Small, Independent
Samples)
HAWKES LEARNING SYSTEMS
Hypothesis Testing (Two or More Populations)
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11.2 Hypothesis Testing – Two Means
(Small, Independent Samples)
Criteria for hypothesis testing with small samples:
• The samples are independent.
• Each population from which the samples were drawn
has a normal distribution.
• n < 30 for one or both samples.
•  is unknown for both populations.
HAWKES LEARNING SYSTEMS
Hypothesis Testing (Two or More Populations)
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11.2 Hypothesis Testing – Two Means
(Small, Independent Samples)
Test Statistic for Small Samples, n < 30:
If the samples can be pooled when variances are equal:
with d.f. = n1 + n2 – 2
Formula for pooled standard deviations:
s 2p
=
2
(n1  1) s1
2
 (n2  1) s2
n1  n2  2
Formula for the value of t:
t=
X1  X 2
2
s p 
1
1 
 
 n1 n2 
HAWKES LEARNING SYSTEMS
Hypothesis Testing (Two or More Populations)
math courseware specialists
11.2 Hypothesis Testing – Two Means
(Small, Independent Samples)
Test Statistic for Small Samples, n < 30:
If the samples cannot be pooled when variances are not equal:
where the degrees of freedom is the smaller of n1 – 1 and n2 – 1
HAWKES LEARNING SYSTEMS
Hypothesis Testing (Two or More Populations)
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11.2 Hypothesis Testing – Two Means
(Small, Independent Samples)
To determine if the test statistic calculated from the
sample is statistically significant we will need to look at
the critical value.
The critical values for n < 30 are found from the
t-distribution.
HAWKES LEARNING SYSTEMS
Hypothesis Testing (Two or More Populations)
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11.2 Hypothesis Testing – Two Means
(Small, Independent Samples)
Rejection Regions:
• For a left-tailed test, reject H0 if t ≤ –t
• For a right-tailed test, reject H0 if t ≥ t
• For a two-tailed test, reject H0 if |t | ≥ t/2
HAWKES LEARNING SYSTEMS
Hypothesis Testing (Two or More Populations)
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11.2 Hypothesis Testing – Two Means
(Small, 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 determining the values of
the test statistic that would lead to rejecting
the null hypothesis.
3. Gather data and calculate the necessary
sample statistics.
4. Draw a conclusion.
HAWKES LEARNING SYSTEMS
Hypothesis Testing (Two or More Populations)
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11.2 Hypothesis Testing – Two Means
(Small, Independent Samples)
Draw a conclusion:
A home improvement warehouse claims that people interested in home
improvement projects can save time by attending their workshops.
Specifically, the warehouse claims that people who attend their tiling
workshops take less time to complete comparable projects than those who do
not. To test the claim, a group of 10 people who attended their workshops is
later surveyed about the time it took to finish the project. This group spent an
average of 14.1 hours to complete their projects with a standard deviation of
2.3 hours. Another 10 people chosen for the study did not attend the
workshops, and they spent on average 15.0 hours with a standard deviation of
2.4 hours. Test the claim at the 0.01 level of significance. Assume that the
populations are normally distributed and that the variances of the populations
are the same.
Solution:
First state the hypotheses:
H0: m1 – m2 ≥ 0
Ha: m1 – m2 < 0
HAWKES LEARNING SYSTEMS
Hypothesis Testing (Two or More Populations)
math courseware specialists
11.2 Hypothesis Testing – Two Means
(Small, Independent Samples)
Solution (continued):
Next, set up the hypothesis test and state the level of
significance:
 = 0.01, d.f. = n1 + n2 – 2 = 10 + 10 – 2 = 18
t0.01 = –2.552
Reject if t ≤ –t , or if t ≤ – 2.552.
Gather the data and calculate the necessary sample statistics:
n1 = 10, 1 = 14.1, s1 = 2.3, n2 = 10, 2 = 15.0, s2 = 2.4
 –0.856
HAWKES LEARNING SYSTEMS
Hypothesis Testing (Two or More Populations)
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11.2 Hypothesis Testing – Two Means
(Small, Independent Samples)
Solution (continued):
Finally, draw a conclusion:
Since t=-0.856 is greater than t =-2.552 we will fail to reject
the null hypothesis. There is not sufficient evidence at the
0.01 level of significance to say that people who complete the
tiling workshops take less time to complete their projects than
those who do not.
–0.856
HAWKES LEARNING SYSTEMS
Hypothesis Testing (Two or More Populations)
math courseware specialists
11.2 Hypothesis Testing – Two Means
(Small, Independent Samples)
Draw a conclusion:
The Smith CPA Firm claims that their clients obtain larger tax refunds than
clients of their competitor, Jones and Company CPA. To test the claim, 15
clients from the Smith firm are selected and found to have an average tax
refund of $942 with a standard deviation of $103. At Jones and Company, 18
clients are surveyed and found to have an average refund of $898 with a
standard deviation of $95. Test the claim at the 0.05 level of significance.
Assume that both populations are normally distributed. Because the two firms
are located in difference parts of a large city, assume the variances for the two
populations are not equal.
Solution:
First state the hypotheses:
H0: m1 – m2 ≤ 0
Ha: m1 – m2 > 0
Next, state the level of significance:
= 0.05, d.f. = 14 d.f. is the smaller of n1 – 1 and n2 – 1
t0.05, 14 = 1.761
Reject if t ≥ t , or if t ≥ 1.761.
HAWKES LEARNING SYSTEMS
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11.2 Hypothesis Testing – Two Means
(Small, Independent Samples)
Solution (continued):
Gather the data and calculate the necessary sample statistics:
n1 = 15, 1 = 942, s1 = 103, n2 = 18, 2 = 898, s2 = 95
1.266
Finally, draw a conclusion:
Since t=1.266 is less than t (1.761),
we will fail to reject the null hypothesis.
There is not sufficient evidence at the
0.05 level of significance to support
Smith CPA’s claim that their clients
receive larger tax refunds than clients
of Jones and Company.
1.266
μ1 - μ2 = 0
μ1 - μ2 ≠ 0
Critical t
Population Means are the same
Population Means are the different
-2.0860
Critical-T (0.025,20) = 2.0860
It resides in the left tail
Thus, Critical-T = -2.0860
t=
X1  X 2
1
1 
s 2p   
 n1 n2 
= -6.1574
s 2p
(n1  1) s12  (n2  1) s22 = 134.80
=
n1  n2  2
1 1
Std .Error = s 2p   
 n1 n2 
Std Error =
SQRT[ 134.80 x ( 1 + 1 )] = 5.0346
9
13
Confidence Interval = (m1 – m2 ) ± (Critical-T x Std Error)
Confidence Interval = (m1 – m2 ) ± (Margin of Error)
Lower = (102 - 133) - (2.086 x 5.0346) = -41.5019
Upper = (102 - 133) + (2.086 x 5.0546) = -20.4980
s 2p
(n1  1) s12  (n2  1) s22
=
= 17.52
n1  n2  2
t=
X1  X 2
1
1 
s 2p   
 n1 n2 
1 1
Std .Error = s 2p    = 1.6466
 n1 n2 
= (77.5 – 84.3) ÷ 1.6466 = -4.1296
Critical-T (0.01,24) = 2.492
It resides in the left tail
Thus, Critical-T = -2.492
μ1 ≥ μ2
μ1 < μ2
Population Mean 1 Greater than Population Mean 2
Population Mean 1 Less than Population Mean 2
Reject if t < -t,d.f. or p-value < 
Reject if |t| > |2.492| or t < -2.492
Computed-t is -4.1296 and it is < -2.492, reject the null hypothesis
p-value is 0.00019 and it is less than 0.01, reject the null hypothesis
Reject the Null Hypothesis. There is enough evidence that μ1 is lower than μ2