155S9.5 - Cape Fear Community College

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MAT 155 Statistical Analysis
Dr. Claude Moore
Cape Fear Community College
Chapter 9
Inferences from Two Samples
9-1
9-2
9-3
9-4
9-5
Review and Preview
Inferences About Two Proportions
Inferences About Two Means: Independent Samples
Inferences from Dependent Samples
Comparing Variation in Two Samples
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Key Concept
This section presents the F test for comparing two population
variances (or standard deviations). We introduce the F distribution
that is used for the F test.
Note that the F test is very sensitive to departures from normal
distributions.
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Notation for Hypothesis Tests with Two Variances or
Standard Deviations
= larger of two sample variances
= size of the sample with the larger variance
= variance of the population from which the sample with the
larger variance is drawn
are used for the other sample and
population
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Requirements
1. The two populations are independent.
2. The two samples are simple random
3.
The two populations are each
distributed.
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samples.
normally
Test Statistic for Hypothesis
Tests with Two Variances
Where s12 is the larger of the two
sample variances
Critical Values: Using Table A-5, we obtain critical F values that
are determined by the following three values:
1. The significance level α
2. Numerator degrees of freedom = n1 – 1
3. Denominator degrees of freedom = n2 – 1
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Properties of the F Distribution
· The F distribution is not symmetric.
·Values of the F distribution cannot be negative.
·The exact shape of the F distribution depends on the
two different degrees of freedom.
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Finding Critical F Values
To find a critical F value corresponding to a 0.05 significance
level, refer to Table A-5 and use the right-tail area of 0.025 or 0.05,
depending on the type of test:
Two-tailed test: use 0.025 in right tail
One-tailed test: use 0.05 in right tail
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Finding Critical F Values
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Properties of the F Distribution - cont.
If the two populations do have equal
variances, then F =
because
and
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will be close to 1
are close in value.
Properties of the F Distribution - cont.
If the two populations have radically different
variances, then F will be a large number.
Remember, the larger sample variance will be s1 .
2
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Conclusions from the F Distribution
Consequently, a value of F near 1 will be
evidence in favor of the conclusion that s1 = s2 .
2
2
But a large value of F will be evidence against the
conclusion of equality of the population variances.
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Example:
Data Set 20 in Appendix B includes weights (in g) of quarters made before
1964 and weights of quarters made after 1964. Sample statistics are listed
below. When designing coin vending machines, we must consider the standard
deviations of pre-1964 quarters and post-1964 quarters. Use a 0.05
significance level to test the claim that the weights of pre-1964 quarters and the
weights of post-1964 quarters are from populations with the same standard
deviation.
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Example:
Requirements are satisfied: populations are independent; simple random
samples; from populations with normal distributions
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Example:
Use sample variances to test claim of equal population variances, still
state conclusion in terms of standard deviations.
Step 1: claim of equal standard deviations is equivalent to claim of
equal variances
Step 2: if the original claim is false, then
Step 3:
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original claim
Example:
Step 4: significance level is 0.05
Step 5: involves two population variances, use F distribution variances
Step 6: calculate the test statistic
For the critical values in this two-tailed test, refer to Table A-5 for the area
of 0.025 in the right tail. Because we stipulate that the larger variance is
placed in the numerator of the F test statistic, we need to find only the
right-tailed critical value.
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Example:
From Table A-5 we see that the critical value of F is between 1.8752 and
2.0739, but it is much closer to 1.8752. Interpolation provides a critical value of
1.8951, but STATDISK, Excel, and Minitab provide the accurate critical value of
1.8907.
Step 7: The test statistic F = 1.9729 does fall within the critical region,
so we reject the null hypothesis of equal variances. There is sufficient
evidence to warrant rejection of the claim of equal standard deviations.
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Example:
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Example:
There is sufficient evidence to warrant rejection of the claim that the
two standard deviations are equal. The variation among weights of
quarters made after 1964 is significantly different from the
variation among weights of quarters made before 1964.
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Recap
In this section we have discussed:
· Requirements for comparing variation in
· Notation.
· Hypothesis test.
· Confidence intervals.
· F test and distribution.
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two samples
Hypothesis Test of Equal Variances. In Exercises 5 and 6, test the given claim. Use a significance level of α
= 0.05 and assume that all populations are normally distributed.
507/5. Zinc Treatment Claim: Weights of babies born to mothers given placebos vary more than weights of
babies born to mothers given zinc supplements (based on data from “The Effect of Zinc Supplementation on
Pregnancy Outcome,” by Goldenberg, et al., Journal of the American Medical Association, Vol. 274, No. 6).
Sample results are summarized below.
Placebo group:
n = 16,
x = 3088 g,
s = 728 g
Treatment group:
n = 16,
x = 3214 g,
s = 669 g
507/5. TI results
Hypothesis Test of Equal Variances. In Exercises 5 and 6, test the given claim. Use a significance level of α
= 0.05 and assume that all populations are normally distributed.
507/6. Weights of Pennies Claim: Weights of pre-1983 pennies and weights of post-1983 pennies have the
same amount of variation. (The results are based on Data Set 20 in Appendix B.)
Weights of pre-1983 pennies: n = 35, x = 3.07478 g, s = 0.03910 g
Weights of post-1983 pennies: n = 37, x = 2.49910 g, s = 0.01648 g
507/6. TI results
Hypothesis Tests of Claims About Variation. In Exercises 9–18, test the given claim. Assume that both
samples are independent simple random samples from populations having normal distributions.
508/10. Braking Distances of Cars A random sample of 13 four-cylinder cars is obtained, and the braking
distances are measured and found to have a mean of 137.5 ft and a standard deviation of 5.8 ft. A random
sample of 12 six-cylinder cars is obtained and the braking distances have a mean of 136.3 ft and a standard
deviation of 9.7 ft (based on Data Set 16 in Appendix B). Use a 0.05 significance level to test the claim that
braking distances of four- cylinder cars and braking distances of six-cylinder cars have the same standard
deviation.
508/10. TI results
Hypothesis Tests of Claims About Variation. In Exercises 9–18, test the given claim. Assume that both
samples are independent simple random samples from populations having normal distributions.
508/12. Home Size and Selling Price Using the sample data from Data Set 23 in Appendix B, 21 homes with
living areas under 2000 ft2 have selling prices with a standard deviation of $ 32,159.73. There are 19 homes with
living areas greater than 2000 ft2 and they have selling prices with a standard deviation of $ 66,628.50. Use a
0.05 significance level to test the claim of a real estate agent that homes larger than 2000 ft 2 have selling prices
that vary more than the smaller homes.
508/12. TI results
Correct F-value because of larger numerator; same P-value.
Incorrect F-value because of smaller numerator; same P-value.
Hypothesis Tests of Claims About Variation. In Exercises 9–18, test the given claim. Assume that both
samples are independent simple random samples from populations having normal distributions.
508/15. Radiation in Baby Teeth Listed below are amounts of strontium-90 (in millibec-querels or mBq per gram
of calcium) in a simple random sample of baby teeth obtained from Pennsylvania residents and New York residents
born after 1979 (based on data from “ An Un-expected Rise in Strontium-90 in U. S. Deciduous Teeth in the 1990s,”
by Mangano, et al., Science of the Total Environment). Use a 0.05 significance level to test the claim that amounts
of Strontium-90 from Pennsylvania residents vary more than amounts from New York residents.
Pennsylvania: 155 142 149 130 151 163 151 142 156 133 138 161
New York:
133 140 142 131 134 129 128 140 140 140 137 143
508/15. TI results
Hypothesis Tests of Claims About Variation. In Exercises 9–18, test the given claim. Assume that both
samples are independent simple random samples from populations having normal distributions.
508/16. BMI for Miss America Listed below are body mass indexes (BMI) for Miss America winners from two
different time periods. Use a 0.05 significance level to test the claim that winners from both time periods have BMI
values with the same amount of variation.
BMI (from recent winners):
19.5 20.3 19.6 20.2 17.8 17.9 19.1 18.8 17.6 16.8
BMI (from the 1920s and 1930s):
20.4 21.9 22.1 22.3 20.3 18.8 18.9 19.4 18.4 19.1
508/16. TI results

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