Transcript 155S9.5 - Cape Fear Community College
MAT 155 Statistical Analysis D
r. Claude Moore Cape Fear Community College
Chapter 9 Inferences from Two Samples
9-1 Review and Preview 9-2 Inferences About Two Proportions 9-3 Inferences About Two Means: Independent Samples 9-4 Inferences from Dependent Samples 9 -5 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
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are used for the other sample and population
Requirements 1. The two populations are independent .
2. The two samples are simple random 3 .
The two populations are each distributed.
samples.
normally
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Test Statistic for Hypothesis Tests with Two Variances
Where
s
1 2 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 = n 1 – 1 3. Denominator degrees of freedom = n 2 – 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 = will be close to 1 because and are close in value.
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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 s
1
.
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 s
1
= s
2
.
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. Copyright © 2010, 2007, 2004 Pearson Education, Inc.
Example:
Requirements are satisfied: populations are independent; simple random samples; from populations with normal distributions Copyright © 2010, 2007, 2004 Pearson Education, Inc.
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: Copyright © 2010, 2007, 2004 Pearson Education, Inc. 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. Copyright © 2010, 2007, 2004 Pearson Education, Inc.
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.
two samples
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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: Treatment group: n = 16, n = 16, x = 3088 g, x = 3214 g, s = 669 g s = 728 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 ft 2 have selling prices with a standard deviation of $ 32,159.73. There are 19 homes with living areas greater than 2000 ft 2 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): BMI (from the 1920s and 1930s): 19.5 20.3 19.6 20.2 17.8 17.9 19.1 18.8 17.6 16.8 20.4 21.9 22.1 22.3 20.3 18.8 18.9 19.4 18.4 19.1
508/16. TI results