1. Data Mining 2016/2017 Fall MIS 331 Chapter 2 Sampliing Distribution
Confidence Interval Estimation
Hypothesis Testing
for Variance of a Population
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2. Outline Sampling Distributio of Sample Variances
Confidence Interval Estimation for the Variance
Tests of the Variance of a Normal Distribution
Tests of Equality of Two Variances

3. Sampling Distributions of Sample Variances
6.4
Sampling Distributions
Sampling Distributions of Sample
Means
Sampling Distributions of Sample Proportions
Sampling Distributions of Sample Variances
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4. Copyright © 2013 Pearson Education, Inc. Publishing as Prentice Hall
Sample Variance
Let x1, x2, , xn be a random sample from a population. The sample variance is
the square root of the sample variance is called the sample standard deviation
the sample variance is different for different random samples from the same population
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5. Sampling Distribution of Sample Variances
The sampling distribution of s2 has mean σ2
If the population distribution is normal, then
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6. Chi-Square Distribution of Sample and Population Variances
If the population distribution is normal then
has a chi-square (2 ) distribution
with n – 1 degrees of freedom
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7. The Chi-square Distribution
The chi-square distribution is a family of distributions, depending on degrees of freedom:
d.f. = n – 1
Text Appendix Table 7 contains chi-square probabilities 2 2 2
d.f. = 1
d.f. = 5
d.f. = 15
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8. Expected value of a chi-square distribution with degree of freedom v is v
E[2v] = v
Variance of achi-square distribution with degree of freedom v is 2v
Var[2v] = 2v

9. Since (n-1)s2/2 has a chi-square distribution with df: n-1
E[(n-1)s2/2] = n-1
((n-1)/2)E[s2] = n-1
E[s2] = 2,
Similarly
Var[(n-1)s2/2] = 2(n-1)
((n-1)2/4)Var[s2] = 2(n-1)
Var[s2] = 24/(n-1)

10. Degrees of Freedom (df)
Idea: Number of observations that are free to vary
after sample mean has been calculated
Example: Suppose the mean of 3 numbers is 8.0
Let X1 = 7
Let X2 = 8
What is X3?
If the mean of these three values is 8.0,
then X3 must be 9
(i.e., X3 is not free to vary)
Here, n = 3, so degrees of freedom = n – 1 = 3 – 1 = 2
(2 values can be any numbers, but the third is not free to vary for a given mean)
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11. Table 7 in Appandix
d.f. versus probabilities for critical values
P(210 < KL) = 0.05
KL = hence
P(210 < 3.940) = 0.05
P(210 > KU) = 0.05
KU = hence
P(210 > 18.31) = 0.05

12. Copyright © 2013 Pearson Education, Inc. Publishing as Prentice Hall
Chi-square Example
A commercial freezer must hold a selected temperature with little variation. Specifications call for a standard deviation of no more than 4 degrees (a variance of 16 degrees2).
A sample of 14 freezers is to be tested
What is the upper limit (K) for the sample variance such that the probability of exceeding this limit, given that the population standard deviation is 4, is less than 0.05?
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13. Finding the Chi-square Value
Is chi-square distributed with (n – 1) = 13 degrees of freedom
Use the the chi-square distribution with area 0.05 in the upper tail:
213
= (α = .05 and 14 – 1 = 13 d.f.)
probability α = .05 2
213
= 22.36
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14. Copyright © 2013 Pearson Education, Inc. Publishing as Prentice Hall
Chi-square Example
(continued)
213
= (α = .05 and 14 – 1 = 13 d.f.)
So: or
(where n = 14) so
If s2 from the sample of size n = 14 is greater than 27.52, there is strong evidence to suggest the population variance exceeds 16.
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15. Confidence Interval Estimation for the Variance
7.5
Confidence
Intervals
Population
Mean
Population
Proportion
Population
Variance
(From a normally distributed population)
σ2 Known
σ2 Unknown
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Ch. 7-15

16. Confidence Intervals for the Population Variance
Goal: Form a confidence interval for the population variance, σ2
The confidence interval is based on the sample variance, s2
Assumed: the population is normally distributed
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Ch. 7-16

17. Confidence Intervals for the Population Variance
(continued)
The random variable
follows a chi-square distribution with (n – 1) degrees of freedom
Where the chi-square value denotes the number for which
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Ch. 7-17

18. P(2n-1 > 2n-1,/2 ) = /2
P(2n-1 > 2n-1,1-/2 ) = 1 - /2 or
P(2n-1 < 2n-1,1-/2 ) = /2
Finally,
P(2n-1,1-/2 < 2n-1 < 2n-1,/2) = 1 - /2 - /2 =1- 

19. two numbers such that probability that chi-square with d. f
two numbers such that probability that chi-square with d.f. 6 is laying between tham is 0.90
P(26, < 26 < 26,0.05) =0.90
The two numbers
26, = 1.635
26,0.05 = hence
P( < 26 < ) =0.90

20. Confidence Intervals for the Population Variance
(continued)
The 100(1 - )% confidence interval for the population variance is given by
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Ch. 7-20

21. Copyright © 2013 Pearson Education, Inc. Publishing as Prentice Hall
Example
You are testing the speed of a batch of computer processors. You collect the following data (in Mhz): Sample size 17 Sample mean 3004 Sample std dev 74
Assume the population is normal.
Determine the 95% confidence interval for σx2
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Ch. 7-21

22. Finding the Chi-square Values
n = 17 so the chi-square distribution has (n – 1) = 16 degrees of freedom
 = 0.05, so use the the chi-square values with area in each tail:
probability α/2 = .025
probability α/2 = .025
216
216
= 6.91
216
= 28.85
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Ch. 7-22

23. Calculating the Confidence Limits
The 95% confidence interval is
Converting to standard deviation, we are 95% confident that the population standard deviation of CPU speed is between 55.1 and Mhz
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Ch. 7-23

24. Tests of the Variance of a Normal Distribution
9.6
Goal: Test hypotheses about the population variance, σ2 (e.g., H0: σ2 = σ02)
If the population is normally distributed,
has a chi-square distribution with (n – 1) degrees of freedom
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Chap 11-24

25. Tests of the Variance of a Normal Distribution
(continued)
The test statistic for hypothesis tests about one population variance is
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Chap 11-25

26. Decision Rules: Variance
Population variance
Lower-tail test:
H0: σ2  σ02
H1: σ2 < σ02
Upper-tail test:
H0: σ2 ≤ σ02
H1: σ2 > σ02
Two-tail test:
H0: σ2 = σ02
H1: σ2 ≠ σ02 a a
a/2
a/2
Reject H0 if
Reject H0 if
Reject H0 if or
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Chap 11-26

27. Newbold 9.47 Test the hypothesis H0:2 <=100 againts H1 2 >100
a) s2 = 165, n=25
b) s2 = 165, n=29
c) s2 = 159, n=25
d) s2 = 67, n=38

28. Solution

29. Solution

30. Newbold 7.48 new safety device random sample for 8 days
management concenrs about variability
test the null hypothesis variance less than 500
at a significance level of 10%

31. Solution

32. Chapter 10 Hypothesis Testing: Additional Topics
Statistics for
Business and Economics 8th Edition
Chapter 10
Hypothesis Testing:
Additional Topics
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33. Copyright © 2013 Pearson Education, Inc. Publishing as Prentice Hall
Chapter Goals
After completing this chapter, you should be able to:
Test hypotheses for the difference between two population means
Two means, matched pairs
Independent populations, population variances known
Independent populations, population variances unknown but equal
Complete a hypothesis test for the difference between two proportions (large samples)
Use the F table to find critical F values
Complete an F test for the equality of two variances
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34. Two Sample Tests Two Sample Tests Population Means, Dependent Samples
Population Means, Independent Samples
Population Proportions
Population Variances
Examples:
Same group before vs. after treatment
Group 1 vs. independent Group 2
Proportion 1 vs. Proportion 2
Variance 1 vs.
Variance 2
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Dependent Samples
10.1
Tests of the Difference Between Two Normal Population Means: Dependent Samples
Dependent Samples
Tests Means of 2 Related Populations
Paired or matched samples
Repeated measures (before/after)
Use difference between paired values:
Assumptions:
Both Populations Are Normally Distributed
di = xi - yi
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36. Test Statistic: Dependent Samples
The test statistic for the mean difference is a t value, with n – 1 degrees of freedom: where
Population Means, Dependent Samples
For tests of the following form:
H0: μx – μy  0
H0: μx – μy ≤ 0
H0: μx – μy = 0
sd = sample standard dev. of differences
n = the sample size (number of pairs)
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37. Decision Rules: Matched Pairs
Matched or Paired Samples
Lower-tail test:
H0: μx – μy  0
H1: μx – μy < 0
Upper-tail test:
H0: μx – μy ≤ 0
H1: μx – μy > 0
Two-tail test:
H0: μx – μy = 0
H1: μx – μy ≠ 0 a a
a/2
a/2
-ta ta
-ta/2
ta/2
Reject H0 if t < -tn-1, a
Reject H0 if t > tn-1, a
Reject H0 if t < -tn-1 , a/2
or t > tn-1 , a/2
has n - 1 d.f.
Where
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38. Copyright © 2013 Pearson Education, Inc. Publishing as Prentice Hall
Matched Pairs Example
Assume you send your salespeople to a “customer service” training workshop. Has the training made a difference in the number of complaints? You collect the following data:  di
d =
Number of Complaints: (2) - (1)
Salesperson Before (1) After (2) Difference, di
C.B
T.F
M.H
R.K
M.O
-21 n
= - 4.2
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39. Critical Value = ± 2.776 d.f. = n − 1 = 4
Matched Pairs: Solution
Has the training made a difference in the number of complaints (at the  = 0.05 level)?
Reject
Reject
H0: μx – μy = 0
H1: μx – μy  0
/2
/2
 = .05
d =
- 4.2
- 1.66
Critical Value = ± d.f. = n − 1 = 4
Decision: Do not reject H0
(t stat is not in the reject region)
Test Statistic:
Conclusion: There is not a significant change in the number of complaints.
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40. Independent Samples
10.2
Tests of the Difference Between Two Normal Population Means: Dependent Samples
Population means, independent samples
Goal: Form a confidence interval for the difference between two population means, μx – μy
Different populations
Unrelated
Independent
Sample selected from one population has no effect on the sample selected from the other population
Normally distributed
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41. Difference Between Two Means
(continued)
Population means, independent samples
σx2 and σy2 known
Test statistic is a z value
σx2 and σy2 unknown
σx2 and σy2 assumed equal
Test statistic is a a value from the Student’s t distribution
σx2 and σy2 assumed unequal
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42. * σx2 and σy2 Known Assumptions: Population means, independent samples
Samples are randomly and
independently drawn
both population distributions
are normal
Population variances are
known *
σx2 and σy2 known
σx2 and σy2 unknown
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43. σx2 and σy2 Known
(continued)
When σx2 and σy2 are known and both populations are normal, the variance of X – Y is
Population means, independent samples *
σx2 and σy2 known
…and the random variable
has a standard normal distribution
σx2 and σy2 unknown
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44. Test Statistic, σx2 and σy2 Known
Population means, independent samples
The test statistic for
μx – μy is: *
σx2 and σy2 known
σx2 and σy2 unknown
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45. Hypothesis Tests for Two Population Means
Two Population Means, Independent Samples
Lower-tail test:
H0: μx  μy
H1: μx < μy
i.e.,
H0: μx – μy  0
H1: μx – μy < 0
Upper-tail test:
H0: μx ≤ μy
H1: μx > μy
i.e.,
H0: μx – μy ≤ 0
H1: μx – μy > 0
Two-tail test:
H0: μx = μy
H1: μx ≠ μy
i.e.,
H0: μx – μy = 0
H1: μx – μy ≠ 0
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46. Decision Rules a a a/2 a/2 -za za -za/2 za/2
Two Population Means, Independent Samples, Variances Known
Lower-tail test:
H0: μx – μy  0
H1: μx – μy < 0
Upper-tail test:
H0: μx – μy ≤ 0
H1: μx – μy > 0
Two-tail test:
H0: μx – μy = 0
H1: μx – μy ≠ 0 a a
a/2
a/2
-za za
-za/2
za/2
Reject H0 if z < -za
Reject H0 if z > za
Reject H0 if z < -za/2
or z > za/2
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47. Newbold 10.8
A screening procedure - measure attitudes toward minorities
high scores indicate negative attitudes
low scores indicate positve attitudes
Independent random samples
151 male 108 female financial analysits
for males
sample mean. 85.8, std dev: 19.13
for females
sample mean. 71.5, std dev: 12.2

48. Newbold 10.8 Test the null hypothesis that
the two population meand are equal
against the alternative that
the true mean score is higher for male then for female financial analysts

49. Solution

50. σx2 and σy2 Unknown, Assumed Equal
Assumptions:
Samples are randomly and
independently drawn
Populations are normally
distributed
Population variances are
unknown but assumed equal
Population means, independent samples
σx2 and σy2 known
σx2 and σy2 unknown *
σx2 and σy2 assumed equal
σx2 and σy2 assumed unequal
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51. σx2 and σy2 Unknown, Assumed Equal
(continued)
The population variances
are assumed equal, so use
the two sample standard
deviations and pool them to
estimate σ
use a t value with
(nx + ny – 2) degrees of
freedom
Population means, independent samples
σx2 and σy2 known
σx2 and σy2 unknown *
σx2 and σy2 assumed equal
σx2 and σy2 assumed unequal
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52. Test Statistic, σx2 and σy2 Unknown, Equal
The test statistic for
H0 :μx – μy = 0 is:
σx2 and σy2 unknown *
σx2 and σy2 assumed equal
σx2 and σy2 assumed unequal
Where t has (n1 + n2 – 2) d.f.,
and
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53. Decision Rules a a a/2 a/2 -ta ta -ta/2 ta/2
Two Population Means, Independent Samples, Variances Unknown
Lower-tail test:
H0: μx – μy  0
H1: μx – μy < 0
Upper-tail test:
H0: μx – μy ≤ 0
H1: μx – μy > 0
Two-tail test:
H0: μx – μy = 0
H1: μx – μy ≠ 0 a a
a/2
a/2
-ta ta
-ta/2
ta/2
Reject H0 if
t < -t (n1+n2 – 2), a
Reject H0 if
t > t (n1+n2 – 2), a
Reject H0 if
t < -t (n1+n2 – 2), a/2 or
t > t (n1+n2 – 2), a/2
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54. Pooled Variance t Test: Example
You are a financial analyst for a brokerage firm. Is there a difference in dividend yield between stocks listed on the NYSE & NASDAQ? You collect the following data:
NYSE NASDAQ Number
Sample mean
Sample std dev
Assuming both populations are
approximately normal with equal variances, is there a difference in average yield ( = 0.05)?
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55. Calculating the Test Statistic
H0: μ1 - μ2 = 0 i.e. (μ1 = μ2)
H1: μ1 - μ2 ≠ 0 i.e. (μ1 ≠ μ2)
The test statistic is:
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56. Copyright © 2013 Pearson Education, Inc. Publishing as Prentice Hall
Solution
Reject H0
Reject H0
H0: μ1 - μ2 = 0 i.e. (μ1 = μ2)
H1: μ1 - μ2 ≠ 0 i.e. (μ1 ≠ μ2)
 = 0.05
df = − 2 = 44
Critical Values: t = ±
Test Statistic:
.025
.025
2.0154 t
2.040
Decision:
Conclusion:
Reject H0 at a = 0.05
There is evidence of a difference in means.
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57. σx2 and σy2 Unknown, Assumed Unequal
Assumptions:
Samples are randomly and
independently drawn
Populations are normally
distributed
Population variances are
unknown and assumed
unequal
Population means, independent samples
σx2 and σy2 known
σx2 and σy2 unknown
σx2 and σy2 assumed equal *
σx2 and σy2 assumed unequal
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58. σx2 and σy2 Unknown, Assumed Unequal
(continued)
Forming interval estimates:
The population variances are
assumed unequal, so a pooled
variance is not appropriate
use a t value with  degrees
of freedom, where
Population means, independent samples
σx2 and σy2 known
σx2 and σy2 unknown
σx2 and σy2 assumed equal *
σx2 and σy2 assumed unequal
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59. Test Statistic, σx2 and σy2 Unknown, Unequal
The test statistic for
H0: μx – μy = 0 is:
σx2 and σy2 unknown
σx2 and σy2 assumed equal *
σx2 and σy2 assumed unequal
Where t has  degrees of freedom:
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60. Two Population Proportions
10.3
Tests of the Difference Between Two Population Proportions (Large Samples)
Population proportions
Goal: Test hypotheses for the difference between two population proportions, Px – Py
Assumptions:
Both sample sizes are large,
nP(1 – P) > 5
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61. Two Population Proportions
(continued)
The random variable
has a standard normal distribution
Population proportions
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62. Test Statistic for Two Population Proportions
The test statistic for
H0: Px – Py = 0
is a z value:
Population proportions
Where
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63. Decision Rules: Proportions
Population proportions
Lower-tail test:
H0: Px – Py  0
H1: Px – Py < 0
Upper-tail test:
H0: Px – Py ≤ 0
H1: Px – Py > 0
Two-tail test:
H0: Px – Py = 0
H1: Px – Py ≠ 0 a a
a/2
a/2
-za za
-za/2
za/2
Reject H0 if z < -za
Reject H0 if z > za
Reject H0 if z < -za/2
or z > za/2
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64. Example: Two Population Proportions
Is there a significant difference between the proportion of men and the proportion of women who will vote Yes on Proposition A?
In a random sample, 36 of 72 men and 31 of 50 women indicated they would vote Yes
Test at the .05 level of significance
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65. Example: Two Population Proportions
(continued)
The hypothesis test is:
H0: PM – PW = 0 (the two proportions are equal)
H1: PM – PW ≠ 0 (there is a significant difference between
proportions)
The sample proportions are:
Men: = 36/72 = .50
Women: = 31/50 = .62
The estimate for the common overall proportion is:
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66. Example: Two Population Proportions
(continued)
Reject H0
Reject H0
The test statistic for PM – PW = 0 is:
.025
.025
-1.96
1.96
-1.31
Decision: Do not reject H0
Conclusion: There is not significant evidence of a difference between men and women in proportions who will vote yes.
Critical Values = ±1.96
For  = .05
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67. Tests of Equality of Two Variances
10.4
Tests of Equality of Two Variances
Tests for Two
Population
Variances
Goal: Test hypotheses about two
population variances
H0: σx2  σy2
H1: σx2 < σy2
Lower-tail test
F test statistic
H0: σx2 ≤ σy2
H1: σx2 > σy2
Upper-tail test
H0: σx2 = σy2
H1: σx2 ≠ σy2
Two-tail test
The two populations are assumed to be independent and normally distributed
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68. Hypothesis Tests for Two Variances
(continued)
The random variable
Tests for Two
Population
Variances
F test statistic
Has an F distribution with (nx – 1) numerator degrees of freedom and (ny – 1) denominator degrees of freedom
Denote an F value with 1 numerator and 2 denominator degrees of freedom by
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69. Copyright © 2013 Pearson Education, Inc. Publishing as Prentice Hall
Test Statistic
Tests for Two
Population
Variances
The critical value for a hypothesis test about two population variances is
F test statistic
where F has (nx – 1) numerator degrees of freedom and (ny – 1) denominator degrees of freedom
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70. Decision Rules: Two Variances
Use sx2 to denote the larger variance.
H0: σx2 = σy2
H1: σx2 ≠ σy2
H0: σx2 ≤ σy2
H1: σx2 > σy2
/2  F F
Do not
reject H0
Reject H0
Do not
reject H0
Reject H0
rejection region for a two-tail test is:
where sx2 is the larger of the two sample variances
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71. Copyright © 2013 Pearson Education, Inc. Publishing as Prentice Hall
Example: F Test
You are a financial analyst for a brokerage firm. You want to compare dividend yields between stocks listed on the NYSE & NASDAQ. You collect the following data: NYSE NASDAQ Number Mean Std dev Is there a difference in the variances between the NYSE & NASDAQ at the  = 0.10 level?
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72. F Test: Example Solution
Form the hypothesis test:
H0: σx2 = σy2 (there is no difference between variances)
H1: σx2 ≠ σy2 (there is a difference between variances)
Find the F critical values for  = .10/2:
Degrees of Freedom:
Numerator
(NYSE has the larger standard deviation):
nx – 1 = 21 – 1 = 20 d.f.
Denominator:
ny – 1 = 25 – 1 = 24 d.f.
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73. F Test: Example Solution
(continued)
The test statistic is:
H0: σx2 = σy2
H1: σx2 ≠ σy2
/2 = .05 F
Do not
reject H0
Reject H0
F = is not in the rejection region, so we do not reject H0
Conclusion: There is not sufficient evidence of a difference in variances at  = .10
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