1. Data Mining 2018/2019 Fall MIS 331 Chapter 7-A Sampliing Distribution,
Confidence Interval Estimation and
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

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. Defined Chi-square distribution defined as: 2v = vi=1Zi2
sum of squares of v standard normal distributions
Zi = N(0,1)

9. Expected value of a chi-square distribution with v degrees of freedom is v
E[2v] = v
Variance of a chi-square distribution with v degrees of freedom is 2v
Var[2v] = 2v

10. 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, unbiesd estimation of popultion variance
Similarly
Var[(n-1)s2/2] = 2(n-1)
((n-1)2/4)Var[s2] = 2(n-1)
Var[s2] = 24/(n-1)

11. Examples of Squares of Distributions
a: side of a square plate – distributed normally with a mean and std.
area of the plate: A = a2,
square of a normal distribution

12. Discrete Distribution Example
X has values -2,-1,0,1,2 with equal probabilities of 1/5
a discrete uniform distribution
what is pdf of X2?
X2 can take values 0,1,4
p(0) = 1/5, p(1) = 2/5, p(2) = 2/5
X2 not symetric and skewed

13. E(xi-)2 = 2 definition of variance
E[ni=1(xi-)2] = n2 expected value of n independent identical distributed (iid) random variables or
E[ni=1(xi-)2]/n = 2 unbiesd estimation of population variance
when population mean is known

14. if  is known:
(xi-)2/2 = z2i =. 21 by definition of the chi-square distribution
zi = xi-) /,
ni=1(xi-)2 /2 ] = (1/2)ni=1(xi-)2]  2n by definition of chi-square as each of these terms in the sumation are standard normal squares
E[2n] = n

15. if  is not known – estimate by xbar,
E[ni=1(xi-xbar)2 ] = (n-1)2
shown in Appandix of Chapter 6 of Newbold 8
independnet of distribution of Xi.
sum of n quantities on the left makes only n-1 2.whan mean of the distribution is etimated by xbar

16. Exercise Show with n = 2 E[2i=1(xi-xbar)2 ] = 2
where xbar = (x1+x2)/2

17. if  is not known – estimate by xbar,
for a normally distributred Xi,
ni=1(xi-xbar)2 /2 = 2n-1 without proof
taking expected values of both sides
E[ni=1(xi-xbar)2 /2] = E[2n-1] = (n-1)
E[ni=1(xi-xbar)2 ] = (n-1)2
dividing by n-1.
E[ni=1(xi-xbar)2 /(n-1) ] = 2 unbiesd

18. ni=1(xi-xbar)2 /2 = 2n-1
dividing by n-1 and multiplying by 2
ni=1(xi-xbar)2 /(n-1) =  22n-1 /(n-1)
s2 = 22n-1 /(n-1) or
2n-1 = (n-1)s2 /2
n-1 times sample variance over population variance is distributed as chi-square with n-1 degree of freedom

19. 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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20. 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

21. 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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22. 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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23. 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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24. 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. 7-24

25. 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. 7-25

26. 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. 7-26

27. 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-

28. Example
Find two numbers such that probability that chi-square with d.f. 6 is laying between tham is 0.90
1-  = 0.90
P(26,0.95 < 26 < 26,0.05) =0.90
The two numbers
26,0.95 = 1.635
26,0.05 = hence
P( < 26 < ) =0.90

29. Confidence Intervals for the Population Variance
(continued)
The 100(1 - )% confidence interval for the population variance is given by
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Ch. 7-29

30. Derivation P(2n-1,1-/2 < 2n-1 < 2n-1,/2) = 1 - 
2n-1 = (n-1)s2/2.
substituting for 2n-1,
P(2n-1,1-/2 < (n-1)s2/2 < 2n-1,/2) = 1 - 
rearranging:
P(2n-1,1-/2/(n-1)s2 < 1/2 < 2n-1,/2 /(n-1)s2)=1-
P((n-1)s2/2n-,/2 < 2 < (n-1)s2/2n-1,1-/2) = 1-

31. 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. 7-31

32. 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. 7-32

33. 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. 7-33

34. 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 11-34

35. 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 11-35

36. 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 11-36

37. 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

38. Solution

39. Solution

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

41. Solution