Post on 09-Feb-2017
Kingdom of Saudi Arabia The Royal Commission at Yanbu Yanbu University College
11 Hypothesis Tests and Estimation for Population VariancesSTAT 311 - Academic Year (1437/1438 H) (2016/2017 G) Semester – I (161)
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Mr. Osama A. Alwusaidi | Yanbu University College | GS Dept. | Saudi Arabia | instructor(B.Ed.), Mathematics. (Ed.M.), Testing, Measurement and Statistics (Psychology)
Mobile: +966-544115001 | Phone: +966-43932961 | Cisco:1574 | Fax: +966-43925394Email: alwusaidio@rcyci.edu.sa | Site: www.rcyci.edu.sa | Address: P.O.Box 31387
Yanbu Industrial City 41912 Saudi Arabia
Hypothesis test and
Estimation for σ
11.2
Test for a Single Population Varianceχ 211.1.1 Hypothesis for a Single σ
Test for a Single Population Varianceχ 211.1.2 Confidence Interval Estimate for a σ
11.1
Test for a Single Population VarianceF11.2 Hypothesis Tests for σ1 , σ2 (continued)
SampelPopulation
Η0 :σ = sΗA :σ ≠ s
n−1( )s2χu
2 ≤σ 2 ≤ n−1( )s2χu2
χ 2 =n−1( ) s2
σ 2
Η0 :σ 1 =σ 2
ΗA :σ 1 ≠σ 2
Population1 Population2
F =s12
s22
Chapter 11 Exercises: Hypothesis Tests and Estimation for Population Variances
! Solution Q2 :
11.1.1 Hypothesis for a Single Population Variance
Q1 : 11-1 Exercises (11-4/p-480) :
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α/2α/2
Confidence Interval
χU2 = χα 2
2 = χ0.052 = 38.8851
χ L2 = χ1−α /2
2 = χ0.052 =15.3792
χU2 = χα 2
2 = χ0.0252 = 28.8453
χ L2 = χ1−α /2
2 = χ0.0252 = 6.9077
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6
Chapter 11 Exercises: Hypothesis Tests and Estimation for Population Variances
! Solution Q2 :
No statistical method exists for developing a confidence interval estimate for a population standard deviation directly
Instead we must first convert to variances. This, we get a sample variance equal to .
�
- Thus, at the 95% confidence level, we conclude that the population variance will fall in the range 74,953.7 to 276,472.2.
- By taking the square root, you can convert to an interval estimate of the population standard deviation as the interval 273.78 to 525.81.
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11.1.2 Confidence Interval Estimate for a Population Variance
Q2 : 11-1 Exercises (11-1/p-480) :
s2 = 3602 =129,600
n = 20 , s = 360 , α = 0.05 ⇒ α / 2( ) = 0.025 , df = 20−1= 19
⇒ 1−α / 2( ) = 0.975(n−1)s2
χU2 ≤σ 2 ≤
(n−1)s2
χ L2
(20−1)129,60032.8523
≤σ 2 ≤(20−1)129,600
8.9065
74,953.7 ≤σ 2 ≤ 276,472.2
273.78 ≤σ ≤ 525.81
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χU2 = χα 2
2
= χ0.0252 = 32.8523
χ L2 = χ1−α /2
2
= χ0.9752 = 8.9065
Chapter 11 Exercises: Hypothesis Tests and Estimation for Population Variances
Given the following null and alternative hypotheses
and the following sample information :
a. If α = 0.05, state the decision rule for the hypothesis. b. Test the hypothesis and indicate whether the null hypothesis should be rejected.
Solution Q3 :
!
Using Appendix H: If the calculated F > 2.278, reject H0, otherwise do not reject H0
Since 1.0985 < 2.278 do not reject H0
11.2 Hypothesis Tests for Two Population Variances
Q3 : 11-2 Exercises (11-19/p-491) :
Η0 :σ 12 ≤σ 2
2
ΗA :σ 12 >σ 2
2
Assume : F 0.05,12,20( ) = 2.278( )
F
α
Reject H0Do not reject H0
Fα
0
Fcal = 14501320
= 1.0985
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Chapter 11 Exercises: Hypothesis Tests and Estimation for Population Variances
Given the following null and alternative hypotheses
and the following sample information :
a. If α = 0.05, state the decision rule for the hypothesis. b. Test the hypothesis and indicate whether the null hypothesis should be rejected.
Solution Q4 :
!
Using Appendix H: If the calculated F > 4.05, reject H0, otherwise do not reject H0
Since 4.84 < 4.405 reject H0
Q4 : 11-2 Exercises (11-20/p-491) :
Η0 :σ 12 =σ 2
2
ΗA :σ 12 ≠σ 2
2
Assume : F 0.025,20,10( ) = 4.405( )
F Reject H0Do not
reject H0Fα/2
0
Fcal = 332
152 = 4.84
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Chapter 11 Exercises: Hypothesis Tests and Estimation for Population Variances
Given the following null and alternative hypotheses
and the following sample information :
If α = 0.05, state the decision rule for the hypothesis. and Test the hypothesis and indicate whether the null hypothesis should be rejected.
Solution Q4 :
!
Using Appendix H: If the calculated F > 4.099, reject H0, otherwise do not reject H0
Since 0.46390 < 4.099 do not reject H0
Q4 : 11-2 Exercises (11-22/p-491) :
Η0 :σ 12 ≤σ 2
2
ΗA :σ 12 >σ 2
2
Assume : F 0.05,11,20( ) = 4.4099( )
F
α
Reject H0Do not reject H0
Fα
0
Fcal = 345.7745.2
= 0.46390
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