ch11 exercises hypothesis tests and estimation for population variances

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: [email protected] | 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


! 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


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 :

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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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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 :

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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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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 :

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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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ch11 exercises hypothesis tests and estimation for population variances

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