GPCO 453: Quantitative Methods I Sec 10: Hypothesis Testing, III - - PowerPoint PPT Presentation

GPCO 453: Quantitative Methods I

Sec 10: Hypothesis Testing, III Shane Xinyang Xuan1 ShaneXuan.com November 30, 2017

1Department of Political Science, UC San Diego, 9500 Gilman Drive #0521. ShaneXuan.com 1 / 11

Contact Information

Shane Xinyang Xuan xxuan@ucsd.edu My office hours for the rest of the quarter 12/4 M 1100-1230 (SSB 332) 12/4 M 1330-1400 (SSB 332) 12/12 Tu 1100-1130 (SSB 332)

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Interval Estimate of a Population Variance

◮ χ2 are based on n − 1 d.f. and (1 − α) confidence level

(n − 1)s2 χ2

α/2

≤ σ2 ≤ (n − 1)s2 χ2

1−α/2

(1)

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Test Statistic for Hyp. Tests about a Population Variance

◮ χ2 follows a chi-square distribution with n − 1 d.f.

χ2 = (n − 1)s2 σ2 (2)

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Hypothesis Tests about a Population Variance, One-tailed

◮ A company produces metal pipes and claims that the standard

deviation of the length is at most 1.2 cm. One of its clients decides to test this claim by taking a sample of 25 pipes. They found that the standard deviation of the sample is 1.5 cm.

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Hypothesis Tests about a Population Variance, One-tailed

◮ A company produces metal pipes and claims that the standard

deviation of the length is at most 1.2 cm. One of its clients decides to test this claim by taking a sample of 25 pipes. They found that the standard deviation of the sample is 1.5 cm.

◮ H0 : σ2 ≤ 1.2; HA : σ2 > 1.2

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Hypothesis Tests about a Population Variance, One-tailed

◮ A company produces metal pipes and claims that the standard

deviation of the length is at most 1.2 cm. One of its clients decides to test this claim by taking a sample of 25 pipes. They found that the standard deviation of the sample is 1.5 cm.

◮ H0 : σ2 ≤ 1.2; HA : σ2 > 1.2 ◮ Calculate the test statistics

χ2 = (n − 1)s2 σ2 = (25 − 1)(1.5)2 (1.2)2 = 37.5

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Hypothesis Tests about a Population Variance, One-tailed

◮ A company produces metal pipes and claims that the standard

deviation of the length is at most 1.2 cm. One of its clients decides to test this claim by taking a sample of 25 pipes. They found that the standard deviation of the sample is 1.5 cm.

◮ H0 : σ2 ≤ 1.2; HA : σ2 > 1.2 ◮ Calculate the test statistics

χ2 = (n − 1)s2 σ2 = (25 − 1)(1.5)2 (1.2)2 = 37.5

◮ Find the critical value

χ2-table → χ2

α = 36.415

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Hypothesis Tests about a Population Variance, One-tailed

◮ A company produces metal pipes and claims that the standard

deviation of the length is at most 1.2 cm. One of its clients decides to test this claim by taking a sample of 25 pipes. They found that the standard deviation of the sample is 1.5 cm.

◮ H0 : σ2 ≤ 1.2; HA : σ2 > 1.2 ◮ Calculate the test statistics

χ2 = (n − 1)s2 σ2 = (25 − 1)(1.5)2 (1.2)2 = 37.5

◮ Find the critical value

χ2-table → χ2

α = 36.415 ◮ Note that χ2 > χ2 α → We reject the null

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Hypothesis Tests about a Population Variance, Two-tailed

◮ Given n = 25, s = 17.7, test against σ2 = 225

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Hypothesis Tests about a Population Variance, Two-tailed

◮ Given n = 25, s = 17.7, test against σ2 = 225 ◮ H0 : σ2 = 225; HA : σ2 = 225

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Hypothesis Tests about a Population Variance, Two-tailed

◮ Given n = 25, s = 17.7, test against σ2 = 225 ◮ H0 : σ2 = 225; HA : σ2 = 225 ◮ Calculate the test statistic

χ2 = (n − 1)s2 σ2 = (25 − 1)(17.7)2 225 = 33.42

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Hypothesis Tests about a Population Variance, Two-tailed

◮ Given n = 25, s = 17.7, test against σ2 = 225 ◮ H0 : σ2 = 225; HA : σ2 = 225 ◮ Calculate the test statistic

χ2 = (n − 1)s2 σ2 = (25 − 1)(17.7)2 225 = 33.42

◮ Find the critical values at 95% CI

χ2-table

χ2

0.025 = 39.364

χ2

0.975 = 12.401

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Hypothesis Tests about a Population Variance, Two-tailed

◮ Given n = 25, s = 17.7, test against σ2 = 225 ◮ H0 : σ2 = 225; HA : σ2 = 225 ◮ Calculate the test statistic

χ2 = (n − 1)s2 σ2 = (25 − 1)(17.7)2 225 = 33.42

◮ Find the critical values at 95% CI

χ2-table

χ2

0.025 = 39.364

χ2

0.975 = 12.401 ◮ Note that χ2 0.975 < χ2 < χ2 0.025 → We fail to reject the null

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Hypothesis Tests about Two Population Variances

◮ Denoting the population with the larger sample variance as

population 1, the test statistic has an F-distribution with n1 − 1 degrees of freedom for the numerator and n2 − 1 degrees of freedom for the denominator.

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Hypothesis Tests about Two Population Variances

◮ Denoting the population with the larger sample variance as

population 1, the test statistic has an F-distribution with n1 − 1 degrees of freedom for the numerator and n2 − 1 degrees of freedom for the denominator.

◮ The test statistic (when σ2 1 = σ2 2) is

F = s2

1

s2

2

(3)

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Example: Two Population Variances

◮ Suppose n1 = 31, and n2 = 26, and

F = s2

1

s2

2

= 120 80 = 1.5 (4) We want to test at 95% confidence level: H0 : σ2

1 ≤ σ2 2

HA : σ2

1 > σ2 2

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Example: Two Population Variances

◮ Suppose n1 = 31, and n2 = 26, and

F = s2

1

s2

2

= 120 80 = 1.5 (4) We want to test at 95% confidence level: H0 : σ2

1 ≤ σ2 2

HA : σ2

1 > σ2 2 ◮ We find that F distribution with 30 numerator degrees of

freedom and 25 denominator degrees of freedom has F.05 = 1.92

F distribution ShaneXuan.com 8 / 11

Example: Two Population Variances

◮ Suppose n1 = 31, and n2 = 26, and

F = s2

1

s2

2

= 120 80 = 1.5 (4) We want to test at 95% confidence level: H0 : σ2

1 ≤ σ2 2

HA : σ2

1 > σ2 2 ◮ We find that F distribution with 30 numerator degrees of

freedom and 25 denominator degrees of freedom has F.05 = 1.92

F distribution

◮ Since the test statistic F is less than the critical value F.05,

we conclude that we fail to reject H0.

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Appendix: χ2−table

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Appendix: χ2−table

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Appendix: F distribution

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