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Variance-Test-1 Inferences about Variances (Chapter 7) Develop point estimates for the population variance Construct confidence intervals for the population.

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Presentation on theme: "Variance-Test-1 Inferences about Variances (Chapter 7) Develop point estimates for the population variance Construct confidence intervals for the population."— Presentation transcript:

1 Variance-Test-1 Inferences about Variances (Chapter 7) Develop point estimates for the population variance Construct confidence intervals for the population variance. Perform one-sample tests for the population variance. Perform two-sample tests for the population variance. In this Lecture we will: Note: Need to assume normal population distributions for all sample sizes, small or large! If the population(s) are not normally distributed, results can be very wrong. Nonparametric alternatives are not straightforward.

2 Variance-Test-2 The point estimate for  2 is the sample variance: What about the sampling distribution of s 2 ? (I.e. what would we see as a distribution for s 2 from repeated samples). If the observations, y i, are from a normal ( ,  ) distribution, then the quantity has a Chi-square distribution with df = n-1. Point Estimate for  2

3 Variance-Test-3 Non-symmetric. Shape indexed by one parameter called the degrees of freedom (df). Chi Square (  2 ) Distribution

4 Variance-Test-4 Chi Square Table Table 8 in Ott and Longnecker

5 Variance-Test-5 If has a Chi Square Distribution, then a 100(1-  )% CI can be computed by finding the upper and lower  /2 critical values from this distribution..8312 12.83.95 df=5 3.247 20.48.95 df=10 Confidence Interval for  2

6 Variance-Test-6 Background Data: A 95% CI for background population variance Consider the data from the contaminated site vs. background. s 2 = 1.277

7 Variance-Test-7 What if we were interested in testing: Test Statistic: Rejection Region: 1.Reject H 0 if  2 >  2 df,  2.Reject H 0 if  2 <  2 df,1-  3.Reject H 0 if either  2  2 df,  /2 Example: In testing H a :  2 > 1: Reject H 0 if  2 >  2 6,0.05 =12.59 Conclude: Do not reject H 0. Hypothesis Testing for  2

8 Variance-Test-8 Objective: Test for the equality of variances (homogeneity assumption). has a probability distribution in repeated sampling which follows the F distribution. F(2,5) F(5,5) The F distribution shape is defined by two parameters denoted the numerator degrees of freedom (ndf or df 1 ) and the denominator degrees of freedom (ddf or df 2 ). Tests for Comparing Two Population Variances

9 Variance-Test-9 Can assume only positive values (like  2, unlike normal and t). Is nonsymmetrical (like  2, unlike normal and t). Many shapes -- shapes defined by numerator and denominator degrees of freedom. Tail values for specific values of df 1 and df 2 given in Table 9. df 1 relates to degrees of freedom associated with s 2 1 df 2 relates to degrees of freedom associated with s 2 2 F distribution:

10 Variance-Test-10 Note this table has three things to specify in order to get the critical value. Numerator df = df 1. Denominator df = df 2. Probability Level 4.28 5.82 F Table Table 9

11 Variance-Test-11 versus Test Statistic: For one-tailed tests, define population 1 to be the one with larger hypothesized variance. Rejection Region: 1.Reject H 0 if F > F df1,df2, . 2.Reject H 0 if F > F df1,df2,  /2 or if F < F df1,df2,1-  /2. In both cases, df 1 =n 1 -1 and df 2 =n 2 -1. Hypothesis Test for two population variances

12 Variance-Test-12 Background Samples Study Site Samples  = 0.05, F 6,6,0.05 = 4.28 One-sided Alternative Hypothesis T.S. R.R. Reject H 0 if F > F df1,df2,  where df 1 =n 1 -1 and df 2 =n 2 -1 Example Reject H 0 if F > F df1,df2,  /2 or if F < F df1,df2,1-  /2  = 0.05, F 6,6,0.025 = 5.82, F 6,6,0.975 = 0.17 Two-sided Alternative Conclusion: Do not reject H 0 in either case.

13 Variance-Test-13 Note: degrees of freedom have been swapped. Example (95% CI): Note: not a  argument! (1-  )100% Confidence Interval for Ratio of Variances

14 Variance-Test-14 Conclusion While the two sample test for variances looks simple (and is simple), it forms the foundation for hypothesis testing in Experimental Designs. Nonparametric alternatives are: Levene’s Test (Minitab); Fligner-Killeen Test (R).

15 Variance-Test-15 Software Commands for Chapters 5, 6 and 7 MINITAB Stat -> Basic Statistics -> 1-Sample z, 1-Sample t, 2-Sample t, Paired t, Variances, Normality Test. -> Power and Sample Size -> 1-Sample z, 1-Sample t, 2-Sample t. -> Nonparametrics -> Mann-Whitney (Wilcoxon Rank Sum Test) -> 1-sample Wilcoxon (Wilcox. Signed Rank Test) R t.test( ): 1-Sample t, 2-Sample t, Paired t. power.t.test( ): 1-Sample t, 2-Sample t, Paired t. var.test( ): Tests for homogeneity of variances in normal populations. wilcox.test( ): Nonparametric Wilcoxon Signed Rank & Rank Sum tests. shapiro.test( ), ks.test( ): tests of normality.

16 Variance-Test-16 Example It’s claimed that moderate exposure to ozone increases lung capacity. 24 similar rats were randomly divided into 2 groups of 12, and the 2 nd group was exposed to ozone for 30 days. The lung capacity of all rats were measured after this time. No-Ozone Group: 8.7,7.9,8.3,8.4,9.2,9.1,8.2,8.1,8.9,8.2,8.9,7.5 Ozone Group: 9.4,9.8,9.9,10.3,8.9,8.8,9.8,8.2,9.4,9.9,12.2,9.3 Basic Question: How to randomly select the rats? In class I will demonstrate the use of MTB and R to analyze these data. (See “Comparing two populations via two sample t-tests” in my R resources webpage.)


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