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 transcript:

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.

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

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

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

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 df= df=10 Confidence Interval for  2

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

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

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

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:

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 F Table Table 9

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 Hypothesis Test for two population variances

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.

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

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).

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.

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.)