Inferences about Variances (Chapter 7) In this Lecture we will: 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. 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 will be presented later. Variance-Test-1 Point Estimate for 2 The point estimate for 2 is the sample variance: n 1 2 s2 ( y y ) i n 1 i1 What about the sampling distribution of s2? (I.e. what would we see as a distribution for s2 from repeated samples).

If the observations, yi, are from a normal (,) distribution, then the quantity 2 (n 1) s 2 has a Chi-square distribution with df = n-1. Variance-Test-2 Chi Square (2) Distribution df2 2 0.5 2 df 5 0.4 df2 10 0.0 0.1 0.2 x2 0.3

2 df 15 0 5 10 15 20 25 30 y Non-symmetric. Shape indexed by one parameter called the degrees of freedom (df). Variance-Test-3 Table 7 in Ott and Longnecker Chi Square Table Variance-Test-4 Confidence Interval for 2 (n 1)s 2 If

2 df ,1 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. 2 Pr( 2df ,1 2 2 (n 1)s2 2 ) 1 2 df , 2 12.83 2 20.48 3.247 0.15 0.10

.8312 2df , df=10 0.05 0.04 x5 x10 0.06 0.10 0.08 df=5 .95 0.0 0.0 0.02 .95 0 5 10

15 y 20 25 30 0 5 10 15 20 25 y Variance-Test-5 30 Pr( 2df ,1 2 (n 1)s2 2

) 1 2 df , 2 2 (n 1)s2 ( n 1 ) s 2 2 2 df , df ,1 2 2 Consider the data from the contaminated site vs. background. Background Data: n 7 y 2.48 s 1.13 A 95% CI for background population variance 2 (6)1.13 2 (

6 ) 1 . 13 2 2 2 6,0.025 6,0.975 0.53 2 6.193 s2 = 1.277 Variance-Test-6 Hypothesis Testing for 2 What if we were interested in testing: Test Statistic: 2 ( n 1 ) s 2 02 H0 : 1 Ha : 2 3 2 02 ( 02 is specified)

2 02 2 02 2 02 Rejection Region: 1. Reject H0 if 2 > 2df, 2. Reject H0 if 2 < 2df,1- 3. Reject H0 if either 2 < 2df,1-/2 or 2 > 2df,/2 Example: n 7 y 2.48 s 1.13 2 ( 6 ) 1 . 13 2 7.66 1 In testing Ha: 2 > 1: Reject H0 if 2 > 26,0.05 =12.59 Conclude: Do not reject H0. Variance-Test-7 Tests for Comparing Two Population Variances Objective: Test for the equality of variances (homogeneity assumption). 0.6 0.8

F(2,5) F(5,5) 0.4 0.2 0.0 The F distribution shape is defined by two parameters denoted the numerator degrees of freedom (ndf or df1 ) and the denominator degrees of freedom (ddf or df2 ). 1.0 has a probability distribution in repeated sampling which follows the F distribution. x25 s12 s22 12 22 0 2 4 6 8

10 y Variance-Test-8 F distribution: 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 df1 and df2 given in Table 8. df1 relates to degrees of freedom associated with s21 df2 relates to degrees of freedom associated with s22 Variance-Test-9 Numerator df = df1. Table 8 Note this table has three things to specify in order to get the critical value. Denominator df = df2. 4.28 5.82 Probability Level F Table Variance-Test-10 Hypothesis Test for two population variances

2 1 H0 : 2 2 versus 2 1 2 2 s Test Statistic: F s Ha : 1. 12 22 2. 12 22 For one-tailed tests, define population 1 to be the one with larger hypothesized variance. For any 0 1 : Rejection Region: Fdf1 ,df 2 ,1 1 Fdf 2 ,df1 , 1. Reject H0 if F > Fdf1,df2,.

2. Reject H0 if F > Fdf1,df2,/2 or if F < Fdf1,df2,1-/2. In both cases, df1=n1-1 and df2=n2 -1. Variance-Test-11 Background Samples Example T.S. R.R. 2 1 2 2 Study Site Samples n1 7 n2 7 y1 2.48 y 2 4.82 s1 1.13 s2 0.89 2 s 1.13 1.2769 F

1.612 2 s 0.89 0.7921 Reject H0 if F > Fdf1,df2, where df1=n1-1 and df2=n2-1 = 0.05, F6,6,0.05 = 4.28 One-sided Alternative Hypothesis Reject H0 if F > Fdf1,df2,/2or if F < F df1,df2,1-/2 = 0.05, F6,6,0.025 = 5.82, F6,6,0.975 = 0.17 Two-sided Alternative Conclusion: Do not reject H0 in either case. Variance-Test-12 (1-)100% Confidence Interval for Ratio of Variances s12 1 12 s12 F 2 2 2 df2 ,df1 , 2

s2 Fdf ,df , 2 s2 1 2 2 Note: degrees of freedom have been swapped. Example (95% CI): F6, 6, 0.025 5.82 1.132 1 12 1.132 F6,6,0.025 2 2 2 0.89 F6, 6, 0.025 2 0.89 12 0.277 2 9.282 2 Note: not a argument! Variance-Test-13 Conclusion While the two sample test for variances looks simple (and is simple), it forms the foundation for hypothesis testing in Experimental Designs (ANOVA). Nonparametric alternatives are: Levenes Test (Minitab); Fligner-Killeen Test (R). Variance-Test-14 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-15 Example Its 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.) Variance-Test-16