Chapter 7: Inference about Variances Sampling Distribution of Variances: Say you want to know the variance of a population. What sample quantity have we seen before that might be a good estimator of the true population variance, σ2? We need the sampling distribution of the sample variance, s2, in order to make inference about the population variance. Assume the population is normally distributed with mean µ and standard deviation σ. Let y denote an observation from the population. Recall that the sample variance is given by 22()1iiyysn−=−∑. Then, sampling distribution of (n-1)s2/σ2=χ is given by 21nχχ−∼ or, chi-square with n-1 degrees of freedom. We can use this distribution to model the sampling distribution of s2 and hence to construct confidence intervals and conduct hypothesis test. The Chi-square Distribution Usually, we use a table for the chi-square distribution or a computer program to get the probabilities we need for inference. Examples: Suppose Χ has chi-square distribution with df = 4. Then P(Χ > 11.1433) = .025. In general, as with the normal distribution, we look up values in the table for ()2,kPXαχα>= 1Examples: 2.025,411.143χ= 2.025,1020.48χ= 2.975,103.247χ= Hence, a 95% Confidence Interval for a Variance is given by: 222.975 .0252(1).95nsPχχσ⎛⎞−<<=⎜⎟⎝⎠ 22222.025 .975(1) (1).95ns nsPσχχ⎛⎞−−<< =⎜⎟⎝⎠ 95% confidence interval for σ2: 2222.025 .975(1) (1),nsnsχχ⎛⎞−−⎜⎟⎝⎠ Inference for the Variance of One Population Test of H0: σ2= σ02 vs. Ha: σ2> σ02 Test statistic: Χ = (n-1)s2/ σ02 Large values are “significant” Example: Consider a data set that consists of the hand-spans of men and women. The summary statistics are given below. Construct 95% confidence interval for σ2 for the hand-span for women 2The summary statistics for the hand-span data are my = 8.99 fy = 8.27 sm2 = 0.470 sf2 = 0.682 nm = 22 nf = 13 2222 2 2.025 .975 12,.025 12,.975( 1) ( 1) (13 1).682 (13 1).682,,8.184 8.184, (0.351,1.858)23.34 4.404nsnsχχ χ χ⎛⎞⎛⎞−− − −=⎜⎟⎜⎟⎜⎟⎝⎠⎝⎠⎛⎞==⎜⎟⎝⎠ Inference for Comparing the Variances of Two Populations Suppose we would like to compare the variances of two populations. We again use the sample estimate from each population. Because the test statistic F = s1 2/s22 has a known distribution, we can use this distribution to conduct inference. This distribution is called the F distribution. Again, we can read probabilities from an F table: The F distribution has numerator and denominator degrees of freedom. We write F~Fdf1,df2 . This means that F has df1 (numerator) and df2 (denominator) degrees of freedom and ,1,2()df dfPF Fαα>= Example: .05,12,212.25F = An Important Property of the F distribution: , 2,1 ,1,2(1/ )( )df df df dfPF F PF Fααα<=>= Result: 1,1,2 ,2,11/df df df dfFFαα−= 3Example: .95,21,121/ 2.25 0.44F == Hypothesis Tests for Variances from Two Populations Hypothesis Test of H0: σ1 2= σ22 vs Ha:σ1 2> σ22 Test statistic: F = s1 2/s22 Large values are “significant” Example: Test H0: σm 2= σf2 versus Ha: σm 2> σf2 F = .470/.682=.689 <1 implies non-significant Hypothesis Test of H0: σ1 2= σ22 versus Ha: σ1 2 ≠ σ22 Test statistic: F = max(s1 2,s22)/min(s1 2,s22) Large values are “significant” Example: Test H0: σm 2= σf2 versus Ha: σm 2≠ σf2 F = .682 /.470 = 1.45, df1 = 12, df2 = 21 F12,21,.25 = 1.38, F12,21,.10 = 1.87 → .10<p<.25