2 and F Distributions. Barrow, Statistics for Economics, Accounting and Business Studies, 4 th edition Pearson Education Limited 2006

Transcription

1 and F Distributions Lecture 9

2 Distribution The distribution is used to: construct confidence intervals for a variance compare a set of actual frequencies with expected frequencies test for association between variables in a contingency table It is asymmetric and dependson the degrees of freedom

3 F Distribution The F distribution is used to test the hypothesis of equality of two variances conduct an analysis of variance (ANOVA), comparing means across several samples Itisas asymmetric mmetricand depends on the degrees of freedom

4 Tails of the 19 Distribution.5%.5%

5 Critical Values of the Chi squared Distribution NB chi squared is not symmetric so table will give different values for thelower anduppertails Excerpt from Table A4: : : : : : :

6 Case 1: Estimating a Variance A random sample of size n = 0 yields a standard deviation of s = 5. How do we estimatethe the population variance? Point estimate: use s = 5 = 65 which is unbiased (E(s ) = ) Interval estimate: we need the sampling distribution of s...

7 The Sampling Distribution of s n 1 s n-1 gives the degrees of freedom for the distribution, 19 in this example. ~ n1

8 Limits to the Confidence Interval For the 95% CI, we need the values cutting off.5% in each tail of the distribution Excerpt from Table A4: : : : : : :

9 Tails of the 19 Distribution (cont.) We can be 95% confident that (n 1)s / lies between 8.91 and 3.85 (for n = 0) n 1s Rearranging: 8 n 1s n Substituting s = 65 and n = 0: s ,33.8 gives the 95% CI estimate

10 Case : Comparing Actual vs Expected Frequencies 7 rolls of a dice yield: Score on dice Frequency From a fair dice one would expect each number to come up 1 times. Isthisevidence of a biased dice?

11 H 0 : the dice is fair fi H 1 : the dice is biased This can be tested using Test Statistic O E which has a distribution with k 1 degrees of freedom, k = 6 in this case because we have 6 outcomes. E

12 Calculating the Test Statistic Score Observed Expected O E (O E) (O E) frequency (O) frequency (E) E Totals

13 Calculating the Test Statistic (cont.) The test statistic, 7.66, is less than the critical value of with = 5, 11.1 Hence the null isnot rejected, the difference between observed and expected outcomes is random Note the critical value cuts off 5% (not 5%)in.5%) the upper tail of the distribution. Only large values of the test statistic reject H 0

14 Case 3: Contingency Tables The association between two variables can also be analysed via the distribution Voting behaviour based on a sample of 00: Social class Labour Conservative Liberal Democrat Total A B C Total

15 Are Social Class and Voting Behaviour Related? H 0 : no association between social class and voting behaviour H 1 : some association Expected values are calculated, based on the null of no association E.g. if there is no association: 40% (80/00) of every social class should vote Labour, i.e. 16 from class A, 40 from B and 4 from C

16 Observed (and Expected) Values Social class Labour Conservative Liberal Democrat Total A 10(16) 15(14) 15(10) 40 B 40(40) 35(35) 5(5) 100 C 30(4) 0(1) 10(15) 60 Total

17 Calculating the Test Statistic For = (rows 1) (columns 1) = 4, the critical value of the distribution is 9.50, so the null of no association is not rejected at the 5% significance level.

18 Testing Two Variances the F Distribution Do two samples have equal variances (i.e. come from populations with the same variance)? Data: n 1 = 30 s 1 = 5 n = 30 s = 0

19 Testing Two Variances the F Distribution (cont.) H 0 : 1 = H 1 : 1 = or, equivalently H 0 : 1 / =1 H 1 : 1 / 1

20 The test statistic is Evaluating this: The Test Statistic s s F 1 ~ F n 1 1, n F* 9,9 =.09 >1.565, so the null is not rejected. The variances may be considered equal

21 Excerpt tfrom Tbl Table A5(b): the F Distribution : : : : : : (Using 1 = 30 (rather than 9) makes little practical difference.)

22 One or Two Tailed Test? As long as the larger variance is made the numerator of the test statistic, only large values of F reject the null. The smallest possible value of F is 1, which occurs if the sample variances are equal. H 0 should not be rejected din this case. So, despite the in H 1, this is a one tailed test.

23 Case 5: Analysis of Variance (ANOVA) A test for the equality of several means, not just two as bf before In our example we test for the equality of output of three factories, i.e. are they equally productive, on average, or not?

24 Example: Daily Output of Three Factories Observation Factory 1 Factory Factory

25 Chart of Output Factory 1 Factory Factory

26 Hypothesis Test H 0 : 1 = = 3 H 1 : 1 3 Principle of the test: break down the total variance of all observations into the within factory variance and the between factory variance If the between variance component is large relative to the within variance component, reject H 0

27 Sums of Squares Rather than variances, work with ihsums of squares s x x n 1 Sum of squares Variance

28 Three Sums of Squares Total sum of squares (TSS) Sum of squares of all deviations from the overall average Between sum of squares (BSS) Sum of squares of deviations of factory means from overall average Within sum of squares (WSS) Sum of squares of deviations within each factory, from factory average

29 Test Statistic F BSS WSS k n 1 k The F statistic is the ratio of BSS to WSS, each adjusted by their degrees of freedom (k 1 and n k) Large values of F BSS large relative to WSS between factories deviations large reject H 0

30 Calculations TSS = x ij x j i (j indexes factories, i indexes observations) = ( ) + ( ) + + ( ) +( ) =, is the overall, or grand, average

31 Calculations (cont.) BSS = x i x j i where is the average output of factory i x i = 6 ( ) + 7 ( ) + 5 ( ) 11) = 1, , , are the three averages, respectively.

32 Calculations (cont.) WSS = TSS BSS =, , = 1, Alternatively, ti l WSS = x ij x i j i = ( ) ( ) ) + ( ) + + ( ) + ( ) + + ( ) ) = 1,

33 Result of the Test F BSS WSS k n k F*,15 = 3.68 (5% significance level) F > F* hence we reject H 0. There are significant differences between the factories. Easiest to do in EXCEL...

34 ANOVA Table (Excel Format) SUMMARY Groups Count Sum Average Variance Factory Factory Factory ANOVA Source of Variation SS df MS F P-value F crit Between Groups Within Groups Total

35 Summary Use the distribution to Cl Calculate lt the CI for a variance Compare actual and expected values Analyse a contingency table Usethe F distribution to Test for the equality of two variances Test tfor the equality of several means