UD DSC 211, 11.2 Lecture Notes, ASW 7th Ed

DSC 211 11.2 Worksheet F- DISTRIBUTION 11.2: Inferences About Two Population Variances Our goal in this section is to compare two population variances to each other. Just as before when we compared two population mean values in chapter 10, we will assume that we have 2 well-defined populations. Letting σ 1 2 represent the true value of the variance from population #1, and σ 2 2 represent the true value of the variance from population #2, we would like to determine which of the following 3 possibilities seems to be true when comparing the value of σ 1 2 to the value of σ 2 2 : 1. σ 1 2 = σ 2 2 . That is, there is no significant difference between the 2 population’s variances. 2. σ 1 2 < σ 2 2 . That is, the variance of population #1 is significantly less than the variance of population #2. 3. σ 1 2 > σ 2 2 . That is, the variance of population #1 is significantly greater than the variance of population #2. A follow-up question to #2 or #3 (when we find that one of them is significantly bigger than the other) is “by how much?”, or “what is the size of the difference?”. There are CIs that provide the answer to this question, but they are beyond the scope of our text, and we will not cover them. The authors take a unique approach to setting up the data to test whether (1), (2), or (3) above is supported by the data in such a way that we don’t actually have to consider the 2 nd possibility. For the hypothesis tests presented in this section to be mathematically and statistically valid, the following 2 conditions/assumptions are required: 1. You have independent SRSs from the correct target populations, and; 2. Both target populations are normally distributed. Recall that these are the first 2 assumptions for the alternative approach to CIs and hypothesis testing on μ 1 − μ 2 presented at the end of the section 10.2 worksheet. The 3 rd assumption there was that σ 1 2 = σ 2 2 , which provides motivation for the hypothesis tests developed in this section. Or, we may be interested in comparing σ 1 2 and σ 2 2 for their own sakes. Q1: Calculate the sample variance for the following 5 daily low temperatures: 5 , − 12 , 1 , 0 , 13 =s qrt (stDEV of numbers) --- 3.01196

It can be shown that the ratio, s 1 2 s 2 2 follows an F -distribution, with n 1 − 1 numerator degrees of freedom ( Ndf ), and n 2 − 1 denominator degrees of freedom ( Ddf ) when the preceding 2 conditions are true. This fact gives rise to the hypothesis tests that we will set up to compare σ 1 2 to σ 2 2 . Hypothesis Tests to Compare σ 1 2 and σ 2 2 Before running any of the following hypothesis tests you should check that the following conditions/assumptions are met: 1. You have independent SRSs from the correct target populations, and; 2. Both target populations are normally distributed (this one is crucial). If the above conditions are satisfied, then you may conduct the following hypothesis tests to compare σ 1 2 and σ 2 2 : Two Tailed Version ** One Tailed Version * Upper Tail H O : σ 1 2 = σ 2 2 H O : σ 1 2 ≤ σ 2 2 H A : σ 1 2 ≠ σ 2 2 H A : σ 1 2 > σ 2 2 α α TS: F O = s 1 2 s 2 2 , with TS: F O = s 1 2 s 2 2 , with n 1 − 1 Ndf , n 2 − 1 Ddf n 1 − 1 Ndf , n 2 − 1 Ddf RR: If F O > F ( α 2 ,Ndf , Ddf ) , R H O RR: If F O > F ( α, Ndf ,Ddf ) , R H O D/C: R H O , or FTR H O , D/C: R H O , or FTR H O , followed by C followed by C PV: 2P[ F ( Ndf ,Ddf ) > F O ] PV: P[ F ( Ndf ,Ddf ) > F O ] ** In the Two Tailed Hypothesis Test described above, before you label the data sets “group #1” and “group #2”, calculate their respective sample variances, and whichever group yields the larger variance will be called “group #1” with the other group being called “group #2”, and then s 1 2 , n 1 , s 2 2 , and n 2 follow. * In the One Tailed Hypothesis Test described above, you will have an idea of which group you think has the significantly larger population variance before you collect any data (which is why you’re running the test). This group should be called “group #1”. In the one tailed hypothesis test there is no need to calculate the sample variances first 2

d. P[ 1.98 ≤ F ( 25 , 13 ) ≤ 3.57 ] = F.DIST(3.57,25,13,TRUE)-F.DIST(1.98,25,13,TRUE) = 0.08961942 Q3: Use the appropriate Excel commands to determine the value of F ¿ that makes each of the following probabilities true: a. P[ F ( 9 , 28 ) ≤ F ¿ ] = 0.95 F.INV(0.95,9,28) = 2.23598166 b. P[ F ( 2,3 ) > F ¿ ] = 0.025 F.INV.RT(0.025,2,3)= 16.0441064 Q4: Given the following male and female GPAs is there evidence that the population variances are significantly different? Do a full hypothesis test with a 5% level of significance. Clearly state your conclusion. males 3.4 3.9 3.0 3.6 3.3 3.4 3.5 females 3.1 3.8 2.9 2.6 3.1 3.3 3.9 Use Excel’s Data Analysis feature to run this test: 1. Using the criteria at the bottom of page 2, determine which group is Group 1 and which group is Group 2. 2. Enter your data into an Excel spreadsheet, with Group 1 data in one column (say Column A) and Group 2 data in another column (say Column B). 3. Click the Data tab on the Ribbon. 4. Click on Data Analysis. 5. Choose F-Test Two-Sample for Variances. 6. Enter the Group 1 cell references into the Variable 1 Range box and the Group 2 cell references into the Variable 2 Range box. 7. If you used Labels at the top of each column, then check the Labels box. 4

8. Enter the appropriate value in the Alpha box. 9. Check the radio-button in fron t of Output Range and with your cursor in the Output Range window, click in an empty cell to identify to Excel where you want it to put the upper right-hand corner of the output table. Use the Excel-generated output at the botom of page 4 to complete the following Hypothesis Test mentiond in Q4 (how do you get the 2-tailed PV?) H O : H O : σ 1 2 = σ 2 2 H A : H A : σ 1 2 ≠ σ 2 2 α : 0.05 TS: F 0 = 2.88125. with DF of 6 RR: If F 0 > F( 5.81975658), THEN REJECT Ho D/C: since 2.88125 is not > than 5.819756 I fail to reject PV: = 2 * 0.11174685 (p value from excel) = 0.22349369 Q5: In 10.3 n 1 and n 2 HSD TO BE EQUAL. Is that a requirement in 11.2? 5

PV: 7