chapter 1

Chapter 9: Inferences Based on Two Samples1d. F. 95,8,5= = . 271F.05,5,8e. F 4. 30. 01,10,12=1 1f.. 99,10,12= = = . 212F 4.71F .g. 6. 16. 05,6,4=.01,12,10F , so ( F ≤ 6 .16) = . 95P .h. Since F = . 177 ,1. 99,10,5=5.64P . 177 ≤ F ≤ 4.74 = P F ≤ 4.74 − P F ≤ .177 . 95 −.01= . 94( ) ( ) ( )= .58.a. Since the given f value of 4.75 falls between 3. 33F and F 5. 64. 05,5,10=can say that the upper-tailed p-value is between .01 and .05.. 01,5,10=, web. Since the given f of 2.00 is less than F 2. 52 , the p-value > .10.. 10,5,10=c. The two tailed p-value = 2 ( F ≥ 5.64) = 2(.01) = . 02P .d. For a lower tailed test, we must first use formula 9.9 to find the critical values:11F. 90,5,10= = .3030 , . 2110F. 95,5,10= = FF.10,10,51. 99,5,10= = F.01,10,5obviously closer to .05)..0995F ,.05,10,5. Since .0995 < f = .200 < .2110, .01 < p-value < .05 (bute. There is no column for numerator d.f. of 35 in Table A.9, however looking at both df =30 and df = 40 columns, we see that for denominator df = 20, our f value is between F .01and F .001 . So we can say .001< p-value < .01.281