Question

Average yield of strain A was measured as 0.905 by 20 experiments and average yield of strain B of the same organism under same experimental conditions was measured as 0.873 by 15 experiments. Standard deviation of yields of strain A was 17/1000 and standard deviation of yields of strain B was (32)/1000 a) Construct 95% confidence interval for the ratio of population variance of yields of strain A to population variance of yields of strain B. b) Construct 99% confidence interval for the difference between population means of yields of strain A and strain B (HA-uB) c) Test the null hypothesis of equality of population means of yields of strain A and strain B against the alternative hypothesis: A B (use p-value approach).

Answer 1

(a) 100 (1-a)% confidenceion interval for the ratio of two population variances és calculated by this formula: S,2 2 1 S₂² Fa/219,-1,9,-) S₂² Foxf219,-1, 92-) Besides a one-sided 100(1-2)% upper confidence limit is calculated by: si? S² Fa ihalnit a one sided 100(1-63% lower confidence limit és calcutaket by 1 Si Sz? Fa, n. pl, ha-1 2 32 2 SB (1000)? SA= 172 (1000) 2 2 & SB 0.001024 = 0.000289 So, 100(1-0.050) % -95% confidence limit is . 0.000289 0.001024 - F0.025,14,19 0.000289 Fo.025,19,14 0.001024 - (0-2822265625). 2-64692795 (0.2822265625) 2.86072154 [(0-2922265027 Food2s (19,14) = 2.86072154 can be found in F-lable at d=0.025

similarly, Foooos! Forczy(14,19) (14,19) = 2.64692795 can be found in F-table at d=0.095 So, 95% confidence limit for GA for a two tailed dest GB is 0.106624 4029] [ 0.03848 0.09865572673, (6) Formula for (1-9)% confidence limits for MA-leg when Sample Standard deviations are known: (6), (SP) L '12 ma - MA-MBE (5,- K2) - tnith 2 0.032 Here SĄ = 0.017 0.032, S8 = Xo = 0.873 TA= 0.905, 20, 15 nB = na = sp²= (Ma-1) set (18-11529 -2 natno- (From t-table for two-tailed) 2 tn, th-2(a)= teori1r-260001) = tz(0-1) = 2.733277 Spa 19 (0-017) + 1410.032) = 0.0006008181818 23 (1) ~ 10.0245 =) Sp = 0.0245115928 2. PLEASE DO UPVOTE-THANK YOU LIT MOTIVATES ME

So, A-TB) - tzza +1 ha ng = 10.905-0.873) - (2.733277 I -15 20 = -0.9015918283 And (XA-TB) - +33 (0.01) (0.01)+1 ha ng = (0.905 - 0-873)+(2.733297) 20 = 0.9655918283 are So, 99% confidence limits for MA-MB when GA 2 GB unknownt SA & SB are known) MA-Ust [-o. 0.9015918263, 0.9655918283 (C) 2 2 6 We want to lest Ho & MA= MB vs Hii MAtle Here since sample standard deviations are given do we shall use t-test for defference of means. Further there is always confusion whether to use pooled t-test or unpoiled t-test. When both samples are drawn from the same population we use pooled t-test because in that case 6A²= 66 - population variance. When there is nothing mentioned about the samples Whether they come from same population or defferent we should conduct hypothesis test for the oti equality of population variances using F-test, So, it the test results in equality of variances we can use pooled t-test else we should use unpooled t-test.

A more informal way of saying this is that if the ratio of two sample st.deviations lies in the interval (1/12) , we can use pooled t-test else we should use unpooled it-test, Here SA 17 00017 0.032 0.53.125€(112) SB So we can use pooted -t-test: Under Ho: MA=MB ta XA-XB a +83) Spltti NA NB by (1) Sp = 0.0245 Oogos -00873 3.82393498 ~ 3.82 39 So, ta こ (0.0245) ++ b-value Plt > taal) ; For right tailed test Plte teal) ; For left failed test 2P (txtcal) ; For two-tailed test Here our case is right -tailed test, Plts tcal) = P(+> 3•62 39) = 0.000 277 (From t-table) at dof=33 p-value Clearly p-value <x le. 0.000277<0.01, we reject Ho and accept H, i.e. we conclude that Ma> us. PLEASE UPVOTE 47