Question

BIOL 3680 Name STUDENT T-TEST PRACTICE PROBLEM & WORKSHEET -1.8 12.7 M -15.5 -0.7 -0.4 1.0 American sweetgum trees (genus Liquidamber) were found on campus. One grew under a closed canopy in deep si en in deep shade; the other sapling grew in direct sunlight. A blindfolded investigator sampled 10 leaves from each tree, and the widths of the leaves were measured X (1 ) (x-7) 8.8 2.9 1.73 Sun Leaves Shade Leaves x (x-x) 2 (x-2) 4.2 17.04 -1.79 10.3 10.5 3:17 -14.1 .98 3.24 9.8 15.5 10 5 -1 LT 123 -1.0 16.6 -13.5 11.2 -39 0.14 11.8 15.6 .49 [1.3 -28 0.08 -13.8 11.7 13.8 . 1.7 .12 0.14 10.3 11.2 12.7 11.8 11.3 .22 DOS 12.2 15.8 1.21 12.2 L 1.10 15.4 6.38 8.8 17.1 1213 072 L 13.8 1. 30 1-69 0.52 16.2 13.5 IV 4.62 21.34 . 2.10 2 4.41 6.76 Assignment (work in pairs): Means: 11-58 14.5 Medians: _1.3 141 Modes: 10.12 13,15 Std. Dev. 21 34 / (16-1) 2.37 4.76/(10-1) = 0.75 Null Hyp: The width of the leaves are offected by Sunlight Alt. Hyp: The width of the leone, affects the photosynthetic production inplents. Determine if the mean widths of the sun & shaded leaves are significantly different using a two-tailed. two-sample t-test & assume variances are equal at the nominal level of significance (alpha the appropriate degrees of freedom: 0.05) & Why 2-tailed test? T-test numerator: T-test denominator: (numerator divided by denominator) Observed t-value: Degrees of freedom: Critical t-value: Statistical conclusion: (table in Molles text, use a = 0.05) Biological conclusion:

Answer 1

Given Samples are Sun leaves shade 1 (xi-x)² leaves (4) (4;-)? 10.5 016 14.1 1.1664 3.1684 9.8 15.5 16.6 15.6 13.8 12.3 11.8 11.7 11-2 11.3 12.2 8.8 16.2 0.5184 0.0484 0.0144 0.1444 0.0784 0.3844 7.7284 21.3444 4:41. 1:21 0.49 17.64 3.24 10:3 12.7 15.8 17.1 13.5 - 1.69 - 6.76 Exi=115.8 Eyi= 1455 34.5956 | 37,6 7= EXi = 11:5.8 - 11:58 L3=11:58) ©= 31c 1452 145 [5=14,5) D Sa ng [(xi-7)+= 34.5956 n-1 = 3.8939 TS & 3.8439] biz 5(4-5)= 35,3 4.1478 5774.1778) n-1

To test the mean widths of the sun & Shaded leaves are significantly different using a two-tailed - two- Sample twor. t-test 955arne Vaziances are equal. - Here we use 2-tailed test because we went to check mean widths of sun & chanded hores are significantly diffent. To test Hoi 4-NL HEM, - Test statistic teat= tritha-2 under Ho - - I Sedatit hi se= (n-1) s, ² + (n-1) 32² nith₂-2 34.5956 + 37.6 = 18 Sć = 4:01086 1. Se= 2.0027 " To test numerator = 7-9 = 11:58 - 14.5 = -2.92 To test denominator Set l ini ha = 2.0027 + VIO 10 0.895634

observed t-values. Totest numerador T- test denominator = -2.92 - 0.895 634 Jobserved t-value = – 3.26026 = 1010m 2 _ Degrees of freedom a nithe ag Degrees of freedom = 18) Critical t-value= t(18,0.05) = +/18, 0.025) Critical t-value= 2.101 Critical region= { Itcall > temethac2,)} Here Itcall > tnitha-2, then we reject Ho at 5% 10.s.

statistical Conclusion : Two types of leaves different mean widths. have Biological conclusion I mean widths of sun & shaded leaves are significantly different