Question

Note. For each of the following problems, state the appropriate null hypothesis and the alternative hypoth- esis, conduct the appropriate test, and state a conclusion. Reject the Ho if p < 0.05. Be sure to include the p-value. Assume the variable in question to be approximately normally distributed unless instructed otherwise. Part I: Solve the following problems by hand 1. Random samples of largemouth bass (Micropterus salmoides) and smallmouth bass (M. dolomieu) were taken from a lake and their lengths (in millimeters) were determined. We wish to know if the mean standard length differs between the two species in this lake. The results were as follows (data from J. Kagel). Largemouth Bass Smallmouth Bass 272.8 164.8 40.0 125 97 96.4 2. Seven tomato plants (Solanum lycopersicum) were treated with chlorogenic acid to determine if this would influence the activity of the enzyme o-diphenol oxidase in the leaves. Enzyme activity was measured as ul O2/mg protein/minute. A control group of seven plants were not treated. We do not know if this variable is approximately normally distributed, nor is it possible to determine this with a sample of this small a size. Does this treatment affect activity of the enzyme? Treated Untreated

Answer 1

ygol Given data t=272.8 Si= 96.4 ni=125 22=164.8 S2=40.0 n2=97 1. Null Hypotheris Ho: Ni=12 2. Alternative Hypothers HiiMIIN 3. Level of significancello 0.05(51) Awume 4. The teet statististe in z a li- , Sets ni na · Ze 272.8-164.8 -11.331 96.42, 402 | 125 + 97 The tabulated value of z at 5y level of significa ! nee is 1096 The tabulated value of z<the calculated

3.58 12.81 value of z. sowe Reject the Null Hypothelis to. and We conclude that the rela significant difference between them. 9) Sd calculation of sample means and so's m (x-ati (x-78 l y 114-8) (4-9 )? 1.43 2.04 lo -4042 19.53 11.43 | 130.64 | 18 2.43 5.90 -22.57 509.40 -6.42 141.21 7.43 55,20 14:58 212.57 -4.57 20.88 2.586.656 4:43 19.62 -6.42 41.21 |-3.42 11.69 235 743.68 101 345.676 a=- 335 -33.57 19= sy. 101 .14.44 sanith-X (ECM-)2+(4-92] (743.687345.676] => 90.77 S = 9.597 1. Null Hypothelis Ho : Me=M₂ 2. Alternative Hypotheen thi Mit M2 3. level of significance a: 0.05 (51. Allume 4. The text statistics tax-9 -777-2 silni na t= 33.57 - 11:48 **9.597 Eta =4.349 The tabulated value of tat 54. level of significance with dif is 7+7-2=12 is 1.785

The tabulated value of t < the calculated u- alue of t so te Reject the null typothelit Ho. and we conclude that there is a bitter ence between them.
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