Question

5. Maximum photosynthetic rates were measured in leaves from diploid (2N) and tetraploid (4N) tall fescue grass. Test the null hypothesis of no difference in the photosynthetic rates for diploids and tetraploids at the 5% level of significance. (4 points) Photosynthetic Rate (umol/m/s) Diploids 18 Tetraploids 25 21 23 26 28 16 19 20 23

Answer 1

Diploids: 18, 22, 16, 19, 20, 23 We calculate sample mean and std deviation from given data Sample Mean, z-20 2-118 Sample Variance, s2 Sample std dev, s -s26.6666672.581989 Tetraploids: 25, 21, 23, 26, 28 We calculate sample mean and std deviation from given data Sample Mean, π--.-_- Sample Variance, 2 Sample std dev, s- Vs73 2.701851 19.666667 , X(a -33.333333 - 6.666667 123 Test: 2 using equal variance assumption Hypothesis test (Null Hypothesis) Alternative Hypothesis, also called H) This is two tailed test

sample 1 19.666667T2- 24.6 sample 2 T s1 2.581989s2 2.701851 Significance level, α-0.05 Degrees of freedom: = df = ni + n2-2-9 The pooled estimate of Variance is: ( ni _ 1 )si + (n2-1)s.. 6.948148 Test statistic, tobervedni +s/m 1-D2 19.66666724.6 -3.091 6.948148/6 6.948148/5 Using excel we find p-value-Plt| > 3.091) = 2*t·dist(-3.091, 9,1) = 0.0129128915 Critical Value, teriticalta/2df to.025,df-9 2.262 (from t-table, two-tails, Rejection criteria: Rejection region is in direction of Alternative hypothesis, since Ha : ,41-12 0, rejection rule is: Reject Ho if Ito> te or equivalently if p-value < α Decision : Since Ito! > tc or equivalently since p-value < α, we reject the null hypothesis. Hence at 5% significance level, we have sufficient evidence to conclude that μι is different from @seoThank ,youJ0户