Problem 3 help Ecologists are interested in whether the presence of arbuscular mycorrhizalfungi (AMF) in the soil affects the mass of seeds from plants growing in the soil. They grow upplants in 12 pots: six with AMF added to the soil and six without AMF. They report the seedbiomass and summary statistics shown below:Seed biomass (g)DifferenceWith AMFNo AMFin columns150.6153.2-2.6147.9145.82.1153.4163.3-9.9140.3146.5-6.2146.8149.8-3.0147.5159.8-12.3H1 = 147.8si = 19.31S1 = 4.39d = -5.3s = 27.66 s:Sd = 5.26Y2 = 153.1s = 51.33= 35.3282 = 7.16Test the null hypothesis that AMF does not affect seed mass. You can assume that the data ineach sample follow a normal distribution. Report the P-value and a circle the correct conclusion.(a) This study design is (CIRCLE ONE):PAIREDUNPAIRED(b) What can you say about the P-value?(c) CIRCLE the appropriate conclusion.• Data are consistent with null hypothesis.• We should reject the null hypothesis; AMF increases the seed biomass.• We should reject the null hypothesis, AMF decreases the seed biomass.• We should reject the null, but we cannot say whether AMF decreases or increases the seedbiomass.

Solution: Given information: n = 6 observation. d = -5.3 Difference of sample mean sd2= 27.66 Difference of sample varaince sd = 5.26 Difference of sample stadnard ddeviation a) This is an experiment of paired study design. This study design is Paired. To test the hypothesis Ho:μd= 0 Vs Ha:μd ≠0 Test statistic t =dsd/nt =-5.35.26/6t= -5.35.26/2.449490t=-5.32.147386= -2.468117t= -2.4681 Test statistic t = -2.4681 b) To find p value p value = 2×P(t>-2.4681)p value = 0.0567 From Excel The p value for the test is 0.0567 Assumed α=0.10 Decision: p value < α=0.10 Reject the null hypothesis. c) Conclusion: We should reject the null hypothesis but we cannot say whether AMF decrease or increase the seed biomass.