Question

Velvetleaf is a particularly annoying weed in corn fields. It produces lots of sceds, and the seeds wait in the soil for years until conditions are right. The velvetleaf seed beetle feeds on the seeds and might be a natural weed control. Here are the total seeds, seeds infected by the beetle, and percent of seeds infected for 28 velvetleaf plants: Seeds 2450 2504 21141110 2137 8015 1623 1531 2008 1716 Infected 135 101 Percent 5.6 3.8 3.62.45.5 2.6 1.83.03.8 0.6 12 76 24 121 18931 44 73 Seeds Infected Percent3.9 4.5 3.63.0 .3.9 3.90.0 3.6 4.2 721 863 1136 2819 1911 2101 1051 218 1711 164 798285 42 27 40 41 0 64 Seeds2228 363 5973 1050 1961 1809 130 880 Infected 156 31 240 91 137 92 Percent 7. 8.3 3.88.9 6.8 5.2 3.6 2.5 5 23 In R, do a complete analysis of the percent of seeds infected by the beetle #R Code Percent 5.6, 3.8, 3.6, 2.4, 5.5, 2.6, 1.8, 3.0, 3.8, 0.6, 3.9, 4.5, 3.6, 3.0, 4.1, 3.9, 3.9, 0.0, 3.6, 4.2, 7.2, 8.3, 3.8, 8.9, 6.8, 5.2, 3.6, 2.5) hist(Percent) mean(Percent) sd (Percent) t.test (Percent, mu 5.0, conf.level 0.90) Researchers would like to test whether the average percent of seeds infected is different than 5 percent. What is the null and alternative hypothesis for this test? Ho: μ > 5 percent Ha: μ 5 percent Ho: p-5 percent Ha:p 5 percent Ho: μ 5 percent Ha: 5 percent Ho: <5 percent Ha: μ-5 percent Ho: p 5 percent Ha: p 5 percent What is the average and standard deviation for the percent of seeds that are infected? (Round your answers to four decimal places.)

Answer 1

> Percent<-с (5 . 6, 3. 8, 3. 6,2.4,5. 5,2.6,1.8,3.0,3.8,0. 6, 3. 9,4. 5, 3. 6, 3. 0,4.1,3.9,3.9,0·0,3·6,4.2,7.2,8.3,3.8,8.9,6. 8, 5.2 ,3.6,2.5) > length(Percent) [1] 28 » hist(Percent) > mean(Percent [1] 4.060714 > sd(Percent) [1] 2.009116 > t.test (Percent,mu-5.0,conf. level0.90) one samp le t-test data Percent t2.4738, df27, p-value0.01995 alternative hypothesis:true mean is not equal to 5 90 percent confidence interval: 3.413997 4.707431 sample estimates: mean of> 4.060714

0 : 11-5 percent Ha : μ 5 percent

= 4.060714 s 2.009

t 2.4738

p - value 0.01995

Given a significance level of 0.10, we fail to reject the null hypothesis.

> error <-qt(0. 90 , df-length(Percent)-1),sd(Percent)/sqrt(Tength(percent) > left<-mean(Percent)-error #1 ower limit > right<-mean(Percent)-error #upper limit > (conf-Int-A=c (1 eft , right)) # confidence interval [1] 3. 561918 4.559511 # calulating Error

90% confidence interval is

( 3.561918, 4.559511)

The beetle infects less than 5% of seeds, so it's unlikely to be effective in controlling velvetweed.