Question

Velvetleaf is a particularly annoying weed in corn fields. It produces lots of seeds, 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 Infected Percent 2450 2504 135 101 5.3 4.0 2114 76 3.5 1110 24 2.2 2137 121 5.9 8015 189 2.3 1623 31 2.0 1531 44 2.7 2008 73 3.7 1716 12 0.7 Seeds Infected Percent 721 27 3.9 863 40 4.8 1136 41 3.4 2819 79 2.9 1911 82 4.4 2101 85 4.2 1051 42 4.2 218 0 0.0 1711 64 3.6 164 7 4.2 91 880 23 2.4 Seeds 2228 363 5973 1050 1961 1809 130 Infected 156 31 240 137 92 5 Percent 7.2 8.4 4.1 8.5 6.8 5.1 3.7 In R, do a complete analysis of the percent of seeds infected by the beetle. #R Code Percent 5.3, 4.0, 3.5, 2.2, 5.9, 2.3, 2.0, 2.7, 3.7, 0.7, 3.9, 4.8, 3.4, 2.9, 4.4, 4.2, 4.2, 0.0, 3.6, 4.2, 7.2, 8.4, 4.1, 8.5, 6.8, 5.1, 3.7, 2.4) = c( hist(Percent) mean(Percent) sd(Percent) t.test(Percent, mu = 5.0, conf.level = 0.90)

Answer 1

Ho :   µ =   5
Ha :   µ ╪   5

Sample Mean,    x̅ = ΣX/n =    4.0704
sample std dev ,    s = √(Σ(X- x̅ )²/(n-1) ) =   2.0418

Standard Error , SE = s/√n =   2.0418/√27=   0.3930
t-test statistic= (x̅ - µ )/SE =    (4.0704-5)/0.393=   -2.3658

p-Value   =   0.0257

reject the null hypothesis

Level of Significance ,    α =    0.1      
degree of freedom=   DF=n-1=   26      
't value='   tα/2=   1.706   [Excel formula =t.inv(α/2,df) ]  
              
Standard Error , SE = s/√n =   2.0418/√27=   0.3930      
margin of error , E=t*SE =   1.706*0.393=   0.6702      
              
confidence interval is               
Interval Lower Limit = x̅ - E =    4.0704-0.6702=   3.4001      
Interval Upper Limit = x̅ + E =    4.0704+0.6702=   4.7406      
90% Confidence interval is   3.40   < µ <   4.74   )

The bettle infects less than 5% of seeds , so it is likely to be effective in controlling