Q1 <Sample 1> n1=9, x bar=16.18, S1=1.54 <Sample 2> n2=6, ybar=4.22, S2=1.37 Tes
ID: 2953984 • Letter: Q
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Q1 <Sample 1> n1=9, x bar=16.18, S1=1.54 <Sample 2> n2=6, ybar=4.22, S2=1.37 Test the null hypohesis 1-2=10 against thealthernative hypothesis 1-2 > 10 with=.01 Q2 In a study of insect response to pesticides, two strains offruitfly were bred. One was resistant to DDT; the other wassusceptible to DDT. Fecundity (number of effgs laid) was measuredon 25 randomly selected females from each strain. Summarystatistics for each group are: Resistant n1: 25, mean :25.26, sd: 7.77 Susceptible n1: 25, mean: 23.63, sd:9.77 Compute a 95% C.I. for the difference in means. Give apractical interpretation of your result. Q1 <Sample 1> n1=9, x bar=16.18, S1=1.54 <Sample 2> n2=6, ybar=4.22, S2=1.37 Test the null hypohesis 1-2=10 against thealthernative hypothesis 1-2 > 10 with=.01 Q2 In a study of insect response to pesticides, two strains offruitfly were bred. One was resistant to DDT; the other wassusceptible to DDT. Fecundity (number of effgs laid) was measuredon 25 randomly selected females from each strain. Summarystatistics for each group are: Resistant n1: 25, mean :25.26, sd: 7.77 Susceptible n1: 25, mean: 23.63, sd:9.77 Compute a 95% C.I. for the difference in means. Give apractical interpretation of your result.Explanation / Answer
(Q1) The test hypothesis isHo:1-2=10 (Q2) =0.05, Z(0.025)=1.96(check normal table)
The 95% CI is
(xbar1-xbar2)±Z*([s1^2/n1] +[s2^2/n2])
--> (25.26-23.63)±1.96*sqrt(7.77^2/25 +9.77^2/25)
-->(-3.26, 6.52)
Since this interval includes 0, we can not conclude that these twomean are not equal.
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