Question

Sample mean chlorophyll concentrations for the four Jerusalem artichoke varieties were 0.29, 0.25, 0.40, and 0.32, with corresponding sample sizes of 5, 5, 4, and 6, respectively. In addition, MSE = 0.0130. Calculate the 95% T-K intervals. (Use Table 7 in Appendix A. Round all answers to four decimal places.)

μ1-μ2: ( , )

μ1-μ3: ( , )

μ1-μ4: ( , )

μ2-μ3: ( , )

μ2-μ4: ( , )

μ3-μ4: ( , )

Answer 1

Given :

X₁ = 0.29, X₂ = 0.25 , X₃ = 0.40, X₄ = 0.32

n₁= 5, n₂ = 5, n₃ = 4, n₄ = 6

MSE = 0.0130

Confidence interval for difference between the means is given by ,

t is critical value follows t distribution with degrees of freedom N - k ;

N is total number of observations and k is number of groups

N = n₁ + n₂ + n₃ + n₄ = 5 + 5 + 4 + 6 = 20

k = 4

degrees of freedom = 20 - 4 = 16

Therefore t(0.05,16) = 2.12 ---- ( from t table )

Confidence interval :

μ1-μ2: ( -0.1129,0.1929 )

μ1-μ3: ( -0.2721,0.0521 )

μ1-μ4: ( -0.1764,0.1164 )

μ2-μ3: ( -0.3121,0.0121 )

μ2-μ4: ( -0.2164,0.0764 )

μ3-μ4: ( -0.076,0.236 )