Question

Sample mean chlorophyll concentrations for the four Jerusalem artichoke varieties were 0.31, 0.25, 0.41, 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-μ_{4 (,)}

μ₃-μ₄: (,)

Answer 1

To FInd : Choose the group of means which do not have significant diffrences.

To calculate this, we apply the formula:

From the critical value of standardized range, Q for alpha =0.05, c=4 and n-c=20-4=16

thus required value of

Obtain critical range: group 1 and group 2

MSE=0.0130

Comparision	Absolute Mean difference	Critical range	Conclusion
	|0.31-0.25|=0.06	0.2061	Mean are not Different
	|0.31-0.41|=0.10	0.2190	Mean are Different
	|0.31-0.32|=0.01	0.1986	Mean are not Different
	|0.25-0.41|=0.16	0.2190	Mean are not Different
	|0.25-0.32|=0.07	0.1986	Mean are not Different
	|0.41-0.32|=0.09	0.2108	Mean are not Different

From the above table, it is clear that all the pairs of absolute mean difference is lesser than critical range. Thus, it can be concluded that there is no significant difference.

The T-K interval is