Answer to Question #222177 in Statistics and Probability for pesky

Question

Given that the least significant difference is 3.62, separate the following means:

I. 75.30

ii. 68.95

iii. 67.40

iv. 72.35

v. 70.75

vi. 48.50

vii. 39.52

Answer 1

Given, the means are:

x1=75.30 

x2=68.95 

x3=67.40

 x4=72.35

 x5=70.75

 x6=48.50

 x7=39.52

Now, the absolute difference of means are:

$|x 1-x 2|=|75.30-68.95|=6.35 \geq L . S . D . \\|x|-x 3|=| 75.30-67.40 \mid=7.9 \geq L . S . D . \\|x 1-x 4|=|75.30-72.35|=2.95 \not L . S . D \\|x|-x 5|=| 75.30-70.75 \mid=4.55 \geq L . S . D . \\|x 1-x 6|=|75.30-48.50|=26.8 \geq L . S . D \\|x 1-x 7|=|75.30-39.52|=35.78 \geq L . S . D \\|x 2-x 3|=|68.95-97.40|=1.55 \ngeqslant L . S . D \\|x 2-x 4|=|68.95-72.35|=3.4 \ngeqslant L . S . D \\|x 2-x 5|=|68.95-70.75|=1.8 \ngeqslant L . S . D \\|x 2-x 6|=|68.95-48.50|=20.45 \geq L . S . D \\|x 2-x 7|=|68.95-39.52|=29.43 \geq L . S . D \\|x 3-x 4|=|67.40-72.35|=4.95 \geq L . S . D \\|x 3-x 5|=|67.40-70.75|=3.35 \ngeqslant L . S . D \\|x 3-x 6|=|67.40-48.50|=18.9 \geq L . S . D \\|x 3-x 7|=|67.40-39.52|=27.88 \geq L . S . D \\|x 4-x 5|=|72.35-70.75|=1.6 \ngeqslant L . S . D \\|x 4-x 6|=|72.35-48.50|=23.85 \geq L . S . D . \\|x 4-x 7|=|72.35-39.52|=32.83 \geq L . S . D \\|x 5-x 6|=|70.75-48.50|=22.25 \geq L . S . D \\|x 5-x 7|=|70.75-39.52|=31.23 \geq L . S . D \\|x 6-x 7|=|48.50-39.52|=8.98 \geq L . S . D$