Question

The following statistics are calculated by sampling from four normal populations whose variances are equal: (You may find it useful to reference the t table and the q table.) x⎯⎯1 = 153, n1 = 6; x⎯⎯2 = 164, n2 = 6; x⎯⎯3 = 159, n3 = 6; x⎯⎯4 = 153, n4 = 6; MSE = 50.5

Use Fisher’s LSD method to determine which population means differ at α = 0.01. (Negative values should be indicated by a minus sign. Round intermediate calculations to at least 4 decimal places. Round your answers to 2 decimal places.)

b. Use Tukey’s HSD method to determine which population means differ at α = 0.01. (If the exact value for n_T – c is not found in the table, then round down. Negative values should be indicated by a minus sign. Round intermediate calculations to at least 4 decimal places. Round your answers to 2 decimal places.)

Answer 1

N = 24, k = Number of groups = 4. Therefore DF error = 24 - 4 = 20. Given MS error = 50.5

SQRT(MSerror / n) = SQRT(50.5 / 6) = 2.901

M1 = 153, M2 = 164, M3 = 159, M4 = 153

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Using Fishers LSD

LSD = t critical * SQRT(MS error / n)

Where, t critical is the 2 tailed critical value at = 0.01, and DF = DF error = 20.

tcritical = 2.845 and SQRT(MS error / n) = 2.901

LSD = 2.845 * 2.901 = 8.25

If the Absolute value of Differences is greater than LSD (8.25), then there is a significant difference.

M1 - M2 = ABSOLUTE (153 - 164) = 11 > 8.25. There is a significant difference between the means.

M1 - M3 = ABSOLUTE (153 - 159) = 6 < 8.25. There isn't a significant difference between the means.

M1 - M4 = ABSOLUTE (153 - 153) = 0 < 8.25. There isn't a significant difference between the means.

M2 - M3 = ABSOLUTE (164 - 159) = 5 < 8.25. There isn't a significant difference between the means.

M2 - M4 = ABSOLUTE (164 - 153) = 11 > 8.25. There is a significant difference between the means.

M3 - M4 = ABSOLUTE (159 - 153) = 6 < 8.25. There isn't a significant difference between the means.

Putting it as a Table

Mi	Mj	Mi - Mj	LSD	Greater/Lesser	Significant (Yes/No)
153	164	11	8.25	Greater	Yes
153	159	6	8.25	Lesser	No
153	153	0	8.25	Lesser	No
164	159	5	8.25	Lesser	No
164	153	11	8.25	Greater	Yes
159	153	6	8.25	Lesser	No

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Using Tukeys HSD

HSD = q critical * SQRT(MS error / n)

Where, q critical is the critical value at = 0.01, and DF = DF error = 20 and k = 4.

qcritical = 3.96 and SQRT(MS error / n) = 2.901

LSD = 3.96 * 2.901 = 11.49

If the Absolute value of Differences is greater than HSD (11.49), then there is a significant difference.

M1 - M2 = ABSOLUTE (153 - 164) = 11 < 11.49. There isn't a significant difference between the means.

M1 - M3 = ABSOLUTE (153 - 159) = 6 < 11.49. There isn't a significant difference between the means.

M1 - M4 = ABSOLUTE (153 - 153) = 0 < 11.49. There isn't a significant difference between the means.

M2 - M3 = ABSOLUTE (164 - 159) = 5 < 11.49. There isn't a significant difference between the means.

M2 - M4 = ABSOLUTE (164 - 153) = 11 < 11.49. There isn't a significant difference between the means.

M3 - M4 = ABSOLUTE (159 - 153) = 6 < 11.49. There isn't a significant difference between the means.

Putting it as a Table

Mi	Mj	Mi - Mj	LSD	Greater/Lesser	Significant (Yes/No)
153	164	11	11.49	Lesser	No
153	159	6	11.49	Lesser	No
153	153	0	11.49	Lesser	No
164	159	5	11.49	Lesser	No
164	153	11	11.49	Lesser	Yes
159	153	6	11.49	Lesser	No

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