Question

For the squid abundance data set, conduct an appropriate test to determine if squid abundance differs significantly among reefs. If you find that the reefs differ in the abundance of squids, then conduct a posteriori tests to determine which reefs are different from which others using two different multiple comparison procedures: the Tukey HSD test and the Newman-Keuls test. Indicate whether the results from these tests are in agreement. Show all steps taken.

Red Rock Reef = 7, 8, 15, 11, 9, 10

Ray's Reef = 12, 17, 13, 18, 19, 15

Shark's Cove = 14, 18, 19, 17, 16, 18

Mollusc Heaven = 19, 25, 22, 23, 18, 20

Answer 1

The Hypothesis:

H0: The Means for all the samples are same.

Ha: At least 1 mean differs from the others.

The Summary statistics is as below

	Red Rock	Rays	Sharks	Mollusc
Total	60	94	102	127
n	6	6	6	6
Mean	10.0	15.67	17.0	21.17
SS	40	39.33	16	34.83
Variance	8	7.8667	3.200	6.9667
SD	2.8284	2.8048	1.7889	2.6395

k = Number of groups = 4

n = samples per group = 6

N = Total Observations = 24

DF1 = DF Numerator = DF Between = k - 1 = 4 - 1 = 3

DF2 = DF Denominator = DF Error = N - k = 24 - 4 = 20

Overall Mean = (60 + 94 + 102 + 127) / 24 = 15.96

SS Between = SUM [n * (Mean - Overall mean)² = 6 * (10 - 15.96)² + 6 * (15.67 - 15.96)² + 6 * (17 - 15.96)² + 6 * (21.17 - 15.96)² = 382.79

MS Between = SS Between / DF between = 382.79 / 3 = 127.6

SS error = SUM (SS) = 40 + 39.33 + 16 + 34.83 = 130.17

MS Error = SS error / DF error = 130.17 / 20 = 6.51

F = MS Between / MS error = 127.6 / 6.51 = 19.61

Source	SS	DF	Mean Square	F
Between	382.79	3.00	127.60	19.61
Within/Error	130.17	20.00	6.51
Total	512.96	23.00

___________________________________________________________

(b) The F critical at = 0.05, DF between (DF1) = 3 and DF error (DF2) = 20 ; Fcv = 3.1

Also the p value for F = 19.61, DF between (DF1) = 3 and DF error (DF2) = 20 ; p value = 0.000

The Decision Rule: If F observed is > Fcv, Then Reject H0.

Also, if p value is < , Then Reject H0.

The Decision: Since F observed (19.61) is > Fcv (3.1), We Reject H0.

Also since p value (0.0000) is < (0.05, We Reject H0.

The Conclusion: There is sufficient evidence at the 0.05 significance level to conclude that at least 1 mean differs from the others.

_________________________________________________________

TUKEYS HSD

From the q table, for DF error = 20 and k = 4 for = 0.05, the q value = 3.96 (From Tukeys critical value tables)

The HSD = q value * SQRT(MS error / n)

MS error (MS within treatments) from the ANOVA = 6.51 and n = 6

Therefore HSD = 4.65 * SQRT(6.51/6) = 4.125

Thus the mean difference between any 2 samples must be at least 4.125 to be significant.

Red Rock - Rays = ABSOLUTE (10 - 15.67) = 5.67 which is > 4.125. There is a significant difference.

Red Rock - Shark = ABSOLUTE (10 - 17) = 7 which is > 4.125. There is a significant difference. .

Red Rock - Mollusc = ABSOLUTE (10 - 21.17) = 11.17 which is > 4.125. There is a significant difference.

Rays - Sharks = ABSOLUTE (15.67 - 17) = 1.33 which is < 4.125. There is not a significant difference.

Rays - Mollusc = ABSOLUTE (15.67 - 21.17) = 5.5 which is > 4.125. There is a significant difference.

Sharks - Mollusc = ABSOLUTE (17 - 21.17) = 4.17 which is > 4.125. There is a significant difference.

_______________________________________________________________

Newman Keuls Test

Arranging the means in ascending order

Red Rock = 10

Rays = 15.67

Shark = 17

Mollusc = 21.17

The Method: We take the largest mean vs the Smallest mean

We get the critical value from the studentized distribution, with p = Number of means (for eg mean 4 - Mean1 means 4 means are being taken), df = df error

The q value = (Difference in means) / SQRT (MS error / n)

If q value is > q critical, then there is a significant difference between means, else there is no difference between the means.

If there is no difference between Mean 1 and Mean 4, it will also be considered that there is no difference between mean 4 and mean 2 and Mean 4 and Mean 3.

Then we take the second largest mean vs smallest mean and repeat the process.

(i) For Mollusc - Red Rock = 21.17 - 10 = 11.17, p = Number of means in between = 4, = 0.05, DF error = 20

From the studentised range, q critical = 3.96

q = 11.17 / SQRT(6.51/6) = 10.72

Therefore q is > 3.96 and hence there is a significant difference between the means

(ii) For Mollusc - Rays = 21.17 - 15.67 = 5.5, p = Number of means in between = 3, = 0.05, DF error = 20

From the studentised range, q critical = 3.58

q = 5.5 / SQRT(6.51/6) = 5.28

Therefore q is > 3.58 and hence there is a significant difference between the means

(iii) For Mollusc - Shark = 21.17 - 17 = 4.17, p = Number of means in between = 2, = 0.05, DF error = 20

From the studentised range, q critical = 2.95

q = 4.17 / SQRT(6.51/6) = 4

Therefore q is > 2.95 and hence there is a significant difference between the means

(iv) For Shark - Red Rock = 17 - 10 = 7, p = Number of means in between = 3, = 0.05, DF error = 20

From the studentised range, q critical = 3.58

q = 7 / SQRT(6.51/6) = 6.72

Therefore q is > 3.58 and hence there is a significant difference between the means

(v) For Shark - Rays = 17 - 15.67 = 1.33, p = Number of means in between = 2, = 0.05, DF error = 20

From the studentised range, q critical = 2.95

q = 1.33 / SQRT(6.51/6) = 1.28

Therefore q is < 2.95 and hence there is not a significant difference between the means

(vi) For Rays - Red Rock = 15.67 - 10 = 5.67, p = Number of means in between = 2, = 0.05, DF error = 20

From the studentised range, q critical = 2.95

q = 1.33 / SQRT(6.51/6) = 5.44

Therefore q is > 2.95 and hence there is a significant difference between the means

_______________________________________________________________________

We see that results from both the tests are in agreement with each other