Question

Iconic memory is a type of memory that holds visual information for about half a second (0.5 seconds). To demonstrate this type of memory, participants were shown three rows of four letters for 50 milliseconds. They were then asked to recall as many letters as possible, with a 0-, 0.5-, or 1.0-second delay before responding. Researchers hypothesized that longer delays would result in poorer recall. The number of letters correctly recalled is given in the table.

Delay Before Recall
0	0.5	1
9	5	2
11	4	5
5	9	7
7	7	3
6	3	4
10	8	3

(a) Complete the F-table. (Round your values for MS and F to two decimal places.)

Source of Variation	SS	df	MS	F
Between groups
Within groups (error)
Total

(b) Compute Tukey's HSD post hoc test and interpret the results. (Assume alpha equal to 0.05. Round your answer to two decimal places.)

The critical value is  for each pairwise comparison.

Answer 1

The table below gives the data which has been calculated from the given figures

	0	0.5'	1'
Total	48	36	24
n	6	6	6
Mean	8	6	4
Sum Of Squares	124	28	18.223778

The Hypothesis:

H0: There is no difference between the mean recall rates for the three different delays.

Ha: There is a difference between the mean recall rates for the three different delays.

Source	SS	DF	Mean Square	F
Between	48	2	24	2.12
Within/Error	170.23	15	11.35
Total	218.23	17

the p value is calculated for F = 2.12 for df1 = 2 and df2 = 15 , p value = 0.1552

The Fcritical is calculated at = 0.05 for df1 = 2 and df2 = 15; Fcritical = 3.6823

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Tukeys Post Hoc Table

Where Mi and Mj are the 2 means being compared and their positive (absolute) differences is taken.

n = number of replicates in each sample. Here n = 6, MSerror = 10.39

The Rule is that if Tukeys observed is > Tukeys critical, then there is a significant difference between groups.

Tukeys critical is found from the critical value tables for = 0.05, for k (# of columns) = 3 on the horizontal and df error = 14 on the vertical.

The Critical Value = 3.70

M1 = 8, M2 = 6, M3 = 4

The Tabular Format is as below, taking 2 groups at a time.

Tire	Mean
A	8
B	6
C	4
Mserror	11.35
n	6
Sqrt(Msbetween/n)	1.375

	Absolute Difference	Observed	Critical	Obs > Crit
M1 - M2	2	1.4545	3.7	No
M1 - M3	4	2.9091	3.7	No
M2 - M3	2	1.4545	3.7	No

Pairwise comparison shows that there is no significanct difference between any 2 means and hence, we would be failing to reject the null hypothesis, that all the 3 means are the same.
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Calculations For the ANOVA Table:

Overall Mean = [(6*8) + (6*6) + (6 * 4) + (15*40.2) ] / 18 = 108 / 18 = 6

SS treatment = SUM n* ( x̅_i - overall mean)² = 6 * (8 - 6)² + 6 * (6 - 6)² + 6 * (4 - 6)² = 48

df1 = k - 1 = 3 - 1 = 2

MSTR = SS treatment/df1 = 48 / 2 = 24

SSerror = SUM [SS] = 124 + 28 + 18.23 = 170.23

df2 = N - k = 18 - 3 = 15

Therefore MS error = SSerror/df2 = 170.23 / 15 = 11.35

F = MSTR/MSE = 24 / 11.35 = 2.12