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
10	4	1
11	9	5
9	6	4
7	5	2
6	8	8
5	4	4

(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.

(c) Which of the comparisons had significant differences? (Select all that apply.)

Recall following no delay was significantly different from recall following a one second delay.

The null hypothesis of no difference should be retained because none of the pairwise comparisons demonstrate a significant difference.

Recall following no delay was significantly different from recall following a half second delay.

Recall following a half second delay was significantly different from recall following a one second delay.

Answer 1

Answer(a):

	Delay Before Recall		Squared observation
	0	0.5	1		0	0.5	1
	10	4	1		100	16	1
	11	9	5		121	81	25
	9	6	4		81	36	16
	7	5	2		49	25	4
	6	8	8		36	64	64
	5	4	4		25	16	16
Total	48	36	24	GT=108	412	238	126	Σx2=776
Mean	8	6	4

We have to test

H₀: μ₀ = μ_0.5 = μ₁

H₁: at least one μ_i is different.

We have n= 18

            t=3

            r=6

Correction factor=

Correction factor=

Correction factor=648

Total sum of squares= sum of squares of every observation - correction factor

Total sum of squares= 776 – 648

Total sum of squares= 128

Treatment sum of squares=

Treatment sum of squares=

Treatment sum of squares=

Treatment sum of squares=48

Error sum of squares= Total SS – Treatment SS

Error sum of squares= 128 – 48

Error sum of squares= 80

ANOVA Table

Source of Variation	SS	df	MS	F
Between Groups	48	3-1=2	24	4.5
Within groups (Error)	80	15	5.33
Total	128	18-1=17

Answer(b):

The critical value for Tukey’s HSD can be obtained as below:

We have q=q_(0.95,3,17)= 3.627963

EMS =5.33

r=6

Hence the critical value is

Answer(c): we have pairwise mean difference for all treatment groups which is presented below. If the pairwise difference is greater than critical value, it is considered to be significantly different.

pair	Pair wise difference of means	S/NS
0-0.5	2	NS
0-1	4	S
0.5-1	2	NS

We can see that the treatment groups no delay and 1 second delay is significantly difference from each other.

Hence the correct option is 1. Recall following no delay was significantly different from recall following a one second delay.