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- delay before responding

0.5 10 Total (b) C Tukey's HSD post hoc test and interpret the results. (Assume alpha equal to 0.05. Round your answer to two for each pairwise comparison. The null hypothesis of no difference should be retained because none of the pairwise co

Answer 1

ANOVA

Since we have to compare mean of three different treatments , we have to perform ANOVA analysis .

We can perform ANOVA using excel .

Go to Data menu ---> data analysis ----> Select Anova: Single Factor

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For Input range select the entire data table ( with headings 0,0.5,1 ).

Check the box for "Labels in the first Row" . ( If you did not selected the column heading 0,0.5,1 then do not need to check this box )

Then click on OK.

..

0.5 Anova: Single Factor 7Input OK 5 Input Range: 2 10 Cancel 4 Columns Grouped By: Help 2Labels in First Row -Alpha: 0.05 Output options O Qutput Range: O New Worksheet Ply: O New Workbook 10 12 14

The analysis output will be as follows :

Anova: Single Factor SUMMARY Groups Count Sum Average Variance 5.6 6.4 4 0 36 24 0.5 ANOVA Source of Variation Between Groups Within Groups MS P-value F crit 24 4.5 0.029451 3.682320344 80 15 5.333333 Total 128 17

a)

(a) Complete the F-table. (Round your values for MS and F to two decimal places) Source of Varlation 48 witin groups Cerror) 8015 5.33 Between groups 2 24 4.5 128 | 17 Total ma ainha

b)Tukey’s HSD Post hoc test critical value formula is as follows,

Critical value = $q\sqrt{\frac{MS_{error}}{n}}$

q is table value for given level of significance α

n is smallest sample size of a group.

We are given α = 0.05 ,we have MS_error = 5.33 df_error = 15 and n = 6 , k = number of groups = 3

Therefore q = 3.67 ( from table )

Critical value =    $3.67*\sqrt{\frac{5.33}{6}}$

Critical value = 3.46

Decision rule : If the difference between group means is greater than critical value then we reject H₀ and we conclude that there is significant difference between corresponding groups.

If the difference between group means is less than critical value then we fail to reject H₀ and we conclude that there is no significant difference between corresponding groups.

We have $\bar{X}$₁ = mean of group 0 sec = 8 ,$\bar{X}$₂ = mean of group 0.5 sec = 6 and $\bar{X}$₃ = mean of group 1 sec = 4  

| $\bar{X}$₁ -$\bar{X}$₂ | = 2 < 3.46

| $\bar{X}$₂ -$\bar{X}$₃ | = 2 < 3.46

| $\bar{X}$₁ -$\bar{X}$₃ | = 4 > 3.46

As difference between mean of group 0 sec and 1 sec is greater than critical value 3.46, they are significantly different.

Option 2) Recall following no delay was significant different from recall following a one second delay.