In: Math
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 |
6 | 5 | 7 |
13 | 2 | 2 |
10 | 10 | 5 |
7 | 5 | 5 |
8 | 8 | 4 |
10 | 6 | 1 |
(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.
Which of the comparisons had significant differences?
(Select all that apply.)
A.) The null hypothesis of no difference should be retained because none of the pairwise comparisons demonstrate a significant difference.
B.) Recall following no delay was significantly different from recall following a half second delay.
C.) Recall following a half second delay was significantly different from recall following a one second delay.
D.) Recall following no delay was significantly different from recall following a one second delay.
Solution:
its anova one way
Null hypothesis:Allthe group means are equal
Alternative Hypothesis:atleast one of the group means are different
Install analysis toolpak in excel and then go to
data>dataanalysis>Anova:singlefactor
you will get
Anova: Single Factor | |||||
SUMMARY | |||||
Groups | Count | Sum | Average | Variance | |
Delay | 7 | 54 | 7.714286 | 16.90476 | |
Before | 7 | 36.5 | 5.214286 | 10.65476 | |
Recall | 7 | 25 | 3.571429 | 5.285714 | |
ANOVA | |||||
Source of Variation | SS | df | MS | F | P-value |
Between Groups | 60.92857 | 2 | 30.46 | 2.78 | 0.088519 |
Within Groups | 197.0714 | 18 | 10.95 | ||
Total | 258 | 20 |
Intrepretation:
F(2,18)=2.78
p=0.0885
Statistical Decision:
p=0.0885
p>0.05
Do not reject null hypothesis.
Accept Null hypothesis.
Conclusion:
There is sufficient evidence at 5% level of signficnce to conclude that all the three group means are equal.
(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___3.61____ for each pairwise comparison.
for alpha=0.05
k=3 v=19
Q critical=3.61
Which of the comparisons had significant differences? (Select all that apply.)
Treatments pair | Tukey HSD Qstatistic | Tukey HSD p value | Tukey HSD Inference | |
Delay vs Before | 1.999 | 0.355805 | Insignifcant | |
Delay vs Recall | 3.3126 | 0.075148 | Insignifcant | |
Before vs Recall | 1.3136 | 0.62346 | Insignifcant |
mark only option A
A.) The null hypothesis of no difference should be retained because none of the pairwise comparisons demonstrate a significant difference.