Will subjects instructed to complete null matches while viewing the Müller-Lyer illusion make different settings than subjects ask to set the variable component to be the same physical length as the fixed component of the illusion?
Two groups of subjects complete several tasks with a computerized version of the Müller-Lyer illusion using the method of adjustment . The data for one group of subjects is recorded when "null matching" instructions are used (n = 9). Data for the other group of subjects is recorded when "set to same physical length" instructions are used (n = 14).
20 trials are collected from each subject. The average "equality" setting is calculated.
Are the mean settings of the two groups sampled from the same population or different populations? Is µ1 = µ2?
Hypothesis Testing Model
Ho: µ1 = µ2
Ha: µ1 <> µ2
The Test Statistic
The test statistic is the F-statistic (aka "variance ratio")
The Sampling Distribution
If the Ho is true then the F value from the experiment will be drawn from the sampling distribution of F with n1 = 1 and n2 = 21 degrees of freedom.
Type-I Error rate
We'll use a 5% Type-I error rate.
ANOVA Source Table
|Source of Variation||SS||df||MS||F|
|Numerator of F||Between Groups||753.127337||1||753.127337||4.465473381|
|Denominator of F||Within Groups||3541.76875||21||168.6556548|
Since the F-value we got from the experiment (4.465) is greater than the critical value (4.38), we reject Ho. There is evidence the independent variable effected the mean setting.
There is evidence that subjects make different mean settings when the "matching" instructions were altered. Since the mean setting for instructions to set the variable stimulus to "physical equality" are closer to the length of the standard stimulus (130 pixels), we can conclude that subjects who know the manner in which the illusion affects settings can adjust the setting more toward physical equality.
Calculations for ANOVA (optional)
SSTotal: Calculate the est sigma for all 23 scores in the data set. Square the value and multiply by 22 to get the Total SS. [est. sigma = 13.9722; n = 23; SSTotal = (13.9722)2 X (23 - 1 ) = 4294.896]
SSWithin Groups: Calculate the SS for each group. Add the separate SS values together.
[SSW/I = 3003.454 + 538.315 = 3541.769]
SSBetween Groups = SSTotal - SSWithin Groups
MS = SS ÷ df [Remember: mean square (aka "Variance") is SS divided by df ].
F = MSBet ÷ MSW/I