   Chapter 13, Problem 9P Essentials of Statistics for the B...

8th Edition
Frederick J Gravetter + 1 other
ISBN: 9781133956570

Solutions

Chapter
Section Essentials of Statistics for the B...

8th Edition
Frederick J Gravetter + 1 other
ISBN: 9781133956570
Textbook Problem

One of the primary advantages of a repealed-measures design, compared to an independent-measures design, is that it reduces the overall variability by removing variance caused by individual differences. The following data are from a research study comparing three treatment conditions. a. Assume that the data are from an independent- measures study using three separate samples, each with n = 6 participants. Ignore the column of P totals and use an independent-measures ANOVA with σ = .05 to test the significance of the mean differences. b. Now assume that the data are from a repealed- measures study using the same sample of n = 6 participants in all three treatment conditions. Use a repealed-measures ANOVA with α = .05 to test the significance of the mean differences. c. Explain why the two analyses lead to different conclusions. a.

To determine

To test: The significance of the mean difference using independent measure ANOVA with α=0.05.

Explanation

Given info:

The data is provided in the question.

 Treatment 1 Treatment 2 Treatment 3 N=18G=108∑X2=800 M=4 M=6 M=8 T=24 T=36 T=48 SS=42 SS=28 SS=34

Calculation:

Step 1:

The null and alternative hypotheses are:

Null hypothesis:

H0:μ1=μ2=μ3,

Where μ1μ2μ3 are the means of the 3 treatments.

Alternate hypothesis:

Ha: At least one treatment mean is different.

Step 2:

Compute the degrees of freedom (df) for the between treatment effects, within treatment effects and total and corresponding sum of squares (SS).

Now, it is known that, for a given sample size, the degrees of freedom (df) is:

df=(number of units)1.

Thus,

dftotal=N1=181=17

The number of treatments, k=3. Thus,

dfbetween=k1=31=2

As within treatments degrees of freedom, dfwithin=dftotaldfbetweentreatment, thus,

dfwithin=dftotaldfbetweentreatment=172=15

With α=0.05 and df=(2,15) critical value from table is CV=3.68

Step 3:

Compute F-ratio.

The total sum of squares is:

SStotal=X2G2N.

The within treatment sum of squares is:

SSwithintreatment=SSinsideeachtreatment.

The within treatment sum of squares is:

SSbetween=T2nG2N

b.

To determine
The significance of mean difference using repeated-measures ANOVA with α=0.05

c.

To determine

To find: Why the results of analysis differ in part a and part b.

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