Question

2. One personality dimension that seems well recognized by most people is that of introversion-extroversion. Extroverts are described as outgoing, sociable, and fun-loving, whereas introverts are described as reserved and less sociable. Because introverts seem more directed toward their own thoughts and ideas, we might suspect that introverts and extroverts may respond differently to external stimulation such as noise. To more fully investigate this issue, Standing, Lynn, and Moxness (1990) used a 2 ✕ 2 between-subjects design to vary personality type (factor A), introverts and extroverts, and background noise (factor B), quiet and noisy, while subjects performed a reading comprehension task. Suppose you conducted a similar study with 12 subjects per cell, and each person completed a 15 item true-false reading comprehension task. A person’s score was the number of items correctly answered. You obtained the following scores:

	Personality Type (A)
	Introvert	Extrovert
Quiet	10	15
	8	14
	12	12
	15	15
	10	11
	9	8
	14	12
	13	8
	14	13
	7	10
	12	9
	11	11
Background Noise (B)
Noisy	6	13
	8	8
	11	9
	4	14
	5	13
	8	12
	9	15
	10	11
	9	9
	10	11
	7	12
	11	10

2.a.What is the relationship between personality type and the effect of background noise? Use α= .05. If an interaction occurred, use the Tukey HSD test for the simple effects.

2.b.Plot the cell means with the appropriate labelling. You can draw this by hand.

2.c.Report the results as it would appear in a research paper.  

Answer 1

The two factor ANOVA analysis is performed in EXCEL by following these steps;

Step 1: Write the data value in excel. The screenshot is shown below;

A B C 1 Introvert Extrovert 2 Quiet 15 10 14 3 8 12 12 4 5 15 15 6 10 11 7 8 14 12

Step 2: Data>Data Analysis>Two-Factor With Replication>OK. The screenshot is shown below:

E INSERT PAGE LAYOUT FORMULAS DATA REVIEW Data Analysis ZA AZ Data Analysis X Analysis Tools OK Anova: Single Factor Anova: Two-Factor With Replication Anova: Two-Factor Without Replication Cancel Correlation Help Covariance Descriptive Statistics Exponential Smoothing F-Test Two-Sample for Variances Fourier Analysis Histogram

Step 3: Select Input Range: All the data, Rows per sample: 12. For Alpha value = 0.05 at 5% significance level. The screenshot is shown below;

A F 1 Introvert Extrovert 2 Quiet 10 15 8 14 4 12 12 5 Anova: Two-Factor With Replication 6 Input OK Input Range: 8 7 SAS1:SCS25 Cancel Rows per sample: 12 9 Help Alpha: 0.05 10 11 Output options 12 O Qutput Range 13 New Worksheet Ply: 14 New Workbook 15 LL LLI LO LO CT

The ANOVA table is obtained. The screenshot is shown below;

D E G 23 ANOVA 24 Source of Variation df p-value SS MS F crit 25 Sample 26 Columns 30.08333 1 30.08333 5.384407 0.025016 4.061706 36,75 1 36.75 6.577627 0.013815 4.061706 27 Interaction 27 1 27 4.832542 0.033232 4.061706 28 Within 44 5.587121 245.8333 29 30 Total 47 339.6667

F 1 Anova: Two-Factor With Replication 3 SUMMARY Introvert Extrovert Total 4 Quiet 5 Count 6 Sum 7 Average 12 12 24 135 138 273 11.375 11.25 11.5 8 Variance 6.386364 6.090909 5.983696 9 10 Noisy 11 Count 12 Sum 13 Average 14 Variance 12 12 24 98 137 235 8.166667 11.41667 9.791667 5.242424 4.628788 7.476449 15 16 Total 17 Count 18 Sum 19 Average 20 Variance 24 24 233 275 9.708333 11.45833 8.041667 5.128623 LL LO

a)

From the ANOVA table,

ANOVA
Source of Variation	P-value		$\alpha$	Decision
Interaction	0.033232	<	0.05	Significant

The P-value for the interaction between personality type and the effect of background noise is less than 0.05 at 5% significance level. Hence the null hypothesis is rejected. Now, it can be concluded that there is a significant relationship between these two factors.

Tuckey HSD test

Now, to test the significance of simple effects of both the factor, Tuckey HSD test is performed in following steps,

For personality type

The null hypothesis is defined as,

H₀: All the means for personality type are equal

The HSD for each comparison are obtained using the formula,

MSE HSD q

Where,

The q value is obtained from studentized q distribution table for number of groups, k = 2, degree of freedom = n-k = 48 - 2 = 46 and significance level = 0.05.

k-2,df-48-2-46, a-0.05 2.85

The MSE is obtained from the ANOVA for factor personality test,

$MSE=30.0833$

30.0833 MSE HSD 4 2.25625 = 2.85 x 48

The decision rule states that if;

12 HSD Difference is significant

From the ANOVA table,

	Introvert	Extrovert
Average	9.708333	11.45833

9.7083 11.45831.75

1 21.75< HSD-2.2563

Since the absolute difference is less than HSD, the null hypothesis is not rejected. It can concluded that there is no simple effect of personality type.

For background noise

The null hypothesis is defined as,

H₀: All the means for background noise are equal

The HSD for each comparison is,

MSE HSD q

Where,

The q value is obtained from studentized q distribution table for number of groups, k = 2, degree of freedom = n-k = 48 - 2 = 46 and significance level = 0.05.

k-2,df-48-2-46, a-0.05 2.85

The MSE is obtained from the ANOVA for factor personality test,

MSE 36.75

36.75 MSE HSD 4 - 2.4938 48 = 2.85 x

From the ANOVA table,

	Average
Quiet	11.375
Noise	9.791667

11.375 9.7917 1.5833

$|\mu_1-\mu_2|=1.5833<HSD=2.4938$

Since the absolute difference is less than HSD, the null hypothesis is not rejected. It can concluded that there is no simple effect of noise type.

b)

	Introvert	Extrovert	Average
Quiet	11.25	11.5	11.375
Noise	8.166667	11.41667	9.791667
Average	9.708333	11.45833

The plot is obtained in excel. The screenshot is shown below,

interaction plot 14 12 10 8 6 2 0 0.5 1 1.5 2 2.5 Noise Quiet co

c)

Interaction: The parallel plot shown no interaction while non parallel plot shows interaction between two groups. From the plot we can see that there is an interaction between two groups personality type and background noise.

Simple effect: For each line we can see that there is a very small difference between the groups of Quiet and and Noise as well which indicate very little of no simple effect.