We want to find out if advertisers adjust their ads based on the suspected educa
ID: 3360409 • Letter: W
Question
We want to find out if advertisers adjust their ads based on the suspected education level of their intended audience. We randomly selected ads from magazines that are typically read by people with a low level of education, medium level of education or high level education. For each ad we went through and counted the number of words that had 3 or more syllables.
Our worksheets lists the number of 3+ syllable words in each magazine grouped by education level. Use your spreadsheet program to conduct a one-way ANOVA on the data. You can assume the data meets all the assumptions required for a one-way ANOVA.
A. Should we:
reject the null hypothesis - the number of 3+ syllable words was significantly different in at least one of the magazine types.
fail to reject the null hypothesis - magazine type had no effect on the number of 3+ syllable words in the ads.
B. Let's do post-hoc comparisions between all the groups. Use t-tests to contrast the low-medium, medium-high, and low-high groups.
Before you conduct your t-tests you will need to use the "F-Test Two-Sample for Variances" tool. According to the results from this tool, how many of your contrasts will use the t-test that assumes equal variances?
C. To lower our chance of making a type I error we should use Bonferroni's correction to adjust our alpha level. Instead of rejecting the null hypothesis if p < .05, what will be our new requirement?
fail to reject the null hypothesis - magazine type had no effect on the number of 3+ syllable words in the ads.
Explanation / Answer
a)
Low
Medium
High
14
18
16
6
17
20
11
23
18
12
17
16
11
10
15
9
20
16
6
15
16
12
17
17
14
25
11
11
11
13
4
18
22
9
18
16
7
20
10
12
21
18
12
27
9
17
13
19
11
22
16
8
17
12
X1
X2
X3
10.33333333
18.27777778
15.55555556
X
14.72222222
SSA=18(10.33-(14.72))^2……18(15.56-(14.72))^2=586.78
SSW=(2-(4.3))^2….(8-(6.6))^2=722.06
MSA=586.78/(4-1)=293.39
MSW=722.06/(54-3)=14.16
F=293.39/14.16=20.72
Anova: Single Factor
SUMMARY
Groups
Count
Sum
Average
Variance
Low
18
186
10.33333333
10.70588235
Medium
18
329
18.27777778
19.85947712
High
18
280
15.55555556
11.90849673
ANOVA
Source of Variation
SS
df
MS
F
P-value
F crit
Between Groups
586.7777778
2
293.3888889
20.72255136
0.0000
3.178799292
Within Groups
722.0555556
51
14.15795207
Total
1308.833333
53
Reject H0 since F>Fcrit. There is evidence that at least one j differs from the rest
b)
Low
Medium
High
14
18
16
6
17
20
11
23
18
12
17
16
11
10
15
9
20
16
6
15
16
12
17
17
14
25
11
11
11
13
4
18
22
9
18
16
7
20
10
12
21
18
12
27
9
17
13
19
11
22
16
8
17
12
10.70588235
19.85947712
11.90849673
Variance
3.271984467
4.456397326
3.450868982
Std Dev
F Test
H0: 21 = 22 (there is no difference between variances)
H1: 21 22 (there is a difference between variances)
F.025, 17, 17 = 2.67
FSTAT=S1^2/S2^2 (Low/Medium)
=10.71/19.86
=0.54
FSTAT does not lie in the rejection region and hence cannot reject the null hypothesis
FSTAT=S1^2/S2^2 (Medium/High)
=19.86/11.91
=1.67
FSTAT does not lie in the rejection region and hence cannot reject the null hypothesis
FSTAT=S1^2/S2^2 (High/Low)
=11.91/10.71
=1.11
FSTAT does not lie in the rejection region and hence cannot reject the null hypothesis
Hence, we would have to conduct t test for all of the combinations
Low-Medium
H0: 1 - 2 = 0 i.e. (1 = 2)
H1: 1 - 2 0 i.e. (1 2)
Assuming population variances are equal, we would have to calculate pooled-variance t-Test
Sp^2= (n1-1)S1^2+(n2-1)S2^2/(n1-1)+(n2-1)
= (18-1)*3.27^2+(18-1)*4.456^2/17+17
= 182+337.61/34
= 15.28
tSTAT=(X1-X2)-(µ1-µ2)/Sp^2(1/n1+1/n2)
=(10.33-18.28)-0/15.28(1/18+1/18)
=-7.944/1.30
=-6.1
tCRIT is +/- 2.03 and hence reject the null hypothesis
Medium-High
H0: 1 - 2 = 0 i.e. (1 = 2)
H1: 1 - 2 0 i.e. (1 2)
Assuming population variances are equal, we would have to calculate pooled-variance t-Test
Sp^2= (n1-1)S1^2+(n2-1)S2^2/(n1-1)+(n2-1)
= (18-1)*4.456^2+(18-1)*3.45^2/17+17
= 337.61+202.44/34
= 15.88
tSTAT=(X1-X2)-(µ1-µ2)/Sp^2(1/n1+1/n2)
=(18.28-15.5)-0/15.88(1/18+1/18)
=2.72/1.328
=2.049
tCRIT is +/- 2.03 and hence reject the null hypothesis
High-Low
H0: 1 - 2 = 0 i.e. (1 = 2)
H1: 1 - 2 0 i.e. (1 2)
Assuming population variances are equal, we would have to calculate pooled-variance t-Test
Sp^2= (n1-1)S1^2+(n2-1)S2^2/(n1-1)+(n2-1)
= (18-1)*3.45^2+(18-1)*3.27^2/17+17
= 202.44+182/34
= 11.31
tSTAT=(X1-X2)-(µ1-µ2)/Sp^2(1/n1+1/n2)
=(15.56-10.33)-0/11.31(1/18+1/18)
=5.22/1.12
=4.66
tCRIT is +/- 2.03 and hence reject the null hypothesis
c)
In order to lower our chance of making a type I error, we should reduce the size of the error that is allowed (alpha) for each comparison by the number of comparisons and as a result of that we would have overall alpha which does not exceed the desired limit. Since, here there are three comparisions and a=0.05, using Bonferroni's correction, we should get 0.05/3=0.0167 and hence all that should sum up to 0.05.
Low
Medium
High
14
18
16
6
17
20
11
23
18
12
17
16
11
10
15
9
20
16
6
15
16
12
17
17
14
25
11
11
11
13
4
18
22
9
18
16
7
20
10
12
21
18
12
27
9
17
13
19
11
22
16
8
17
12
X1
X2
X3
10.33333333
18.27777778
15.55555556
X
14.72222222
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