Suppose you are interested in studying whether scent affects spatial reasoning abilities. You decide to test spatial reasoning using completion time scores for the paper-folding test with five people, repeating the test on each person with three different scents (cinnamon, lavender, and peppermint). In this experiment, the null hypothesis is that: O There are no differences in the mean completion times among the scents compared O There are no individual differences in the completion time means O The completion time mean for at least one scent is different from another The factor of interest is the , and the dependent variable is the The results of the study are presented in the following data table. All scores are times necessary to complete the paper-folding task, recorded in seconds. Scent Participant Cinnamon Lavender Peppermint Participant Totals 3.41 3.40 3.49 P = 10.30 = 3.29 3.47 3.25 P = 10.01 3 %3D 3.70 3.71 3.63 P = 11.04 15 D 3.68 4.08 3.88 P = 11.64 G 54.70 3.80 4.17 3.74 P = 11.71 ΣΧ2 200.4864 T = 17.88 T = 18.83 T 17.99 SS = 0.1858 SS = 0.4866 SS = 0.2335 The three treatments in the experiment define three populations of interest. You use an ANOVA to test the hypothesis that the three population means are equal. Present the results of your analysis in the following ANOVA table by entering the missing df values and selecting the correct values for the missing SS, MS, and F entries. (Note: For best results, retain at least two additional decimal points throughout your calculations. Depending on the order in which you do these calculations and the number of digits you retain, you may find slight rounding differences in the last digit between your answers and the answer choices.) ANOVA Table df MS F Source Between treatments 0.0755 Within treatments 0.1973 Between subjects Error 14 Total

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SS = 0.1858
The three treatments in the experiment define three populations of interest. You use an ANOVA to test the hypothesis that the three population means
are equal. Present the results of your analysis in the following ANOVA table by entering the missing df values and selecting the correct values for the
missing SS, MS, and F entries. (Note: For best results, retain at least two additional decimal points throughout your calculations. Depending on the
order in which you do these calculations and the number of digits you retain, you may find slight rounding differences in the last digit between your
answers and the answer choices.)
ANOVA Table
df
MS
F
Source
Between treatments
0.0755
Within treatments
Between subjects
0.1973
Error
Total
14
F Distribution
Numerator Degrees of Freedom -26
Denominator Degrees of Freedom -26
5000
5000
1
2
4
8
1.00
F
Use the Distributions tool to find the critical value for a = 0.01. The critical value is F =
At a significance level of a = 0.01, evaluate the null hypothesis that the population means for all treatments are equal. The null hypothesis is
You
conclude that scent affects spatial reasoning abilities.
Transcribed Image Text:SS = 0.1858 The three treatments in the experiment define three populations of interest. You use an ANOVA to test the hypothesis that the three population means are equal. Present the results of your analysis in the following ANOVA table by entering the missing df values and selecting the correct values for the missing SS, MS, and F entries. (Note: For best results, retain at least two additional decimal points throughout your calculations. Depending on the order in which you do these calculations and the number of digits you retain, you may find slight rounding differences in the last digit between your answers and the answer choices.) ANOVA Table df MS F Source Between treatments 0.0755 Within treatments Between subjects 0.1973 Error Total 14 F Distribution Numerator Degrees of Freedom -26 Denominator Degrees of Freedom -26 5000 5000 1 2 4 8 1.00 F Use the Distributions tool to find the critical value for a = 0.01. The critical value is F = At a significance level of a = 0.01, evaluate the null hypothesis that the population means for all treatments are equal. The null hypothesis is You conclude that scent affects spatial reasoning abilities.
Suppose you are interested in studying whether scent affects spatial reasoning abilities. You decide to test spatial reasoning using completion time
scores for the paper-folding test with five people, repeating the test on each person with three different scents (cinnamon, lavender, and peppermint).
In this experiment, the null hypothesis is that:
O There are no differences in the mean completion times among the scents compared
O There are no individual differences in the completion time means
O The completion time mean for at least one scent is different from another
The factor of interest is the
, and the dependent variable is the
The results of the study are presented in the following data table. All scores are times necessary to complete the paper-folding task, recorded in
seconds.
Scent
Participant
Cinnamon
Lavender
Peppermint
Participant Totals
3.41
3.40
3.49
P = 10.30
3.29
3.47
3.25
P = 10.01
%3D
3.70
3.71
3.63
P = 11.04
15
3.68
4.08
3.88
P = 11.64
G
54.70
3.80
4.17
3.74
P = 11.71
ΣΧ2
200.4864
T = 17.88
T= 18.83
T = 17.99
SS = 0.1858
SS = 0.4866
SS = 0.2335
The three treatments in the experiment define three populations of interest. You use an ANOVA to test the hypothesis that the three population means
are equal. Present the results of your analysis in the following ANOVA table by entering the missing df values and selecting the correct values for the
missing SS, MS, and F entries. (Note: For best results, retain at least two additional decimal points throughout your calculations. Depending on the
order in which you do these calculations and the number of digits you retain, you may find slight rounding differences in the last digit between your
answers and the answer choices.)
ANOVA Table
SS
df
MS
F
Source
Between treatments
0.0755
Within treatments
0.1973
Between subjects
Error
14
Total
Transcribed Image Text:Suppose you are interested in studying whether scent affects spatial reasoning abilities. You decide to test spatial reasoning using completion time scores for the paper-folding test with five people, repeating the test on each person with three different scents (cinnamon, lavender, and peppermint). In this experiment, the null hypothesis is that: O There are no differences in the mean completion times among the scents compared O There are no individual differences in the completion time means O The completion time mean for at least one scent is different from another The factor of interest is the , and the dependent variable is the The results of the study are presented in the following data table. All scores are times necessary to complete the paper-folding task, recorded in seconds. Scent Participant Cinnamon Lavender Peppermint Participant Totals 3.41 3.40 3.49 P = 10.30 3.29 3.47 3.25 P = 10.01 %3D 3.70 3.71 3.63 P = 11.04 15 3.68 4.08 3.88 P = 11.64 G 54.70 3.80 4.17 3.74 P = 11.71 ΣΧ2 200.4864 T = 17.88 T= 18.83 T = 17.99 SS = 0.1858 SS = 0.4866 SS = 0.2335 The three treatments in the experiment define three populations of interest. You use an ANOVA to test the hypothesis that the three population means are equal. Present the results of your analysis in the following ANOVA table by entering the missing df values and selecting the correct values for the missing SS, MS, and F entries. (Note: For best results, retain at least two additional decimal points throughout your calculations. Depending on the order in which you do these calculations and the number of digits you retain, you may find slight rounding differences in the last digit between your answers and the answer choices.) ANOVA Table SS df MS F Source Between treatments 0.0755 Within treatments 0.1973 Between subjects Error 14 Total
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