Question
In: Statistics and Probability
10 research participants were administered an anxiety test under both noisy and quiet conditions. The data...
10 research participants were administered an anxiety test under both noisy and quiet conditions. The data are as follows: Using SPSS
|
Participant |
Noisy |
Quiet |
|
1 |
16 |
10 |
|
2 |
5 |
3 |
|
3 |
12 |
10 |
|
4 |
9 |
5 |
|
5 |
23 |
15 |
|
6 |
7 |
8 |
|
7 |
13 |
12 |
|
8 |
21 |
17 |
|
9 |
10 |
15 |
|
10 |
5 |
25 |
- What is the mean and standard deviation for anxiety in the quiet condition?
- What is the mean and standard deviation for anxiety in the noisy condition?
- What is the mean difference score?
- How would you report the finding of this analysis in a written report? Be sure to include both the numerical report and a written interpretation.
- Compute the value of eta-squared and interpret this value.
Solutions
Expert Solution
Solution:
A)
The sample standard deviation for quiet condition can be calculated as:
B)
The sample standard deviation for quiet condition can be calculated as:
C)
The mean difference score between quiet condition and the noisy condition is 0.1 (12.1 - 12)
D)
From the above information, it can be concluded that the mean for quiet and noisy are quite similar which is 12 and 12.1 respectively. Also, the standard deviation for quiet and noisy are 6.28 and 6.378 respectively. it means anxiety test under quiet conditions is pretty well as compare to noisy conditions.
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their vitamin C dosage. Using a=.05, analyze the data using the
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0mg
250mg
500mg
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2000mg
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