Part 2. Use the -Test Data (PRT) for Question 2\" file to answer Questions 6-10.
ID: 2933263 • Letter: P
Question
Part 2. Use the -Test Data (PRT) for Question 2" file to answer Questions 6-10. In this data set, there was one group of patients who was treated with PRT. You recorded pre-treatment and post-treatment measures for the NRS and PSFS, which allowed you to calculate change scores for each outcome measure. You want to determine if there are differences between the patients' pre and post measurement of pain as measured using the NRS). 6, which type of t-test is appropriate for the scenario? 7. What is the computed t statistic? 8. What are the degrees of freedom? 9. What is the p value (significance)? an (e.g., statistical significance, clinical significance, etc.]? How might you write-up the meaning for dissemination?Explanation / Answer
Solution:
Here, we have to use paired t test.
The null and alternative hypothesis for this test is given as below:
Null hypothesis: H0: There is no any statistically significant difference exists between the patient’s pre and post measurement of pain.
Alternative hypothesis: Ha: There is a statistically significant difference exists between the patient’s pre and post measurement of pain.
H0: µpre = µpost
Ha: µpre µpost
This is a two tailed test.
We assume = 0.01 (level of significance).
The test statistic formula is given as below:
t = Dbar / [ Sd / sqrt(n) ]
Calculation table for Dbar and Sd is given as below:
NRS Pre
NRS Post
Di
(Di - DBar)^2
8
3
5
1.8225
4
1
3
0.4225
6
3
3
0.4225
4
2
2
2.7225
7
4
3
0.4225
5
0
5
1.8225
6
1
5
1.8225
4
2
2
2.7225
7
4
3
0.4225
3
0
3
0.4225
8
4
4
0.1225
6
2
4
0.1225
6
3
3
0.4225
5
2
3
0.4225
8
5
3
0.4225
4
0
4
0.1225
5
0
5
1.8225
6
1
5
1.8225
4
1
3
0.4225
7
2
5
1.8225
From the above table, we get
Dbar = 3.6500
Sd = 1.0400
n = 20
df = n – 1 = 20 – 1 = 19
t = 3.65 / [1.04/sqrt(20)]
t = 15.6956
Critical values = -2.093 and 2.093
P-value = 0.0000
P-value < = 0.01
So, reject the null hypothesis that there is sufficient evidence to conclude that there is a statistically significant difference exists between the patient’s pre and post measurement of pain.
Answers:
Question 6
Paired t-test is appropriate for the scenario.
Question 7
Computed t statistic = 15.6956
Question 8
Degrees of freedom = n – 1 = 20 – 1 = 19
Question 9
P-value = 0.0000
Question 10
P-value < = 0.01
So, we reject the null hypothesis
There is sufficient evidence to conclude that there is a statistically significant difference exists between the patient’s pre and post measurement of pain.
NRS Pre
NRS Post
Di
(Di - DBar)^2
8
3
5
1.8225
4
1
3
0.4225
6
3
3
0.4225
4
2
2
2.7225
7
4
3
0.4225
5
0
5
1.8225
6
1
5
1.8225
4
2
2
2.7225
7
4
3
0.4225
3
0
3
0.4225
8
4
4
0.1225
6
2
4
0.1225
6
3
3
0.4225
5
2
3
0.4225
8
5
3
0.4225
4
0
4
0.1225
5
0
5
1.8225
6
1
5
1.8225
4
1
3
0.4225
7
2
5
1.8225
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