The following are pre-treatment and post-treatment systolic blood pressure measu
ID: 3359885 • Letter: T
Question
The following are pre-treatment and post-treatment systolic blood pressure measurements from a study sample of 15 individuals. Assume that systolic blood pressure measurements are approximately normally distributed. Post-treatment measurement Person Pre-treatment measurement 95 97 133 143 102 119 119 98 130 149 123 4 130 155 104 118 105 101 109 140 155 131 121 116 12 13 14 116 Construct a 98% confidence interval for D, the population mean difference between pre-treatment and post- treatment systolic blood pressure measurements. Interpret your interval Using a .01 significance level, test the hypothesis that the treatment has no effect on systolic blood pressure Interpret your test result. a) b)Explanation / Answer
Solution:
First of all we have to find Dbar and Sd required for further calculations. The table for calculations is given as below:
Pre treatment
Post treatment
Di
(Di - DBar)^2
95
99
-4
10.67111111
111
119
-8
0.537777778
97
98
-1
39.27111111
133
130
3
105.4044444
143
149
-6
1.604444444
102
123
-21
188.6044444
119
133
-14
45.33777778
111
109
2
85.87111111
130
140
-10
7.471111111
155
155
0
52.80444444
104
131
-27
389.4044444
118
121
-3
18.20444444
105
116
-11
13.93777778
101
111
-10
7.471111111
117
116
1
68.33777778
From above table, we have
Di = -109
n = 15
df = n – 1 = 15 – 1 = 14
Dbar = Di/n = 109/15 = -7.26667
(Di - DBar)^2 = 1034.9333
Var = (Di - DBar)^2/(n – 1) = 1034.9333/14 = 73.92381
Sd = sqrt(73.92381) = 8.597896
Part a
We have to find 98% confidence interval for µd.
We have
Dbar = -7.26667
Sd = 8.597896
n = 15
df = 15 – 1 = 14
Confidence level = 95%
Critical value = t = 2.6245
Confidence interval = Dbar -/+ t*Sd/sqrt(n)
Confidence interval = -7.26667 -/+ 2.6245*8.597896/sqrt(15)
Confidence interval = -7.26667 -/+ 2.6245*2.219968234
Confidence interval = -7.26667 -/+ 5.8263
Lower limit = -7.26667 - 5.8263 = -13.09
Upper limit = -7.26667 + 5.8263 = -1.44
We are 98% confident that the population mean difference will be lies within -13.09 and -1.44.
Part b
Here, we have to use paired t test.
H0: Treatment has no effect on systolic blood pressure.
Ha: Treatment has significant effect on systolic blood pressure.
H0: µd = 0 versus Ha: µd 0
This is a two tailed test.
We are given
Dbar = -7.26667
Sd = 8.597896
n = 15
df = 15 – 1 = 14
= 0.01
Test statistic formula is given as below:
Test statistic = t = Dbar/[Sd/sqrt(n)]
Test statistic = t = -7.26667/[8.597896/sqrt(15)]
Test statistic = t = -7.2667/2.2200
Test statistic = t = -3.2733
Lower critical value = -2.9768
Upper critical value = 2.9768
P-value = 0.0055
P-value < = 0.01
So, we reject the null hypothesis treatment has no effect on systolic blood pressure.
There is insufficient evidence to conclude that treatment has no effect on systolic blood pressure.
There is sufficient evidence to conclude that treatment has significant effect on systolic blood pressure.
Pre treatment
Post treatment
Di
(Di - DBar)^2
95
99
-4
10.67111111
111
119
-8
0.537777778
97
98
-1
39.27111111
133
130
3
105.4044444
143
149
-6
1.604444444
102
123
-21
188.6044444
119
133
-14
45.33777778
111
109
2
85.87111111
130
140
-10
7.471111111
155
155
0
52.80444444
104
131
-27
389.4044444
118
121
-3
18.20444444
105
116
-11
13.93777778
101
111
-10
7.471111111
117
116
1
68.33777778
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.