A patient who had undergone triple coronary artery bypass was tested in the foll
ID: 3040765 • Letter: A
Question
A patient who had undergone triple coronary artery bypass was tested in the following way. The patient took his pulse at rest and then on a treadmill, exercised at a 10% grade at 4.4 kilometres per hour for 2 minutes. This procedure was repeated on six different days. The pulse rate 6 minutes after the test and the pulse rate before the test were taken. The results are shown in the following table.
Pulse Rate (beats per minute)
Test
Pulse before
Pulse after
Difference in pulse rates
(Before – After)
1
69
85
-16
2
72
79
-7
3
75
83
-8
4
73
84
-11
5
70
87
-17
6
74
78
-4
Sample means
Sample sd
x1 = 72.17
s1 = 2.32
x2= 82.67
s2 = 3.5
x(diff)= -10.5
sdiff= 5.17
Pulse Rate (beats per minute)
Test
Pulse before
Pulse after
Difference in pulse rates
(Before – After)
1
69
85
-16
2
72
79
-7
3
75
83
-8
4
73
84
-11
5
70
87
-17
6
74
78
-4
Sample means
Sample sd
x1 = 72.17
s1 = 2.32
x2= 82.67
s2 = 3.5
x(diff)= -10.5
sdiff= 5.17
Explanation / Answer
H0 : µd =0
H1: µd < 0
t = d/[sdiff/n]
=-10.5/[5.17/6]
=-10.5/0.86167
=-11.837
ttab=t5,0.05 = -2.015 (from t table for df=5 at 5%)
Since, t (-11.837) < ttab(-2.015), hence we reject the null hypothesis and conclude that the population mean heart rate 6 minutes after the test is higher than before the test.
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