A medical researcher believes that the heart rates of people who have had a hear
ID: 3248385 • Letter: A
Question
A medical researcher believes that the heart rates of people who have had a heart attack are higher than those who have never had a heart attack. The researcher tests five people who have had a heart attack, and five who have not. There data is listed below
Heart attack: 115 125 111 128 139
No heart attack: 109 115 110 107 101
a) test the researcher's claim at the 90% confidence level
b) The researcher believes that exercise will reduce heart rates for patients who have had a heart attack. After putting the "heart attack" group above through a 6-week aerobics regimen, thier heart rates were as follows (given in the same order as above): 112 130 105 115 138. Test the researcher's claim at a 95% confidence level.
Explanation / Answer
To test the equality of population variance:
Heart attack/without exercise: x1 = 123.6 , s1 = 11.0815, s12 = 122.8, n1 =5
No heart attack: x2 = 108.4, s2 = 5.0794, s22 = 25.8, n2 =5
Heart attack/with exercise: x3 = 120, s3 = 13.5831, s32 =184.5, n3=5
H0 = The population variances are equal.
H1= The population variances are not equal.
F = s12 /s22 = 122.8/25.8 = 4.7597
F critical value at 10% significance level :
F/2,(4,4) = 6.3882
Since Fcal = 4.7597 < Ftab = 6.3882, we accept the null hypothesis and conclude that the population variances are equal.
To test the equality in population mean:
H0 =There is no significant difference between the heart rates of people who have had heart attack and who have never had a heart attack i.e., µ1 = µ2
H1 = The heart rates of people who have had a heart attack are higher than those who have never has a heart attack, i.e., µ1 > µ2
Since the population variances are equal, the test statistics is given by:
t = x1 - x2 / sp(1/n1 + 1/n2)
where sp is the pooled standard deviation.
Sp2 = (n1-1)s12 + (n2-1)s22 / (n1 + n2 -2)
= (4*122.8)+(4*25.8)/8
=74.3
Sp = 8.6197
t = 123.6-108.4/8.6197*0.6324
= 2.7884
= 1-0.90 = 0.10
d.f = 5+5-2 = 8
t critical value = t0.10,8 = 1.397
p-value = 0.011
since, tcal = 2.7884 > ttab = 1.397 and p-value < =0.10, we reject the null hypothesis and conclude that the heart rates of people who have had a heart attack are higher than those who have never has a heart attack.
Results after exercise
Heart attack/without exercise: x1 = 123.6 , s1 = 11.0815, s12 = 122.8, n1 =5
Heart attack/with exercise: x3 = 120, s3 = 13.5831, s32 =184.5, n3=5
To test the equality in population variance:
H0 = The population variances are equal.
H1= The population variances are not equal.
F = s12 /s32 = 122.8/184.5 = 0.6655
F critical value at 5% significance level :
F/2,(4,4) = 9.6045
Since Fcal = 0.6655 < Ftab = 9.6045, we accept the null hypothesis and conclude that the population variances are equal.
To test the equality in population means:
H0 =There is no significant difference between the heart rates of people after exercise i.e., µ1 = µ3
H1 = The heart rates of people have reduced after excercise, i.e., µ1 > µ3
Since the population variances are equal, the test statistics is given by:
t = x1 - x3 / sp(1/n1 + 1/n3)
where sp is the pooled standard deviation.
Sp2 = (n1-1)s12 + (n2-1)s32 / (n1 + n3 -2)
= (4*122.8)+(4*184.5)/8
=153.65
Sp = 12.39556
t = 123.6-120/12.3956*0.6324
= 0.459244
= 1-0.95 = 0.05
d.f = 5+5-2 = 8
t critical value = t0.05,8 = 1.860
p-value = 0.329
since, tcal = 0.4592 < ttab = 1.860 and p-value > =0.05, we accept the null hypothesis and conclude that there is no significant difference between the heart rates of people after exercise.
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