In: Statistics and Probability
In this assignment, you will use the method of comparing confidence intervals to test for a significant difference between the resting heart rate of males and females. You have observed that the mean rates are not exactly the same but are they significantly different?
Steps
Additional Instructions:
Your assignment should be typed into a Word or other word processing document, formatted in APA style. The assignments must include
Heart rate before and after exercise
M=0 F=1 Resting After Exercise
0 85.9 87.5
0 67.7 79.4
0 80.3 93.4
0 85.2 97.7
0 86.3 99.7
0 76.6 83.7
0 94.4 101.9
0 86.4 100.6
0 83.4 97.4
0 89.8 97.4
0 88.7 97.1
0 78.4 87.2
0 71.3 79.9
0 92.6 104.7
0 86.2 95.9
0 83.9 93.9
0 78.1 90.1
0 64 70.7
0 72.8 86.7
0 72.7 81.2
0 80.2 83.3
0 78.2 86
0 70.6 90.2
0 75.5 84.4
0 82.7 94.2
0 87.7 95.1
0 80 88.7
0 73.4 82.7
0 89.5 94.6
0 77.6 84.6
0 76.6 86.4
0 85.6 96.2
0 74.2 82.1
0 79 91.6
0 74.6 86.7
0 88.8 98.8
0 82.1 85.6
0 77.6 80.6
0 77.9 83.8
0 88.1 93.9
0 81.6 90.3
0 91.2 100.6
0 80.3 88
0 76.7 91.8
0 88.4 103
0 75.2 86.5
0 75.2 84.9
0 73.1 71.9
0 77 84.7
0 59 68.2
0 84.9 96
0 87.5 105.9
0 75.6 84.3
0 84 90.4
0 78.2 94
0 86.6 90.6
0 84.9 95.1
0 78.8 90.4
0 69.4 82.6
0 78.3 91.1
0 76.9 92.3
0 84.2 87.9
0 76.3 85.9
0 86.3 99.7
0 72.3 80.9
0 81.8 93.8
0 92.8 99.8
0 74.8 90.2
0 91.7 99.2
0 71 87
0 96.1 100.2
0 82.5 95.1
0 81.9 97.5
0 89.7 94.8
0 81.4 100.9
0 74.8 94
0 88.1 102.1
0 69.2 81.4
0 78.8 90.9
0 85.3 94.2
0 74.8 81.3
0 77.7 89.9
0 78 89.8
0 80.5 95.3
0 75.4 84.8
0 81.5 84.2
0 73.9 85.2
0 69.4 74.1
0 89.4 96.7
0 70.9 82
0 82.9 90.2
0 89.6 106.7
0 74.5 75.6
0 92.3 102.2
0 87.7 98
0 78.9 89.7
0 79.8 81.5
0 85.5 97.4
0 87.3 94.1
0 77.8 97.8
0 71 80.1
0 82.5 90.7
0 74.8 83.7
0 69.2 79.4
0 80.5 87.4
0 89.4 99.2
0 74.5 88
0 85.5 92
1 76.6 88.2
1 79.2 90.4
1 80.6 101.3
1 75.5 93.1
1 83.9 90.5
1 73.9 89.1
1 76.8 90.8
1 85.2 93.5
1 82.1 93.5
1 76.3 87
1 97 104.5
1 81.5 86.5
1 65.3 86.3
1 80.8 86.7
1 78.5 89.9
1 86.3 97.6
1 89.8 92.9
1 87.8 98.5
1 76.2 89.9
1 74.2 88.8
1 67.4 78.8
1 75.5 80.2
1 80 90.2
1 76.4 88
1 94.9 95.7
1 89.2 96.9
1 83.3 87.7
1 85.8 90.4
1 75.3 84.1
1 77.9 99
1 70 83
1 88 94.2
1 86.9 95
1 87.1 95.9
1 79.3 82.7
1 81.2 90.7
1 82.9 91.9
1 87.4 103.6
1 83 90
1 76.8 83.3
1 76.9 87.7
1 79.8 88.2
1 83.2 93
1 79.5 88.6
1 82.4 89.3
1 80.8 84.2
1 83.2 94.5
1 71.6 81.5
1 82.8 93.1
1 76.8 92.8
1 93.2 100.4
1 91.4 100.9
1 97.3 103.3
1 88.3 90.1
1 80.6 85.2
1 87.4 91.7
1 96.5 99.3
1 77.9 91.6
1 76.1 84.1
1 85.2 89.7
1 68.6 72.8
1 79.4 92
1 85.2 99.2
1 74.3 85.6
1 74.3 89.2
1 78.5 98.5
1 80.4 90.8
1 82.9 85.9
1 78.9 90.7
1 78.6 87
1 87.5 93.9
1 78.9 91.4
1 80 89.1
1 80.4 89.2
1 88.3 93.5
1 80.6 95.9
1 85.8 90.5
1 84.6 93
1 80.5 91.8
1 92.4 101.2
1 84.4 96.7
1 82.3 86.9
1 77.2 85.8
1 83.3 82.1
1 86.2 98.9
1 81.3 97.7
1 90.2 96.4
1 78.4 85.5
1 84.7 101.6
1 89.7 94.3
1 78.4 88
1 79.9 88.5
Soln
From the data, we calculate the mean and standard deviation of the two groups for Resting.
Mean (X bar) = Sum of Values /n
and
Group 1 (Male)
Mean (x1) = 80.39
Std Dev (s1) = 7.05
n1 = 108
Group 2 (Female)
Mean (x2) = 81.77
Std Dev (s2) = 6.27
n2 = 92
alpha = 0.05
ZCritical = 1.96
Null and Alternate Hypothesis
H0: µ1 = µ2 (Mean Heart Rate is same for both Males and Female)
Ha: µ1 <> µ2 (Mean Heart Rate is not same for both Males and Female)
Test Statistic
Assuming, the population std deviation is not same.
t = (X1 – X2 – (µ1 - µ2))/ (s12/n1 + s22/n2 )1/2 = -1.47
95% CI for Difference of Means = (X1 – X2 ) +/- ZCritical *(s12/n1 + s22/n2 )1/2 = {-3.23, 0.46}
P-value = TDIST(1.47,108+92-2,2) = 0.143
Result
Since the test statistic lies in the 95% CI, we fail to reject the null hypothesis.
Conclusion
Rested Mean Heart Rate is same for both Males and Female