In what follows, use any of the following tests/procedures: Regression, multiple
ID: 2907588 • Letter: I
Question
In what follows, use any of the following tests/procedures: Regression, multiple regression, confidence intervals, one-sided t-test or two-sided t-test. All the procedures should be done with a 5% P-value or a 95% confidence interval.
Upload the data HeartRate_Exercise. These data are based on 45 randomly chosen high school students.
SETUP: It is believed that high school students who do not exercise have a heart rate (at rest) above 70. Given the data your job is to confirm or disprove this belief.
1. What test/procedure did you perform?
a. One-sided t-test
b. Two-sided t-test
c. Regression
d. Confidence interval
2. What is the P-value/margin of error?
a. 1.56075E-45
b. 3.1215E-45
c. 6.149555
d. 8.065421
e. None of these
3. Statistical interpretation
a. Since the P-value is very small, we are confident that the average heart rate is above 70.
b. Since the P-value is very small, we are very confident that the averages are different.
c. Since the P-value is very small, we are confident that the slope of regression line is not zero.
d. We are 95% certain that the confidence interval is [70.268, 86.399].
e. None of these
4. Conclusion
a. Yes, I am confident that the above assertion is correct.
b. No, we cannot claim that the above assertion is correct.
Explanation / Answer
A. One-sided t-test
D. 8.065421
D. We are 95% certain that the confidence interval is [70.268, 86.399].
A. Yes, I am confident that the above assertion is correct.
Ho: high school students who do not exercise have a heart rate (at rest) equal to 70. U = 70
H1: high school students who do not exercise have a heart rate (at rest) above 70. U > 70
Mean= 78.33
sd= 7.69
u= 70.00
n= 6.00
alpha= 5%
T(a,n-1)
t(0.05,6-1)
2.015
T = (mean-u)/(sd/sqrt(n))
= (78.333-70)/(7.68548/sqrt(6))
2.656
P-value
(1-P(T<t)
1-P(T<2.656)
t.Dist.Rt(2.656,6-1)
0.022550032
ME = t(a/2,n-1)*(sd/sqrt(n))
= 2.571*(7.68548415304245/sqrt(6))
= 8.07
CI = mean +-t(a/2,n-1)*(sd/sqrt(n))
lower = 78.33-2.571*(7.685/sqrt(6))= 70.27
upper = 78.33 + 2.571*(7.685/sqrt(6))= 86.40
Since the P value is less than 5%, I reject the null hypothesis and conclude that high school students who do not exercise have a heart rate (at rest) above 70. U > 70
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