In: Statistics and Probability
Random samples of resting heart rates are taken from
two groups. Population 1 exercises regularly, and Population 2 does
not. The data from these two samples is given below:
Population 1: 71, 68, 67, 70, 70, 70, 67
Population 2: 69, 70, 73, 69, 76, 79, 75, 78
Is there evidence, at an α=0.02 level of significance,
to conclude that there those who exercise regularly have lower
resting heart rates? (Assume that the population variances are
equal.) Carry out an appropriate hypothesis test, filling in the
information requested.
A. The value of the standardized test
statistic:
Note: For the next part, your answer should use
interval notation. An answer of the form (−∞,a) is expressed
(-infty, a), an answer of the form (b,∞) is expressed (b, infty),
and an answer of the form (−∞,a)∪(b,∞) is expressed (-infty, a)U(b,
infty).
B. The rejection region for the standardized test
statistic:
C. The p-value is
D. Your decision for the hypothesis test:
A. Reject H0.
B. Do Not Reject H0
C. Do Not Reject H1.
D. Reject H1.
t-Test: Two-Sample Assuming Equal Variances | ||
Population1 | Population2 | |
Mean | 69 | 73.625 |
Variance | 2.666667 | 15.98214 |
Observations | 7 | 8 |
Pooled Variance | 9.836538 | |
Hypothesized Mean Difference | 0 | |
df | 13 | |
t Stat | -2.84931 | |
P(T<=t) one-tail | 0.006838 | |
t Critical one-tail | 2.281604 | |
P(T<=t) two-tail | 0.013675 | |
t Critical two-tail | 2.650309 |
A. The value of the standardized test statistic:
=((69-73.625)-0)/sqrt(2.66/6+15.98/7)
=-2.8
The value of the standardized test statistic: =-2.8
B. The rejection region for the standardized test statistic:
we reject Ho if T-stat < -2.28
C. The p-value is
P(T<=-2.8493) one-tail =0.0068
D. Your decision for the hypothesis test:
A. Reject H0.
since p-value < 0.02