In: Statistics and Probability
Answer:
Given that,
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 are given below:
Population 1:
70, 70, 71, 71, 70, 71, 62
Population 2:
69, 75, 75, 72, 75, 68, 70, 72
Is there evidence, at an =0.05 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:
Hypothesis test:
Null hypothesis:
Alternative hypothesis:
Calculation table for population 1:
Mean ():
=485/7
=69.2857
Standard Deviation ():
=(70 - 69.2857)2 + ... + (62 - 69.2857)2/7
=63.42857/7
=9.06122449
=√9.06122449
= 3.010186787
=3.0102 (Approximately)
Population 1 | |
69.2857 | |
3.0102 | |
7 |
Calculation table for population 2:
Mean ():
=576/8
=72
Standard Deviation ():
=(69 - 72)2 + ... + (72 - 72)2/8
=56/8
=7
=√7
= 2.64575131
=2.6458 (Approximately)
Population 2 | |
72 | |
2.6458 | |
8 |
Point estimate= -
=69.2857-72
=-2.7143
Degree of freedom (df)=min (,)-1
=min(7,8)-1
=6
For 0.05 level of significance, left tail and 6 df, the critical value is t=-1.943
Decision rule:
Reject if test statistic t < 1.943
Standard Error (SE):
=1.47295621
=1.4729
Test stat t
[Since, ]
=-2.7143/1.4729
=-1.8428
The p-value:
The p-value is 0.0575.
The result is not significant at p < 0.05.
From above:
(A). Test statistic t =-1.8428
(B) p-value =0.0575
(C). Since p-value < 0.05
Option (D): reject Ho