Question

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: 70, 70, 71, 71, 70, 71, 62

Population 2: 69, 75, 75, 72, 75, 68, 70, 72

Is there evidence, at an ?=0.05

α

=

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.

A. The value of the standardized test statistic:

B. The p-value is

D. Your decision for the hypothesis test:

A. Do Not Reject ?0

H

0

.

B. Reject ?1

H

1

.

C. Do Not Reject ?1

H

1

.

D. Reject ?0

H

0

.

Answer 1

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