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: 71, 67, 61, 62, 67, 70, 68

Population 2: 71, 68, 71, 78, 76, 73, 71, 68

Is there evidence, at an α=0.075, 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)(−∞,a) is expressed (-infty, a), an answer of the form (b,∞)(b,∞) is expressed (b, infty), and an answer of the form (−∞,a)∪(b,∞)(−∞,a)∪(b,∞) is expressed (-infty, a)U(b, infty).

B. The rejection region for the standardized test statistic:

C. The p-value is

Answer 1

For Population 1 :

∑x = 466

∑x² = 31108

n1 = 7

Mean , x̅1 = Ʃx/n = 466/7 = 66.5714

Standard deviation, s2 = √[(Ʃx² - (Ʃx)²/n)/(n-1)] = √[(31108-(466)²/7)/(7-1)] = 3.7796

For Population 2 :

∑x = 576

∑x² = 41560

n2 = 8

Mean , x̅2 = Ʃx/n = 576/8 = 72.0

Standard deviation, s2 = √[(Ʃx² - (Ʃx)²/n)/(n-1)] = √[(41560-(576)²/8)/(8-1)] = 3.5456

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A. Null and Alternative hypothesis:

Ho : µ1 = µ2

H1 : µ1 < µ2

Pooled variance :

S²p = ((n1-1)*s1² + (n2-1)*s2² )/(n1+n2-2) = ((7-1)*3.7796² + (8-1)*3.5456²) / (7+8-2) = 13.3626

Test statistic:

t = (x̅1 - x̅2) / √(s²p(1/n1 + 1/n2 ) = (66.5714 - 72) / √(13.3626*(1/7 + 1/8)) = -2.8694

df = n1+n2-2 = 13

B. Critical value, t crit = T.INV(0.075, 13) = -1.530

Rejection region: (-infinity, -1.530)

C. p-value = T.DIST(-2.8694, 13, 1) = 0.0066