Question

Amber 66 69 70 66 Elizabeth Holly 66 69 70 71 71 71 70 69 69 68 The table gives the resting pulse for 3 different women, taken at multiple times. At a = 0.05, determine whether the women appear to have the same resting pulse. 1. The null hypothesis is Ho: Opi = 2 = 3 = 44 = 45 OH = uy = 43 OH = H2 = H3 = HA 2. This is a right tailed test with: d.f.n = and d.f.d = 3a. The STS (to 2 decimals) is F = 3b. The P-value (to 3 decimals) is: 4a. At a = 0.05, the decision is: O Accept Ha O Accept Ho O Reject Ho Do not reject Ho 4b. The conclusion is: There is sufficient evidence to conclude that the women have the same resting pulse There is insufficient evidence to conclude that the women have the same resting pulse There is sufficient evidence to conclude that the women do not have the same resting pulse There is insufficient evidence to conclude that the women do not have the same resting pulse

Answer 1

$1.~Null~hypothesis:~\mu_1=\mu_2=\mu_3.$

Excel output:

Anova: Single Factor
SUMMARY
Groups	Count	Sum	Average	Variance
Amber	4	271	67.75	4.25
Elizabeth	4	277	69.25	4.916667
Holly	6	417	69.5	1.5
ANOVA
Source of Variation	SS	df	MS	F	P-value	F crit
Between Groups	7.928571	2	3.964286	1.245918	0.325289	3.982298
Within Groups	35	11	3.181818
Total	42.92857	13

2. d.f.n=2, d.f.n.=11

3a. F= 1.25.

3b. P-value=0.325

4a. Do not reject H₀.

4b. There is sufficient evidence to conclude that the women do not have the same resting pulse.