Data:
n1 = 6
n2 = 6
x1-bar = 154.8333333
x2-bar = 396.5
s1 = 15.93005545
s2 = 18.52295873
Hypotheses:
Ho: μ1 ≥ μ2
Ha: μ1 < μ2
Decision Rule:
α = 0.05
Degrees of freedom = 6 + 6 - 2 = 10
Critical t- score = -1.812461102
Reject Ho if t < -1.812461102
Test Statistic:
Pooled SD, s = √[{(n1 - 1) s1^2 + (n2 - 1) s2^2} / (n1 + n2 - 2)] = √(((6 - 1) * 15.9300554508347^2 + (6 - 1) * 18.5229587269421^2) / (6 + 6 - 2)) = 17.27522311
SE = s * √{(1 /n1) + (1 /n2)} = 17.2752231051681 * √((1/6) + (1/6)) = 9.97385471
t = (x1-bar -x2-bar)/SE = (154.833333333333 - 396.5)/9.97385471007965 = -24.23001675
p- value = 1.63272E-10
Decision (in terms of the hypotheses):
Since -24.23001675 < -1.812461102 we reject Ho and accept Ha
Conclusion (in terms of the problem):
There is sufficient evidence that the mean blood pressure is significantly higher at 5 ° C than at 26 ° C
ANSWERS:
(a) t = -24.23
Rejection region t ≤ -1.8125
p- value = 0
Option (d)
(b)
Assumptions: Samples are drawn randomly from the populations The two samples are independent Each population is at least 10 times larger than its respective sample. Each sample is drawn from a normal or near-normal population (Skews are less for both samples). |
(c)
n1 = 6
n2 = 6
x1-bar = 154.83
x2-bar = 396.5
s1 = 15.93
s2 = 18.523
% = 95
Degrees of freedom = n1 + n2 - 2 = 6 + 6 -2 = 10
Pooled s = √(((n1 - 1) * s1^2 + (n2 - 1) * s2^2)/DOF) = √(((6 - 1) * 15.93^2 + ( 6 - 1) * 18.523^2)/(6 + 6 -2)) = 17.27521967
SE = Pooled s * √((1/n1) + (1/n2)) = 17.2752196657524 * √((1/6) + (1/6)) = 9.973852724
t- score = 2.228138842
Width of the confidence interval = t * SE = 2.22813884242587 * 9.97385272433209 = 22.22312866
Lower Limit of the confidence interval = (x1-bar - x2-bar) - width = -241.67 - 22.2231286637194 = -263.8931287
Upper Limit of the confidence interval = (x1-bar - x2-bar) + width = -241.67 + 22.2231286637194 = -219.4468713
The 95% confidence interval is [-263.89, -219.45]
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