An article in IEEE International Symposium on Electromagnetic Compatibility (Vol
ID: 3155193 • Letter: A
Question
An article in IEEE International Symposium on Electromagnetic Compatibility (Vol. 2, 2002, pp. 667-670) describes the quantification of the absorption of electromagnetic energy and the resulting thermal effect from cellular phones. The experimental results were obtained from in vivo experiments conducted on rats. The arterial blood pressure values (mmHg) for the control group (8 rats) during the experiment are x_1 = 90, s_1 = 5 and for the test group (9 rats) are x_2 = 115,s_2 = 10. (a) Is there evidence to support the claim that the test group has higher mean blood pressure? Assume that both populations are normally distributed but the variances are not equal. Answer this question by finding the P-value for this test. The test group has mean blood pressure, since P-value? (b) Calculate 95% one sided CI to answer the claim in (a). Round your answer to 1 decimal place. mu_1 - mu_2 lessthanorequalto.Explanation / Answer
B)
Calculating the means of each group,
X1 = 90
X2 = 115
Calculating the standard deviations of each group,
s1 = 5
s2 = 10
Thus, the standard error of their difference is, by using sD = sqrt(s1^2/n1 + s2^2/n2):
n1 = sample size of group 1 = 8
n2 = sample size of group 2 = 9
Thus, df = n1 + n2 - 2 = 15
Also, sD = 3.773077141
For the 0.95 confidence level, then
alpha = (1 - confidence level) = 0.05
Hence,
t(alpha) = 1.753050356
Thus,
upper bound = [X1 - X2] + t(alpha/2) * sD = -18.38560578
Thus,
u1 - u2 <= -18.38560578 [ANSWER]
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Hi! If you use another method/formula in calculating the degrees of freedom in this t-test, please resubmit this question together with the formula/method you use in determining the degrees of freedom. That way we can continue helping you! Thanks!
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