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 X1 = 90, s₁ = 5 and for the test group (9 rats) are X2 = 115,s₂ = 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 ( higher OR the same OR smaller)  mean blood pressure, since P-value ( > 0.05 OR < 0.05)

(b) Calculate 95% one sided CI to answer the claim in (a). Round your answer to 1 decimal place.

μ₁ - μ₂ ≤ -------

Answer 1

Part a)

To Test :-

H0 :- µ1 = µ2
H1 :- µ1 < µ2

Test Statistic :-


t = -6.6259

Decision based on P value
P - value = P ( t > 6.6259 ) = 0
Reject null hypothesis if P value < α level of significance
P - value = 0 < 0.05 ,hence we reject null hypothesis
Conclusion :- Reject null hypothesis

There is sufficient evidence to support the claim that the test group has higher mean blood pressure.

Part b)

Confidence interval :-
\( \bar{X}_{1} - \bar{X}_{2}) \pm t_{\alpha /2 , DF}\sqrt{ (S_{1}^{2}/n1) + (S_{2}^{2}/n2)}
t(α/2, DF) = t(0.05 /2, 12 ) = 2.179
( 90 - 115 ) \pm t_{ 0.05/2 , 12 }\sqrt{ ( 5^{2}/ 8 ) + ( 10^{2}/ 9 )}
Lower Limit = ( 90 - 115 ) - t_{ 0.05/2 , 12 }\sqrt{ ( 5^{2}/ 8 ) + ( 10^{2}/ 9 )}
Lower Limit = -33.2
Upper Limit = ( 90 - 115 ) + t_{ 0.05/2 , 12 }\sqrt{ ( 5^{2}/ 8 ) + ( 10^{2}/ 9 )}
Upper Limit = -16.8
95% Confidence interval is ( -33.2 , -16.8 )

Since the value in the confidence interval is negative, hence we reject the null hypothesis.