Question

 

A pharmacist wants to test whether a new kind of sleeping pill would be effective to increase the hours of sleep for people who take it. A random sample of 10 persons (Group A) was given the new pills and another random sample of 13 persons (Group B) was given the old pills.

Their sleep in hours were recorded as follows:

Mean of group A = 8.9 and Standard Deviation of group A = 0.8

Mean of Group B = 8.5 and standard Deviation of group B = 0.5

Construct a 95% confidence interval for the difference of the average hours of sleep that would be obtained for the people taking the new pills over the people taking the old pills. State the assumptions made.

Answer 1

Given X1 bar = 8.9, S1 = 0.8 and n1 = 10

X2 bar = 8.5, S2 = 0.5 and n2 = 13

here samples are normally distributed and samples are independent

Equal variance test :

F stat = S1^2/S2^2 = 0.8^2/0.5^2 = 2.56

F critical : Fl = 0.259 and Fu = 3.436

F stat < Fu Critical

Equal variance is not rejected

95% Confidence Interval :

CI = (X1 bar - X2 bar) +/- tc * Se

Se = Sp * SQRT(1/n1 + 1/n2)

Sp = SQRT(((n1-1)S1^2 + (n2-1)S2^2)/(n1+n2-2))

First we have to find Sp = SQRT(((n1-1)S1^2 + (n2-1)S2^2)/(n1+n2-2))

Sp = SQRT(((10-1)0.8^2 + (13-1)0.5^2)/(10+13-2))

Sp = 0.646

Next Se = Sp * SQRT(1/n1 + 1/n2)

Se = 0.646 * SQRT(1/10 + 1/13)

Se = 0.272

and t critical value = 2.08 ( alpha = 0.05 and df = n1+n2-2 = 21)

CI = (X1 bar - X2 bar) +/- tc * Se

CI = (8.9-8.5) +/- 2.08 * 0.272

CI = 0.4 +/- 0.5658

CI = (-0.1658, 0.9658)

95% confidence interval for the difference of the average hours of sleep that would be obtained for the people taking the new pills over the people taking the old pills −0.165<μ1​−μ2​<0.965, Which means 95% confident that the true value of difference population mean contained by the interval