Question

A purchasing manager for a large university is investigating which brand of LCD projector to purchase to equip "smart" classrooms. Of major concern is the longevity of the light bulbs used in the projectors. The purchasing manager has narrowed down the choice of projector to two brands, Infocus and Proxima, and wishes to determine if there is any difference between the two brands in the mean lifetime of the bulbs used. The purchasing manager obtained eight projectors of each brand for testing over the last several academic terms. The number of hours the bulbs lasted on each of the eight machines is given in the table. Lifetimes of light bulbs (hours) Infocus 951, 804,789, 773, 1021, 891, 1109, 1038 Proxima 899,878, 1225, 961, 889, 854,935, 755 X Send data E to Excel Assume that the two populations of lifetimes are normally distributed and that the population variances are equal. Can we conclude, at the 0.05 level of significance, that there is a difference in the mean lifetime of the light bulbs in the two brands? Perform a two-tailed test. Then fill in the table below. Carry your intermediate computations to at least three decimal places and round your answers as specified in the table. The null hypothesis: The alternative hypothesis: H :II c . 8 O=D OO 020 [+O OO O>o The type of test statistic: (Choose one) The value of the test statistic: (Round to at least three decimal places.) The two critical values at the 0.05 level of significance: (Round to at least three decimal places.) and I Can we conclude that there is a difference in the mean lifetimes of the light bulbs in the two brands? Yes No

Please answer neatly and correctly!

Answer 1

a)

Null hypothesis, Ho: u1 = u2

b)

Alternative hypothesis, h1: u1 =/= u2

c)

t test for independent samples with equal variances

d)

test statisitc,

	sample 1	sample 2
n=	8.000	8.000	COUNT
mean=	922.0000000	924.5000000	AVERAGE
s=	127.5270000	136.0231000	STDEV
s^2/n	2032.8920	2312.7855

Sp^2 = {(n1-1)*s1^2 + (n2-1)*s2^2} / {n1+n2-2}
Sp^2 = ((8-1)*127.527^2+(8-1)*136.0231^2)/(8+8-2)
Sp^2 = 17382.7097

Test statistic, t = (Xbar1 - Xbar2)/sqrt(Sp^2*(1/n1+1/n2))
t = (922-924.5)/SQRT(17382.70973*(1/8+1/8))
t = -0.038

e)
df= n1+n2-2 = 8+8-2 = 14

critical value:
t(a/2, df) = t(0.05/2,14)
t(a/2, df) = t.INV.2T(0.05,14)
t(a/2, df) = 2.145

f)

With |t| < t(A/2,df), i fail to reject ho and conclude that u1=u2

No