Question

Acme Products, a manufacturer of low-priced hair care products (as well as frozen dinners and large screen TVs), claims that its hair spray lasts as long as the industry leader’s higher priced product. You take a sample of 60 applications of Acme’s hair spray and 60 applications of the leading hair spray. Results: On average, the Acme applications lasted for 12.7 hours with a standard deviation of 1.4 hours while the leading hairspray lasted 11.8 hours with a standard deviation of 1.8 hours. Use a 5% significance level. Assume that the standard deviations are KNOWN TO BE POPULATION STANDARD DEVIATIONS.

Answer 1

X1 bar	12.7		X2 bar	11.8
σ1	1.4		σ2	1.8
n1	60		n2	60

Hypothesis :		α=	0.05
Ho:	μ1​ = μ2			X1 bar
Ha:	μ1​ > μ2
Z Critical Value :
Zc	1.644853627	NORM.S.INV(1-α)	RIGHT
Zs	>=	Zc	RIGHT	To reject
Test :
Z	3.057147992	(X1 bar-X2 bar )/SQRT(σ1^2/n1+σ2^2/n2)
P value :
P value	0.00111727	1-NORM.S.DIST(z,true)	RIGHT
Decision :
P value	<	α	Reject H0

P value < 0.05, Reject H0

There is enough evidence to conclude that Acme’s hair spray lasts as long as the industry leader’s higher priced product at 5% significance level