Question

In: Statistics and Probability

Acme Products, a manufacturer of low-priced hair care products (as well as frozen dinners and large...

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.

Solutions

Expert Solution

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


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