Question

Allied Corporation is trying to sell its new machines to Ajax. Allied claims that the machine will pay for itself since the time it takes to produce the product using the new machine is significantly less than the production time using the old machine. To test the claim, independent random samples were taken from both machines. You are given the following results.

New Machine Old Machine
Sample Mean 25 23
Sample Variance 27 7.56
Sample Size 45 36
As the statistical advisor to Ajax, would you recommend purchasing Allied's machine? Explain.

Answer 1

z(s) is in the acceptance region. We accept H₀ we did not find a significantly difference in the performance of the two machines therefore we suggest not to buy a new machine

Explanation:

We must evaluate the differences of the means of the two machines, to do so, we will assume a CI of 95%, and as the interest is to find out if the new machine has better performance ( machine has a bigger efficiency or the new machine produces more units per unit of time than the old one) the test will be a one tail-test (to the left).

New machine

Sample mean x₁ = 25

Sample variance s₁ = 27

Sample size n₁ = 45

Old machine

Sample mean x₂ = 23

Sample variance s₂ = 7,56

Sample size n₂ = 36

Test Hypothesis:

Null hypothesis H₀ x₂ - x₁ = d = 0

Alternative hypothesis Hₐ x₂ - x₁ < 0

CI = 90 % ⇒ α = 10 % α = 0,1 z(c) = - 1,28

To calculate z(s)

z(s) = ( x₂ - x₁ ) / √s₁² / n₁ + s₂² / n₂

s₁ = 27 ⇒ s₁² = 729

n₁ = 45 ⇒ s₁² / n₁ = 16,2

s₂ = 7,56 ⇒ s₂² = 57,15

n₂ = 36 ⇒ s₂² / n₂ = 1,5876

√s₁² / n₁ + s₂² / n₂ = √ 16,2 + 1.5876 = 4,2175

z(s) = (23 - 25 )/4,2175

z(s) = - 0,4742

Comparing z(s) and z(c)

|z(s)| < | z(c)|

z(s) is in the acceptance region. We accept H₀ we did not find a significantly difference in the performance of the two machines therefore we suggest not to buy a new machine

The very hight dispersion of values s₁ = 27 is evidence of frecuent values quite far from the mean