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

Step-by-step 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