Answer:
Explanation:
We will develop a test to compare the mean of two population
Population 1.
population mean μ₀₁ = 25 ; Sample variance 27 ; and sample size n = 45
Population 2.
population mean μ₀₂ = 23 ; Sample variance 7,56; and sample size n = 36
As our major interest is to investigate if the new machine uses less time for the same production, the test will be a one tail test ( left test)
Test Hypothesis
Null Hypothesis H₀ ⇒ μ₀₂ - μ₀₁ = 0
Alternative Hypothesis Hₐ ⇒ μ₀₂ - μ₀₁ < 0
We will use confidence of 90 %, therefore α = 10 % α = 0,1
α = 0,1
We get z score of z = 1,28 or z = - 1,28 ( left tail)
And compute z(s) = ( μ₀₂ - μ₀₁ ) /√ (s₁)²/n₁ + (s₂)²/n₂
z(s) = - 2 / √(729/45) + (57,15/36)
z(s) = - 2 / √16,2 + 1,59
z(s) = - 2 / 4,2178
z(s) = - 0,4742
As |z(s)| < |z(c)|
We are in the acceptance region. If we lok at 90 % as Confidencial Interval α = 0,1 and α/2 = 0,05 in this case
₀,₉CI ( μ₀₂ - μ₀₁) = [ -2 ± z(0,05)√ (s₁)²/n₁ + (s₂)²/n₂ )
From z Table z ( 0,05 ) ⇒ z score z = 1,64
And √ (s₁)²/n₁ + (s₂)²/n₂ ) = √(729/45) + (57,15/36) = 4,2178
₀,₉CI ( μ₀₂ - μ₀₁) = [ -2 ± 1,64 *4,2178]
₀,₉CI ( μ₀₂ - μ₀₁) = ( - 8,917 ; 4,917 )
We can see that 0 is a possible value in the ₀,₉CI ( μ₀₂ - μ₀₁) so again we cannot reject H₀. Then as we are not quite sure about the strengths of the new machine over the old one we should not recomend to purchase the new machine