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 your

Step-by-step 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

step-by-step explanation:

we can use this expression to calculate the energy gain as a particle moves with a field or the energy loss as it moves against it. (note that it is also used to define the electron-volt, a unit of energy equal to that gained by an electron as it falls through a pd of 1 v, i.e. w = 1 ev = 1.6 10-19 c 1 v = 1.6 10-19j).in fact this expression is really nothing more than the statement that potential is the potential energy per unit charge. in this case, v= w/qis simply stating that the change in potential energy(or work done) per unit charge is equal to the change in potential (or potential difference), v.student questions: uniform electric fields2electron beamcathode+5 kvanode0 v

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step-by-step explanation:

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