7. A mechanical engineer wishes to compare strength properties of steel beans wi

Question

7. A mechanical engineer wishes to compare strength properties of steel beans with similar beams made with a particular alloy. Each beam will be set in horizontal position with a support on each end, a force of 2500 lbs will be applied at the centre, and the deflection will be measured. From past experience with such beams, the engineer is willing to assume that the true standard deviation of deflection for botlh types of beam is .05 inches. Because the alloy is more expensive, the engineer wishes to test at level = .01 whether or not it has a smaller average deflection than the steel beam. (a) What would be the power if there were 25 beams of each type used and the difference in true average deflection favours the alloy by .04 inches? (b) What size of samples are appropriate if the desired power is .95 when the difference in true average deflection favours the alloy by 04 inches?

Explanation / Answer

Back-up Theory

Let X = strength (expressed in terms of deflection in inches) of beams made from the alloy and Y = strength (expressed in terms of deflection in inches) of beams made from steel.

We assume: X ~ N(µ1, 12) and Y ~ N(µ2, 22).

We are also give that it is fine to assume 1 = 2 = 0.05 inches.

We wish to test the mechanical engineer’s claim that alloy beam has better strength (i.e., lower deflection) than the steel beam or in terms of means, µ1 < µ1.

Thus, we are interested in testing: Null: H0: (µ1 - µ2) = 0 Vs Alternative: HA: (µ1 - µ2) < 0 under the conditions 1 = 2 = (0.05,known) and n1 = n2 = n, say.

Test statistic is:

Z = {(Xbar - Ybar) - (µ1 - µ2)}/{(2/n)} where

Xbar and Ybar are sample averages based on n observations each on X and Y.

and

Under H0, Z ~ N(0, 1). Hence, for level of significance (given as 1)%, Critical Value = lower % point of N(0, 1) = - 2.3264 and

Reject H0 if Zcal < Zcrit.i.e., [(Xbar - Ybar)/{0.05(2/25)}] < 2.3264 [since under H0, (µ1 - µ2) = 0]

=> (Xbar - Ybar) < 0.0329 …………………………………………………………(1)

Now, to work out the solution,

Part (a)

Power of a test = 1 – , where

= probability of Type II error

= P(rejecting H0 when HA istrue)

= P(Zcal > 2.3264 given (µ1 - µ2) = 0.04)

= P{(Xbar - Ybar) < 0.0329 given (µ1 - µ2) = 0.04)} [from (1) above]

= P[Z < (0.0329 – 0.04)/0.01414]

[because under (µ1 - µ2) = 0.04, (Xbar - Ybar) ~ N(0.04, (2/n) ] …………………….. (2)

= P(Z < - 0.5021)

= 0.3078

So, power = 1 – 0.3078 = 0.6922 ANSWER

Part (b)

We want power to be 0.95 => 1 – = 0.95 or = 0.05

i.e., P[Z < (0.0329 – 0.04)/ (2/n)] = 0.05 from (2) above [because under (µ1 - µ2) = 0.04, (Xbar - Ybar) ~ N(0.04, (2/n) ]

or P[Z < - (0.0071/(2/n)] = 0.05

=> - (0.0071/(2/n) = lower 5% point of N(0, 1) = - 1.645

=> (2/n) = 0.0071/1.645 = 0.004316

=> (2/n) = 0.004316/ = 0.004316/0.05 = 0.08632

=> 2/n = 0.007451

=> n = 2/0.007451 = 268.42

Thus, the required sample size is 269 ANSWER

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