authors studied the impact of several factors involving the heat treatment of le

Question

authors studied the impact of several factors involving the heat treatment of leaf springs. In this process, a conveyor system transports leaf spring assemblies through a high temperature furnace. After the spring leaves a high pressure press, an oil quench cools it to near room temperature . An important material property of this process is the resulting free height of the spring, which should ideally be 8 inches. A materials engineer assigned to this process believes that the heating time affects this free height and so carries out 25 tests at two different times - 23 seconds and 25 seconds. The results are shown here:

Spring free height at:

23 seconds 25 seconds

7.12 7.33

7.33    7.54

7.34    7.77

7.37    8.07

7.45    8.1

7.47    8.14

7.51    8.23

7.51    8.26

7.52    8.27

7.53    8.31

7.56    8.31

7.57    8.36

7.57    8.36

7.61    8.36

7.65    8.37

7.66    8.37

7.67    8.39

7.67    8.42

7.67    8.42

7.67    8.45

7.71    8.45

7.71    8.46

7.71    8.48

7.71    8.49

7.75    8.53

messures are calculated independently.

QUESTION 10 Suppose it is known that the random variable X has a log normal distribution, and so Y-In) has a normal distribution with a population mean of 0.5 and a population standard deviation of 0.1 An estimator is considered biased if its %Bias exceeds 1%. Otherwise it is unbiased. A large sample means n>29. Make use of Excel file "sim1" to generate 10,000 samples of specified sizes and then match the following questions to the answer sets provided. A is unbiased in small and large samples B.has a % Bias of less than 1%. C is biased in small samples but diminishes with increasing sample The sample mean The sample variance (obtained using n in the demoninator) The sample standard deviation (obtained using n-1 in the size. denominator) The sample variance (obtained using n in the demoninator) D.is biased in small and large samples. based on a sample of size 5 has a %Bias around-20%. The sample variance (obtained using n-1 in the demoninator) based on a sample of size 5

Explanation / Answer

Let X = free-height of the spring subjected to a heating time of 23 seconds.

      Y = free-height of the spring subjected to a heating time of 25 seconds.

Then, X ~ N(µ1, 12) and Y ~ N(µ2, 22), where 12 = 22 = 2, say but 2 is unknown.

Claim:

Heating time impacts the free-height.

Hypotheses:

Null: H0: µ1 = µ2 Vs Alternative: HA: µ1 µ2

Test Statistic:

t = (Xbar - Ybar)/{s(2/n)} where Xbar and Ybar are sample averages and s1, s2are sample standard deviations based on n observations each on X and Y.

Calculations

Summary of Excel calculations is given below:

n =

25

Xbar =

7.2788

Ybar =

8.2496

s1 =

1.376388

s2 =

0.296613

s^2 =

0.991212

s =

0.995596

tcal =

3.447479

=

0.05

tcrit =

2.010635

p-value =

0.001187

Distribution, Critical Value and p-value:

Under H0, t ~ t2n - 2. Hence, for level of significance %, Critical Value = upper (/2)% point of t2n - 2 and p-value = P(t2n - 2 > | tcal |).

Using Excel Functions, the above are found to be as follows:

tcrit =

2.010635

p-value =

0.001187

Decision Criterion (Rejection Region):

Reject H0 if | tcal | > tcrit or p-value <

Decision:

Since | tcal | > tcrit, H0 is rejected. Since p-value < , H0 is rejected.

Conclusion:

There is sufficient evidence to suggest that the claim ‘Heating time impacts the free-height.’

is valid.

DONE

n =

25

Xbar =

7.2788

Ybar =

8.2496

s1 =

1.376388

s2 =

0.296613

s^2 =

0.991212

s =

0.995596

tcal =

3.447479

=

0.05

tcrit =

2.010635

p-value =

0.001187

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