Two different analytical tests can be used to determine the impurity level in st

Question

Two different analytical tests can be used to determine the impurity level in steel alloys. Five specimens are tested using both procedures, and the results are shown in the following tabulation. (a) Is there sufficient evidence to conclude that tests differ in the mean impurity level, using alpha = 0.01? (b) Is there evidence to support the claim that test 1 generates a mean difference 0.1 units lower than test 2? Use alpha = 0.05 (c) If the mean from test 1 is 0.1 less than the mean from test 2, it is important to detect this with probability at least 0.90. Was the use of five alloys an adequate sample size? If not, how many alloys should have been used?

Explanation / Answer

Solution

Let X = impurity level as assessed by Test 1 and Y = impurity level as assessed by Test 2.

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

Part (a)

We want to test if (µ2 - µ1) = 0 or not.

So, null hypothesis: H0 : (µ2 - µ1) = 0   Vs alternative HA : (µ2 - µ1) 0.

Note: Since both tests are done on the same specimens, the test is a paired t-test and NOT 2-sample t-test.

Test Statistic

t = (n){dbar - (µ2 - µ1)}/sd , where di = Test 2 result - Test 1 result of the ith sample, n is the sample size and dbar and sd are respectively the sample mean and standard deviation of d.

Computtions

x

1.2

1.3

1.5

1.4

1.7

y

1.4

1.7

1.5

1.3

2.0

d = (y – x)

0.2

0.4

0

- 0.1

0.3

(d - dbar)

0.04

0.24

- 0.16

- 0.26

0.4

(d - dbar)2

0.0016

0.0576

0.0256

0.0676

0.0016

dbar = 0.8/5 = 0.16

sd2 = Sum(d - dbar)2/4 = 0.1540/4 = 0.0308 and hence sd = sqrt(0.0308) = 0.1755

Then, tcal = (5)(0.16)/0.1755 = 2.038

Under H0, t ~ tn – 1 i.e., t4

Given = 0.01, tcrit = upper 0.5% point of t4 = 4.604 [read off Standard t-distribution Table]

Since tcal < tcrit, H0 is accepted.

Conclusion

The above decision implies that there is not enough evidence to suggest that the mean levels are different between the two tests. ANSWER

Part (b)

Here, we want to test if (µ2 - µ1) = 0.1 or not.

So, null hypothesis: H0 : (µ2 - µ1) = 0.1   Vs alternative HA : (µ2 - µ1) 0.1.

tcal = (5)(0.06)/ sqrt(0.0308) = 0.7644

Given = 0.05, tcrit = upper 2.5% point of t4 = 2.776 [read off Standard t-distribution Table]

Since tcal < tcrit, H0 is accepted.

Conclusion

The above decision implies that there is not enough evidence to suggest that the mean level of Test 1 is 0.1 lower than that of Test 2. ANSWER

Part (c)

Sample size required = n = (s2 t2n – 1,/2)/E2, where E = error margin = 0.1, in our case and other quantities are as defined earlier. Here = 0.1 and hence t2n – 1,/2 = 2.1322

So, n = (0.0308 x 2.1322)/0.12 = 14 ANSWER

x

1.2

1.3

1.5

1.4

1.7

y

1.4

1.7

1.5

1.3

2.0

d = (y – x)

0.2

0.4

0

- 0.1

0.3

(d - dbar)

0.04

0.24

- 0.16

- 0.26

0.4

(d - dbar)2

0.0016

0.0576

0.0256

0.0676

0.0016

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