Philips Semiconductors is a leading European manufacturer of integrated circuits. Integrated circuits are produced on silicon wafers, which are ground to target thickness early in the production process. The wafers are positioned in different locations on a grinder and kept in place through vacuum decompression. One of the goals of process improvement is to reduce the variability in the thickness of the wafers in different positions and in different batches. Data were collected from a sample of 30 batches. In each batch, the thickness of the wafers on positions 1 and 2 (outer circle), 18 and 19 (middle circle), and 28 (inner circle) was measured. The results are given in the data below. Position 1 Position 2 Position 18 Position 19 Position 281 240 243 250 253 2482 238 242 245 251 2473 239 242 246 250 2484 235 237 246 249 2465 240 241 246 247 2496 240 243 244 248 2457 240 243 244 249 2468 245 250 250 247 2489 238 240 245 248 24610 240 242 246 249 24811 240 243 246 250 24812 241 245 243 247 24513 247 245 255 250 24914 237 239 243 247 24615 242 244 245 248 24516 237 239 242 247 24517 242 244 246 251 24818 243 245 247 252 24919 243 245 248 251 25020 244 246 246 250 24621 241 239 244 250 24622 242 245 248 251 24923 242 245 248 243 24624 241 244 245 249 24725 236 239 241 246 24226 243 246 247 252 24727 241 243 245 248 24628 239 240 242 243 24429 239 240 250 252 25030 241 243 249 255 253a. At the 0.01 level of significance, determine if there is evidence of a difference in the mean thickness of the wafers for the five positions. (Assume that the assumptions for this type of test are satisfied.)Complete the following:1. State H0.2. State H1.3. State the value of a.4. State the value of the test statistic.5. State the p-value.6. State the decision in terms of H0 and why.7. State the decision in terms of the problem. b. Do you think that there was a significant block effect in this experiment? Explain.

Step-by-step explanation:

1) The null hypothesis is,

H_0: The mean thickness of teh wafers for the five positions are equal

i.e,

2)

The alternative hypothesis is,

H_1: There is an evidence of a difference in the mean thickness of the wafers for the five positions

3)

Let us consider the level of significance

from the Minitab outout

One-way ANOVA:C1 versus C2

source         DF            SS                 MS               F            P

C2                  4      1417.73          354.43        51.00       0.00

Error           145     1007.77              6.95

Total           14      2425.50

S = 2.636       R - S = 58.45%     R - Sq(adj) = 57.31%

Individual 95% CIs For Mean Based on Pooled StDev

level         N         Mean            StDev    -,----------,----------,----------,----------

1              30    240.53                2.62    (--,--)

2             30     243.73                2.79             (--,--)

3             30     246.07                2.90                           (--,--)

4             30     249.10                 2.66                                        (--,--)

5             30     247.07                 2.15                                  (--,--)

                                                               -,----------,----------,----------,----------

                                                        240.0   243.0   246.0   249.0

Pooled StDev = 2.64

4)

The test statistic is, F = 51

5)

The P-value is approximately 0

6)

Here, the P - value is less than the level of significance

So, we do not accept our null hypothesis H_0

7)

Therefore, we conclude that there is an evidence of a difference in the mean thickness of the wafers for the five positions at level of significance

b)

chek attachment

we observe that,

The mean thickness of the wafer for position 1 is significant with position 2,

position 18, position 19 and position 28.

The mean thickness of the wafer for position 2 is significant with position 18,

position 19 and position 28.

The mean thickness of the wafer for position 18 is significant with the

position 19.

But the mean thickness of the wafer for position18 is not significant with

position 28.

The mean thickness of the wafer for position 19 is significant with position 28.

it's equivalent, so we can say that its equal.

let's call the denominator x and the numerador is 18 less then denominator, x - 18.

multiply

2.x = 5.(x - 18)

2x = 5x - 90

90 = 5x - 2x

90 = 3x

x = 90/3

x = 30

so: