Question

A new process has been developed for applying photoresist to 125 mm silicon wafers used in the manufacturing of integrated circuits. 10 wafers were tested and the following photoresist thickness measurements (Angstrom x1000) were observed: 13.3987, 13.3957, 13.3902, 13.4015, 13.4001, 13.3918, 13.3965, 13.3925, 13.4002, 13.3946.

a) Test the hypothesis that the mean thickness is 13.4 x 1000 Å. Use α=.05. Assume a 2- sided alternative. What type of test statistic will be used? Why?

b) Find the 95% confidence interval on the mean photoresist thickness. Assume that thickness is normally distributed and use α=.05.

c) Find the 99% confidence interval on the mean photoresist thickness. Assume that thickness is normally distributed and use α=.01.

d) Assume that you are now using a one-sided test where the mean is greater than 13.4 x 1000 Å. Using an α=.01, test this hypothesis.

Answer 1

Sample mean using excel function AVERAGE(), x̅ = 13.3962

Sample standard deviation using excel function STDEV.S, s = 0.0039

Sample size, n = 10

a) we will use student's-t test.

Null and Alternative hypothesis:

Ho : µ = 13.4

H1 : µ ≠ 13.4

Test statistic:

t = (x̅- µ)/(s/√n) = (13.3962 - 13.4)/(0.0039/√10) = -3.0906

df = n-1 = 9

p-value :

Two tailed p-value = T.DIST.2T(ABS(-3.0906), 9) = 0.0129

Decision:

p-value < α, Reject the null hypothesis.

Conclusion:

There is enough evidence to conclude that the population mean thickness is different than 13.4 at 0.05 significance level.

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b) 95% Confidence interval :

At α = 0.05 and df = n-1 = 9, two tailed critical value, t-crit = T.INV.2T(0.05, 9) = 2.262

Lower Bound = x̅ - t-crit*s/√n = 13.3962 - 2.262 * 0.0039/√10 = 13.3934

Upper Bound = x̅ + t-crit*s/√n = 13.3962 + 2.262 * 0.0039/√10 = 13.3990

13.3934 < µ < 13.399

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c) 99% Confidence interval :

At α = 0.01 and df = n-1 = 9, two tailed critical value, t-crit = T.INV.2T(0.01, 9) = 3.250

Lower Bound = x̅ - t-crit*s/√n = 13.3962 - 3.25 * 0.0039/√10 = 13.3922

Upper Bound = x̅ + t-crit*s/√n = 13.3962 + 3.25 * 0.0039/√10 = 13.4002

13.3922 < µ < 13.4002

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d) Null and Alternative hypothesis:

Ho : µ ≤ 13.4 ; H1 : µ > 13.4

Test statistic:

t = (x̅- µ)/(s/√n) = (13.3962 - 13.4)/(0.0039/√10) = -3.0906

df = n-1 = 9

p-value :

Right tailed p-value = T.DIST.RT(-3.0906, 9) = 0.9935

Decision:

p-value > α, Do not reject the null hypothesis.

Conclusion:

There is not enough evidence to conclude that the population mean thickness is greater than 13.4 at 0.01 significance level.