Question

The carbon content (in parts per million) was measured five times for each of two different silicon wafers. The measurements were as follows. [3 Points] Wafer A : 1.12 1.15 1.16 1.10 1.13 Wafer B : 1.21 1.18 1.15 1.18 1.14 Construct a 90% confidence interval for the difference in carbon content between the two wafers. `Assume that the populations are normally distributed and the population variances are equal.

Answer 1

Confidence interval for difference between two population means is given as below:

Confidence interval = (X1bar – X2bar) ± t*sqrt[Sp2*((1/n1)+(1/n2))]

Where Sp2 is pooled variance

Sp2 = [(n1 – 1)*S1^2 + (n2 – 1)*S2^2]/(n1 + n2 – 2)

From given data, we have

X1bar = 1.132

X2bar = 1.172

S1 = 0.023875

S2 = 0.027749

n1 = 5

n2 = 5

df = n1 + n2 – 2 = 8

Confidence level = 90%

Critical t value = 1.8595

(by using t-table)

Sp2 = [(n1 – 1)*S1^2 + (n2 – 1)*S2^2]/(n1 + n2 – 2)

Sp2 = [(5 – 1)* 0.023875^2 + (5 – 1)* 0.027749^2]/(5 + 5 – 2)

Sp2 = 0.0007

(X1bar – X2bar) = 1.132 - 1.172 = -0.04

Confidence interval = (X1bar – X2bar) ± t*sqrt[Sp2*((1/n1)+(1/n2))]

Confidence interval = -0.04 ± 1.8595*sqrt[0.0007*((1/5)+(1/5))]

Confidence interval = -0.04 ± 1.8595*0.0164

Confidence interval = -0.04 ± 0.0304

Lower limit = -0.04 - 0.0304 = -0.0704

Upper limit = -0.04 + 0.0304 = -0.0096

Confidence interval = (-0.0704, -0.0096)