Question

Film 2. Nanoscale ferroelectric experiments find a relationship between film thickness X and flux-closure period Y, both in nm, for a certain film-substrate combination. The sample size is 28, 2x,693, and S-1783.25. Assuming the SLR model, we also know bo-1.01, b,-1.49, SSR = 3980.6, and SST = 4165.5. Compute point and interval estimates for the mean flux-closure period when film thickness is 20 nm. Compute point and interval estimates for the next flux-closure period when film thickness is 20 nm. a) b)

Answer 1

Sqrt(SSE/(n-2))
SSE = SST-SSR= =4165.5-3980.6 = 184.9

Sqrt(SSE/(n-2))
SSE = SST-SSR= =4165.5-3980.6 = 184.9

Standard Error= sqrt(SSE/(n-2)) = 2.666746794

A)

Point estimate, when x= 20
yhat=-1.01+1.49*20
yhat = 28.79

Interval estimate for mean flux closure
t(n-2) = 2.0555
s = 2.666747
1/n = 0.0357
(x*-xbar)^2 = 22.5625 (20-24.75)^2
(n-1)*sx^2 = 85859475.19 (28-1)*1783.25^2

CI
lower= 27.75 =28.79-(2.0555*2.666747*SQRT(0.0357+(22.5625/85859475.1875)))
upper= 29.82570245 = 28.79+(2.0555*2.666747*SQRT(0.0357+(22.5625/85859475.1875)))

b)

A)

Point estimate, when x= 20
yhat=-1.01+1.49*20
yhat = 28.79

Interval estimate for next flux closure
t(n-2) = 2.0555
s = 2.666747
1/n = 0.0357
(x*-xbar)^2 = 22.5625 (20-24.75)^2
(n-1)*sx^2 = 85859475.19 (28-1)*1783.25^2

PI
lower =23.2115141 =28.79-(2.0555*2.666747*SQRT(1+0.0357+(22.5625/85859475.1875)))
upper =34.3684859 =28.79+(2.0555*2.666747*SQRT(1+0.0357+(22.5625/85859475.1875)))