Individual Homework #3Design of Experiment and Analysis

2.22.

By hand:

X = 241.5 vs µo = 225 Normally distributedS = 98.7n = 16

a) Ho: µ = µo H1: µ > 225b) to = (X- µ)/S/sqrt(n) = (241.5-225)/(98.7/sqrt(14)) = 0.6687

t0.05, 17 = 1.740.to = 0.6687 < t0.05, 17 = 1.740. No rejection.

c) P-value > 0.05d) t0.05, 17 * S/sqrt(n) ± X = 1.740*98.8/4 ± 241.5 = 42.93± 241.5

R solution

e1 = c(159, 224, 222, 149, 280, 379, 362, 260, 101, 179, 168, 485, 212, 264, 250, 170)> t.test(e1, alternative = "greater", mu = 225)

One Sample t-testdata: e1 t = 0.6685, df = 15, p-value = 0.257alternative hypothesis: true mean is greater than 225 95 percent confidence interval: 198.2321 Inf sample estimates:mean of x 241.5

2.29a) Yes, there is evidence. Rejection of Null hypothesis based on t-test and p-value of the below test assuming equal and difference variances.

> t95 = c(11.176, 7.089, 8.097, 11.739, 11.291, 10.759, 6.467, 8.315)> t100 = c(5.263, 6.748, 7.461, 7.015, 8.133, 7.418, 3.772, 8.963)

T-test assuming equal variances

> t.test(t95, t100, alternative = "greater", var.equal = TRUE)

Two Sample t-testdata: t95 and t100 t = 2.6751, df = 14, p-value = 0.009059alternative hypothesis: true difference in means is greater than 0

95 percent confidence interval: 0.8608158 Inf sample estimates:mean of x mean of y 9.366625 6.846625

T-test assuming different variances

t.test(t95, t100, alternative = "greater", var.equal = FALSE)

Welch Two Sample t-testdata: t95 and t100 t = 2.6751, df = 13.226, p-value = 0.009423alternative hypothesis: true difference in means is greater than 0 95 percent confidence interval: 0.8539293 Inf sample estimates:mean of x mean of y 9.366625 6.846625

b) p-value = 0.009059 Equal variances. p-value = 0.009423 Unequal variances

c) 2.52 ± 1.771*sqrt(4.4/8+2.69/8) = 2.52 ± 1.66. 95% of the time the difference between the thicknesses at 95 and 100 degrees will fall within this interval.

d) tfile <- read.csv(file="tem.csv",head=TRUE,sep=",")> tfile T95 T1001 11.176 5.2632 7.089 6.7483 8.097 7.4614 11.739 7.0155 11.291 8.1336 10.759 7.4187 6.467 3.7728 8.315 8.963

dotplot(jitter(thickness)~T, data=tfile)

e) Based on normal probability plots normal distribution can be assumed

Normal Probability Plots

f) power.t.test(n=8,delta=2.5,sd=1.88,sig.level=0.05,type="two.sample", alternative = "two.sided", strict=TRUE)

Two-sample t test power calculation n = 8

delta = 2.5 sd = 1.88 sig.level = 0.05 power = 0.6964335 alternative = two.sided NOTE: n is number in *each* group

g.) library(pwr) pwr.t.test(d=(1.5)/1.88,power=.9,sig.level=.05,type="two.sample",alternative="two.sided") Two-sample t test power calculation

n = 34.00074 d = 0.7978723 sig.level = 0.05 power = 0.9 alternative = two.sided

NOTE: n is number in *each* group

2.35.

a)f1=c(206, 188, 205, 187, 193, 207, 185, 189, 192, 210, 194, 178)f2=c(177, 197, 206, 201, 176, 185, 200, 197, 198, 188, 189, 203)qqnorm(f1, ylab = "Formulation 1")qqline(f1)qqnorm(f2, ylab = "Formulation 2")qqline(f2)

Points fall more or less on the lines for both formulations between the 25th and 75th percentile justifying the assumption of normal distribution. Also the lines have similar slopes so the assumption of constant variance is met as well.

b) No, it does not. No Rejection of Null hypothesis. t = 0.3448 < t0.05, 22 = 1.7171 and p-value = 0.3667

t.test(f1, f2, alternative = "greater", var.equal = TRUE)

Two Sample t-test

data: f1 and f2 t = 0.3448, df = 22, p-value = 0.3667alternative hypothesis: true difference in means is greater than 0 95 percent confidence interval: -5.637883 Inf sample estimates:mean of x mean of y 194.5000 193.0833

c) p-value = 0.3667d) No it does not. p-value = 0.6482

t.test(f1, f2, alternative = "greater", mu =3, var.equal = TRUE)

Two Sample t-test

data: f1 and f2 t = -0.3854, df = 22, p-value = 0.6482alternative hypothesis: true difference in means is greater than 3

95 percent confidence interval: -5.637883 Inf sample estimates:mean of x mean of y 194.5000 193.0833