( ) ( ) STAT 5031 Statistical Methods for Quality Improvement. Homework n = 8; x = 127 psi; σ = 2 psi (a) µ 0 = 125; α = 0.

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1 STAT 531 Statistical Methods for Quality Improvement Homework n = 8; x = 17 psi; σ = psi (a) µ = 15; α =.5 Test H : µ = 15 vs. H 1 : µ > 15. Reject H if Z > Z α. x µ Z = = =.88 σ n 8 Z α = Z.5 = Reject H : µ = 15, and conclude that the mean tensile strength exceeds 15 psi. (b) P-value = 1 Φ(Z ) = 1 Φ(.88) = =.34 (c) In strength tests, we usually are interested in whether some minimum requirement is met, not simply that the mean does not equal the hypothesized value. A one-sided hypothesis test lets us do this. (d) x Zα σ n µ µ 15.8 µ 4.1. (a) x ~ N(µ, σ), µ = 1, α =.1 n = 1, x = 1.15, s =.3 Test H : µ = 1 vs. H 1 : µ > 1. Reject H if t > t α. x µ t = = = S n.33 1 t α, n 1 = t.1, 9 =.81 Do not reject H : µ = 1, and conclude that there is not enough evidence that the mean fill volume exceeds 1 oz. (b) α =.5 t α/, n 1 = t.5, 9 =.6 µ α n ( S ) µ ( S ) x t S n x + t S n α /, n 1 /, µ 1.37 (c) We can check this with a normal probability plot fillvol area score fillvol Normal probability plot of fill volume score 1

2 The plotted points fall approximately along a straight line, so the assumption that fill volume is normally distributed is appropriate. For the confidence interval, the standard deviation is.3 oz, and the sample size is n=1, giving (n-1)s =.81. There are 9 degrees of freedom, and the upper and lower ½ % points of this chisquared are.7 and 19. respectively. So a 95% confidence interval is given by ;.46 < σ < < σ <.3. The 95% CI for σ is.1 < σ < x ~ N(µ, σ), n = 16, x = 1.59 V, s =.999 V (a) µ = 1, α =.5 Test H : µ = 1 vs. H 1 : µ 1. Reject H if t > t α/. x µ t = = = S n t α/, n 1 = t.5, 15 =.131 Reject H : µ = 1, and conclude that the mean output voltage differs from 1V. (b) /, n 1 µ α /, n 1 µ x t S n x + t S n α µ 1.79 (c) σ = 1, α =.5 Test H : σ = 1 vs. H 1 : σ 1. Reject H if χ > χ α/, n-1 or χ < χ 1-α/, n-1. ( n 1) S (16 1).999 χ = = = σ 1 χ α/, n 1 = χ.5,16 1 = χ 1 α/, n 1 = χ.975,16 1 = 6.6 Do not reject H : σ = 1, and conclude that there is insufficient evidence that the variance differs from 1. (d) ( n 1) S ( n 1) S χ σ α /, n 1 χ1 α /, n 1 (16 1).999 (16 1).999 σ σ σ Since the 95% confidence interval on σ contains the hypothesized value, σ = 1, the null hypothesis, H : σ = 1, cannot be rejected.

3 =.5; = = 7.69 (e) α χ1 α, n 1 χ.95,15 ( n 1) S σ χ 1 α, n 1 (16 1).999 σ 7.69 σ.6 σ (f) The probability plot looks decently linear, so we are OK with the normality assumption. 4. n =, x = 18, ˆp = x/n = 18/ =.9 (a) p =.1, α =.5. Test H : p =.1 versus H 1 : p.1. Reject H if Z > Z α/. np = (.1) = Since (x = 18) < (np = ), use the normal approximation to the binomial for x < np. This gives Z ( x +.5) np (18 +.5) = = = np (1 p ) (1.1) If we omit the continuity correction, we get x np 18 Z = = =.47 np (1 p ) (1.1) Z α/ = Z.5/ = Z.5 = 1.96 Either way, do not reject H, and conclude that the sample process fraction nonconforming does not differ from.1. P-value = [1 Φ Z ] = [1 Φ.3536 ] = [1.638] =.736 (b) α =.1, Z α/ = Z.1/ = Z.5 = pˆ Z pˆ (1 pˆ ) n p ˆp + Z ˆp (1 ˆp ) n α / α / (1.9) p (1.9).57 p n = 5, x = 65, ˆp = x/n = 65/5 =.13 (a) p =.8, α =.5. Test H : p =.8 versus H 1 : p.8. Reject H if Z > Z α/. np = 5(.8) = 4 Since (x = 65) > (np = 4), use the normal approximation to the binomial for x > np. 3

4 ( x.5) np (65.5) 4 With the continuity correction this gives Z = = = 4.387, and np (1 p ) 4(1.8) x np 65 4 omitting the correction gives Z = = = 4.1 np (1 p ) 4(1.8) Z α/ = Z.5/ = Z.5 = 1.96 Either way, reject H, and conclude the sample process fraction nonconforming differs from.8. (b) P-value = [1 Φ Z ] = [1 Φ ] = [ ] =.6 (c) α =.5, Z α = Z.5 = p pˆ + Z pˆ (1 pˆ ) n α p (1.13) 5 p (a) n 1 =, x 1 = 1, ˆp 1 = x 1 /n 1 = 1/ =.5 n = 3, x =, ˆp = x /n = /3 =.67 (b) Use α =.5. Test H : p 1 = p versus H 1 : p 1 p. Reject H if Z > Z α/ or Z < Z α/ x + x 1 + n + n pˆ = = =.6 1 Z pˆ pˆ = = = pˆ (1 pˆ ) 1 n + 1 n.6(1.6) Z α/ = Z.5/ = Z.5 = 1.96 Z α/ = Do not reject H. Conclude there is no strong evidence to indicate a difference between the fraction nonconforming for the two processes. (c) ˆ ˆ pˆ (1 pˆ ) pˆ (1 pˆ ) 1 1 ( p p ) Z + ( p p ) 1 α / 1 n n 1 pˆ (1 pˆ ) pˆ (1 pˆ ) 1 1 ( pˆ pˆ ) + Z + 1 α / n n 1.5(1.5).67(1.67) (.5.67) ( p p ) 1 3.5(1.5).67(1.67) (.5.67) p p 1.5 ( ) Test H : µ d = versus H 1 : µ d. Reject H if t > t α/, n1 + n. 4

5 t α/, n1 + n = t.5, = d =.417; s d =.13 ( d ) t = d s n = = 1.1 ( t = 1.1) <.8188, so do not reject H. There is no strong evidence to indicate that the two calipers differ in their mean measurements. 4.8 The test is two-sided, so the cutoff value will be z.5 = We want a probability of Type II error of.1 when the mean is actually. We can figure the needed sample size using the formula given in class ( zα / + zβ ) σ ( )3 n = = = µ µ (a) We wish to test H : λ = λ versus H 1 : λ λ. Select random sample of n observations x 1, x,, x n. Each x i ~ POI(λ). In other words, we need n=4 n x ~ POI( nλ ). i Using the normal approximation to the Poisson, if n is large, x = x/n = ~ N(λ, λ/n). Z = x λ λ n. Reject H : λ = λ if Z > Z α/ / (b) x ~ Poi(λ), n = 1, x = 11, x = x/n = 11/1 =.11 Test H : λ =.15 versus H 1 : λ.15, at α =.1. Reject H if Z > Z α/. Z α/ = Z.5 =.5758 Z = ( x λ) λ n = (.11.15).15 1 = 1.38 ( Z = 1.38) <.5758, so do not reject H. Lab portion 1. > x <- c(145, 15, 153, 148, 141, 15, 146, 154, 139, 148) > y <- c(15, 15, 147, 155, 14, 146, 158, 15, 151, 143) > t.test(x, y) Welch Two Sample t-test data: x and y t = -.778, df = , p-value =.459 alternative hypothesis: true difference in means is not equal to 95 percent confidence interval: sample estimates: mean of x mean of y The t test shows no evidence of a difference between the two groups i = 1. > n <- 16 > C <- sqrt(qchisq(.95, n-1) / (n-1)) 5

6 > C [1] So the needed C value is 1.9. Now we ll trace out the OC curve. I also draw a horizontal line at OC=. to help answer the followup question > sig <- seq(1,,.5) > OC <- pchisq((n-1)*(c/sig)^, n-1) > dev.new() > plot(oc~sig, main="oc curve for testing sigma", xlab="sigma", + ylab="oc", type="l") > abline(h=.) Rather than try reading the answer off the curve, I ll get R to tell me. > offset <- abs(oc -.) > sig[offset==min(offset)] [1] 1.55 The σ for which the OC is closest to. (ie the power is closes to 8%) is σ=1.55. As this is just a bit bigger than the 1.5 the question asks about, the needed sample size is going to be a bit above 16. Some searching by getting the right C and then evaluating the OC gives: > for (n in 16:) { + C <- sqrt(qchisq(.95, n-1) / (n-1)) + oc <- pchisq((n-1)*(c/1.5)^, n-1) + print(paste(n, oc)) + } [1] " " [1] " " [1] " " [1] " " [1] " " So we needed n=19 3. The equation for the OC was given in class. Coding it up gives 6

7 mu <- seq(15, 18,.1) > sigma <- > n <- 8 > OC <- pnorm( (mu-15)*sqrt(n)/sigma) > dev.new() > plot(mu, OC, main="oc curve for normal mean", xlab="mu", + ylab="oc", type="l") 4. I ve edited out some unnecessary lines > allin <- read.csv(" > > fitter <- with(allin, lm(satisfaction ~ Severity)) > summary(fitter) Coefficients: Estimate Std. Error t value Pr(> t ) (Intercept) e-1 *** Severity e-5 *** Residual standard error: on 3 degrees of freedom Multiple R-squared:.57, Adjusted R-squared:.5 F-statistic: 5.19 on 1 and 3 DF, p-value: 4.447e-5 The regression is highly significant. Each 1 unit increase in severity corresponds to 1.18 units less in patient satisfaction. The graph of the relationship looks fine; no evidence of the sort of violations that would get you worried. > dev.new() > with(allin, plot(severity, Satisfaction, main="table 4E.11")) > abline(fitter) 7

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