Download IE 446: Problem Set #3 Solutions - Mean Comparison for UIC Mechanical Engineering Students and more Assignments Engineering in PDF only on Docsity! UNIVERSITY OF ILLINOIS AT CHICAGO Mechanical Engineering IE 446 Michael J. Scott Solutions to Problem Set #3 Spring 2000 1. We use the following normally distributed data: Technician 1 1.45 1.37 1.21 1.54 1.48 1.29 1.34 Technician 2 1.54 1.41 1.56 1.37 1.20 1.31 1.27 1.35 Thus n1 = 7, n2 = 8. We can calculate the sample means and variances for each set of data: x̄1 = 1.382857 x̄2 = 1.376250 S21 = 0.013190 S22 = 0.015598 I used matlab and the functions nanmean and nanstd to get these numbers. For example: >> m1 = nanmean([1.45 1.37 1.21 1.54 1.48 1.29 1.34]) m1 = 1.38285714285714 >> s1 = nanstd([1.45 1.37 1.21 1.54 1.48 1.29 1.34])^2 s1 = 0.01319047619048 (a) Since we assume variances are equal but unknown, we will use a t distribution with n1 + n2 − 2 degrees of freedom and the pooled variance S2p = (n1 − 1)S21 + (n2 − 1)S22 n1 + n2 − 2 = 0.014487 The two-sided 95% confidence interval (α2 = 0.025) is given by x̄1 − x̄2 − tα2 ,n1+n2−2Sp √ 1 n1 + 1 n2 ≤ µ1 − µ2 ≤ x̄1 − x̄2 + tα2 ,n1+n2−2Sp √ 1 n1 + 1 n2 We must look up or calculate t0.025,13 = 2.16. It’s in the table, or use matlab: >> -tinv(0.025,13) ans = 2.16036865646279 Applying all the numbers, get: −0.127969 ≤ µ1 − µ2 ≤ 0.141183 (b) The confidence interval on the ratio of variances uses an F distribution: S21 S22 F1−α2 ,n2−1,n1−1 ≤ σ21 σ22 ≤ S 2 1 S22 Fα 2 ,n2−1,n1−1 The table says F0.025,7,6 = 5.7. Matlab says: 1 >> finv(0.975,7,6) ans = 5.69547047368316 (Note that the conventions are different.) Following the note at the bottom of the table: F0.975,7,6 = 1 F0.025,6,7 = 1 5.12 = 0.195312 Matlab says: >> finv(0.025,7,6) ans = 0.19536604962876 which is presumably more accurate. In any event, we can calculate the confidence interval: 0.0131905 0.0155982 0.195366 = 0.165209 ≤ σ 2 1 σ22 ≤ 0.0131905 0.0155982 5.695470 = 4.816318 Since the confidence interval includes 1, the assumption that the two variances were equal was reasonable (to a confidence level of 95%). (c) Need a χ2 distribution for a confidence interval on σ22 (n = 8 for technician 2): (n− 1)S22 χ2 α 2 ,n−1 ≤ σ22 ≤ (n− 1)S22 χ21−α2 ,n−1 (7)0.015598 16.012764 ≤ σ22 ≤ (7)0.015598 1.689869 0.006819 ≤ σ22 ≤ 0.064613 2. (a) Test on the mean, σ = 2 is known, n = 8 sampled, x̄ = 127 is calculated. The 95% confidence interval uses Z0.05 = 1.644854 (use norminv(0.95,0,1) in matlab, or estimate from the table), and is: µ ≥ x̄− Z0.05 σ√ n = 127− 1.644854 2√ 8 = 125.83691 (b) Since the lower confidence interval has a lower bound of 125.83691, we can conclude immediately that we must reject the null hypothesis that µ = 125 and accept the alternative hypothesis that µ > 125. (You can do the algebra if you like.) The P -value is the smallest value of α for which we reject the null hypothesis. Now we do need the test statistic: Z0 = (x̄− µ) √ n σ = (127− 125) √ 8 2 = √ 8 = 2.828427 We want to find α such that Zα = 2.828427, which means that we need to find Φ(2.828427). From the table, we see that Φ(2.82) = 0.99760 and Φ(2.83) = 0.99767, so we can conclude that the P -value is between 0.00233 and 0.00240. We can also use matlab: >> 1 - normcdf(sqrt(8),0,1) ans = 0.00233886749052 A P -value of about 0.00234, or 2.34%, means that we are able to conclude that µ > 125 (i.e., that the alternative hypothesis is correct) if we are prepared to have that conclusion be in error 2.34% (or more) of the time. 2
