Question

Chapter 13 Analysis of Variance Saved Help Save & Exit Sub Check my work mode: This shows what is correct or incorrect for the work you have completed so far. It does not indicate c Return to questio 4 The following statistics are calculated by sampling from four normal populations whose variances are equal: (You may find it useful to reference the t table and the g table.) 10 points z = 153, n1 =6; 2 164, n2 =6; 23= 159, n3 6; 24 153, n4 6; MSE= 58.5 Use Fisher's LSD method to determine which population means differ at α= 0.01. (Negative values should be indicated by a minus sign. Round intermediate calculations to at least 4 decimal places. Round your answers to 2 decimal places.) Population Mean Differences Can we conclude that the population means differ? Confidence Interval b. Use Tukey's HSD method to determine which population means differ at α= 0.01. (lf the exact value for mT-c is not found in the table, then round down. Negative values should be indicated by minus sign. Round intermediate calculations to at least 4 decimal places. Round your answers to 2 decimal places.) Population Mean Differences Can we conclude that the population means differ? Confidence Interval H2 - u3

Answer 1

Here we have 4 groups and total number of observations are 24. So degree of freedom is

df= 24-4 = 20

(a)

Critical value of t for and df = 20 is 2.845. The Fisher's LSD Value is

α 0.01

$LSD=t_{\alpha/2}\sqrt{\frac{MSE\cdot 2}{n}}=2.845\cdot \sqrt{\frac{50.5\cdot 2}{6}}\approx 11.6726$

Formula for confidence interval is

For $\mu_{i}-\mu_{j}$ :

$[\bar{x}_{i}-\bar{x}_{j}]\pm LSD$

Following is the completed table:

xbarj 164 159 153 159 153 153 groups (i-j) mu1-mu2 mu1-mu3 mu1-mu4 mu2-mus mu2-mu4 mu3-mu4 xbari 153 153 153 164 164 159 LSD xbari-xbarj absolute diff Upper limit Significant(Yes/No) No No No No No No Lower limit 22.67 17.67 11.67 6.67 0.67 5.67 ni nj 6 11.6726 6 11.6726 6 11.6726 6 11.6726 6 11.6726 6 11.6726 0.67 5.67 11.67 16.67 22.67 17.67 6 6 0 0 6

(b)

Critical value for , df=20 and k=4 is

α 0.01

q 5,02

So Tukey's HSD will be

$HSD=q\sqrt{\frac{MSE}{2}\left ( \frac{1}{n_{i}}+\frac{1}{n_{j}} \right )}=5.02\sqrt{\frac{50.5}{2}\left ( \frac{1}{6}+\frac{1}{6} \right )}=14.6644$

The confidence intervals is:

For $\mu_{i}-\mu_{j}$ :

$[\bar{x}_{i}-\bar{x}_{j}]\pm HSD$

Following is the completed table:

HSD xbari-xbarjLower limit Upper limit Significant(Yes/No) No No No No No No groups (i-j)xbari 153 153 153 164 164 159 xbarj 164 159 153 159 153 153 ni nj 6 14.564 14.564 14.564 14.564 14.564 14.564 25.56 20.56 14.56 9.56 3.56 8.56 3.56 8.56 14.56 19.56 25.56 20.56 mu1-mu2 mu1-mu3 mu1-mu4 mu2-mu3 mu2-mu4 mu3-mu4 6 0 6 6 6