This preview shows page 1. Sign up to view the full content.
Unformatted text preview: Stat 4120 Spring 2011 Take-Home Assignment 1 Solutions 1: Do exercise 2.11 on page 37 of the textbook. Hint: If you find the correct F-value, the P-value will be 0.0101. Source Treatments Error Total df 4 29-4=25 29 SS 4*10=40 100-40=60 100 MS 10 60/25=2.4 F-Value 10/2.4=4.17 P-Value 0.0101 Given values are shown in Bold. There are 4 degrees of freedom for treatments because there are five treatments. There are 29 total degrees of freedom due to the 30 observations. All other calculations are indicated in the table. 2: Do exercise 2.18 on page 39 of the textbook. In part (c) in the question, use Tukey’s procedure for pairwise comparisons, as usual, but do not use Scheffé’s method for pairwise comparisons. Use Scheffé’s method to test the contrast:
H0 : µ 0.3 + µ 0.4 µ 0.5 + µ 0.6 = 2 2 HA : µ 0.3 + µ 0.4 µ 0.5 + µ 0.6 ≠ 2 2 € Do not attempt to correct any problems you might find with the residuals, but comment on them. € a) The graph of the data values indicates increasing tensile strength as the carbon content increases, perhaps in a nonlinear manner. b) One-way ANOVA: Strength versus Carbon Source Carbon Error Total DF 3 16 19 SS 243960 25640 269600 MS 81320 1603 F 50.75 P 0.000 The small p-value indicates there is an effect of carbon content on the tensile strength of steel. The residual graphs show reasonable Normality and constant variance. c) Grouping Information Using Tukey Method Carbon 0.60% 0.50% 0.40% 0.30% N 5 5 5 5 Mean 1754.00 1594.00 1504.00 1468.00 Grouping A B C C Tukeys’ pairwise comparisons show that 0.60% carbon content leads to the strongest steel with an estimated mean tensile strength of 1754 kilograms per square centimeter. Both 0.30% and 0.40% carbon content lead to the weakest steel with an estimated mean tensile strength between 1468 and 1504 (probably near 1486) kilograms per square centimeter. Testing H 0 :
Sum of Squares Numerator DF Denominator DF €F Ratio Prob > F µ 0.3 + µ 0.4 µ 0.5 + µ 0.6 µ + µ 0.4 µ 0.5 + µ 0.6 = ≠ versus H A : 0.3 yields the result 2 2 2 2
176720 1 16 € 110.27769111 1.3849193e-8 The very small p-value leads to rejection of H0. There is sufficient evidence to conclude that the average of the population mean tensile strength of steel with 0.3 and 0.4 percent carbon is not the same as the average population mean tensile strength of steel with 0.5 and 0.6 percent carbon. In fact, from the sample means, you can see the average population mean tensile strength of steel with 0.3 and 0.4 percent carbon is less than the average population mean tensile strength of steel with 0.5 and 0.6 percent carbon. 3: In a study of four different methods of producing clay slurry, it was found that there is no statistically significant difference among the mean viscosities for the four methods. Consequently, the chemists turned their attention to the standard deviation, because consistency is also an important characteristic of the slurries. The data in the Take-Home 1 Data file are the results of measuring the standard deviation of six batches of five slurries each for each production method. Use the data to test the hypotheses H0: The mean standard deviations of the four methods are the same. HA: not H0 Correct any problems you find with the residuals. For the corrected data, use a pairwise comparison method to determine which method(s) is most consistent. Residual graphs for the ANOVA for the original data show reasonable Normality, but non-constant variance. Transforming the data using the square root transformation yields the ANOVA results
Source Method Error Total DF 3 20 23 SS 0.09506 0.02109 0.11615 MS 0.03169 0.00105 F 30.05 P 0.000 The residuals now appear to have nearly constant variance. The Normality of the residuals was not affected. Tukey’s method used for the square root of the standard deviation indicates the smallest mean standard deviation (most consistent result) will be achieved using method 2 or 4. Grouping Information Using Fisher Method Method 3 1 4 2 N 6 6 6 6 Mean 0.33479 0.21483 0.18917 0.17548 Grouping A B BC C 4: Do exercise 3.13 on pages 68 and 69 of the textbook. In addition to testing for the treatment effect, state whether blocking was a good idea in this experiment. Hint: If you find the correct F-value for treatments, the corresponding P-value will be 0.376. Source df Blocks 5 Treatments 3 Error 15 Total 23 SS 145 200–145–45=10 15*3=45 200 MS 145/5=29 10/3=3.33 3 F-Value P-Value 29/3=9.67 3.33/3=1.11 0.376 Given values are shown in Bold. There are 5 degrees of freedom for blocks because there are 6 blocks. There are 3 degrees of freedom for treatments because there are 4 treatments. There are 23 total degrees of freedom because there must be 24 observations (6*4), because this is a randomized complete block design. All other calculations are indicated in the table. I did not include a p-value for the blocks, because a test for blocks is generally not appropriate. However, because the F-value for blocks is much greater than 1, blocking was helpful in this experiment. 5: Do exercise 3.26 on pages 71 and 72 of the textbook. In part (d), use only Tukey’s method. Does it appear that the requirement of no interaction between the blocks and the treatment is satisfied? Why? Do not attempt to correct any problems you might find with the residuals, but comment on them. a) The subjects are likely to be quite different, contributing a lot of variability that could “mask” the effect of the gas. Blocking on the subjects will take that variability into account, making the test for the gas much more powerful. b) The graph of distance by type of gas does not indicate there are any differences in the mean distances walked for the four gases. Note that this graph does not take into account the subject differences. c) The ANOVA results indicate a difference on mean distance due to the type of gas breathed. The large F-value for the subjects (the blocks) indicates that blocking was a good idea for this experiment. Subjects are a random effect.
Source Gas Subject Error Total DF 3 6 18 27 Seq SS 44827 1471772 16316 1532915 Adj SS 44827 1471772 16316 Adj MS 14942 245295 906 F 16.48 270.61 P 0.000 0.000 The residual graphs show relatively good Normality with a tendency for the variability of the residuals to decrease with increasing predicted value of distance. The variance components below illustrate how much variability is accounted for by the subjects (98.5% of the total variance).
Variance Components, using Adjusted SS Source Subject Error Estimated Value 61097.2 906.5 d) Tukey’s method for pairwise comparisons based on the square root of the distance finds that only Air + CO leads to a lower mean distance walked.
Grouping Information Using Tukey Method and 95.0% Confidence Gas Oxygen Oxygen+CO Air Air+CO N 7 7 7 7 Mean 592.1 571.7 547.1 485.6 Grouping A A A B e) The graph of the residuals versus the gases (as opposed to the fitted values) shows equal variability of distance walked across the four types of gases. ...
View Full Document
This note was uploaded on 02/28/2011 for the course STAT 4120 taught by Professor Vanbrackle during the Spring '11 term at Kennesaw.
- Spring '11