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Unformatted text preview: 1 Homework 10 (9 points) due Dec. 2 (2.5 pts.) 11.12 (p.507) [Please use two different hypotheses tests.] Exercise 23 (Section 3.3) gave SAS output from a regression of attenuation on fracture strength. a) Does the simple linear regression model appear to specific a useful relationship between these two variables? State the relevant hypothesis and carry out a test using two different hypothesis test. using : 1. Let be the true mean slope of attenuation vs. fracture strength. 2. H : = 0 H a : ≠ 0 3. = 0.05 4. ? = ? ? ? = − 0.014711 0.00143620 = − 10.243 ( ¡?¢ ? h £ ¤?¥ ¥??¦£§ ? §£¨?§£© ) 5. SAS = 0.0001 df = 12, P = 2P(t > 10.243) < 0.002 (not required) 6. Reject H because 0.0001 0.05 7. This data does give strong support (P = 0.0001) to the claim that there is a linear relationship between attenuation and fracture strength. using : 1. Let be the true mean correlation coefficient of attenuation vs. fracture strength. 2. H : = 0 H a : ≠ 0 3. = 0.05 4. ? = § ? − 2 1 − § 2 = ( − 0.8974 ) 12 1 − 0.8974 = − 10.243 ( ¡?¢ ? h £ ¤?¥ ¥??¦£§ ? §£¨?§£© ) 5. SAS = 0.0001 df = 12, P = 2P(t > 10.243) < 0.002 (not required) 6. Reject H because 0.0001 0.05 7. This data does give strong support (P = 0.0001) to the claim that there is a linear relationship between attenuation and fracture strength. 2 using F test 1. Let be the true mean slope of attenuation vs. fracture strength. 2. H : = 0 H a : ≠ 0 3. = 0.05 4. ? = ? = 2.29469 0.02187 = 104.917 ( ¡¢? ? h £ ¤¢¥ ¥¢¦§£¨ ¦ ¨£©?¨£ª ) 5. SAS = 0.0001 df 1 = 1 df 2 = 12, P = P(F > 104) < P(F > 9.33) = 0.001 (not required) 6. Reject H because 0.0001 0.05 7. This data does give strong support (P = 0.0001) to the claim that there is a linear relationship between attenuation and fracture strength....
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 Spring '08
 Staff
 Statistics, Linear Regression, Regression Analysis, crude oil, fracture strength, true mean slope

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