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Unformatted text preview: STAT 350 Lecture 26 11.2 Inferences about the slope Final Exam Date: May 3, 2011 (Tuesday) Time: 7 PM 9 PM Where: EE 270 Crib sheet: three pages, 1sided, handwriSen Simple Linear Regression Model The true regression line as a model: yi = + xi + ei In this model: ei is assumed to follow a normal distribuUon with mean 0 and standard devia,on all ei 's are assumed independent of each other Our Goal: EsUmaUng , , and EsUmaUng the slope and intercept A point esUmate of is b A point esUmate of is a EsUmaUng The model assumes ei ~ N(0, ) Recall: Why n 2? We calculaUng two esUmates, a and b, hence we lose 2 df Example Revisited (ex 20 on page 126) Example Revisited (Lecture 21  ex 20 on page 126) 1) Find point esUmates of the slope and intercept of the populaUon regression line. 2) What is the equaUon of the esUmated true regression line? 3) Find a point esUmate of the error standard deviaUon . 4) What proporUon of the observed variaUon in y can be aSributed to the simple linear regression relaUonship between x and y? Example Revisited (Lecture 21  ex 20 on page 126) 1) Find point esUmates of the slope and intercept of the populaUon regression line. 2) What is the equaUon of the esUmated true regression line? y = 15.245 + 0.094x + e Example Revisited (Lecture 21  ex 20 on page 126) 3) Find a point esUmate of the error standard deviaUon . Answer: note: Root MSE entry on the SAS output 4) What proporUon of the observed variaUon in y can be aSributed to the simple linear regression relaUonship between x and y? Answer: Rsquare = 0.4514 11.2 Inferences about the slope NotaUon: : the slope of the populaUon regression line b: the slope of the sample regression line b is a point es,mate of , just like is a point esUmate of p is a point esUmate of Goal: Sta,s,cal Inference A Confidence Intervals for TesUng Hypotheses about Sampling distribuUon of the slope b b follows a normal distribuUon Standardize b: Sampling distribuUon of the slope, b Standard Z distribuUon when is known: When is unknow, replacing b by sb gives us a familiar t distribuUon with df = n 2 Confidence interval for The confidence interval for is: b (t crit)sb Procedures: Calculate esUmates b and sb Determine confidence level Find t crit from the table Interpret the interval for the true slope Example 11.4 (same data as 11.2) Calculate 95% CI for : b (t crit)sb Step 1: Calculate b Example 11.4 (same data as 11.2) Calculate 95% CI for : b (t crit)sb Step 2: Calculate sb Example 11.4 (same data as 11.2) Calculate 95% CI for : b (t crit)sb Step 3: find t* and the CI 95% CI: 0.92 (2.16)*(0.146) 0.920.315 (1.233, 0.603) Hypotheses tesUng for H0: = 0 Ha: 0 (can do < or >) Test staUsUcs is: Based on t distribuUon with df = n 2 Usually want to test H0: = 0 Why? Model UUlity Test H0: =0 (no useful linear relaAonship) Ha: 0(can do < or >) Test staUsUcs is: Based on t distribuUon with df = n 2 Learn to read b and sb from SAS output Example Revisited (ex 20 on page 126) Example Revisited (ex 20 on page 126) Perform a model uUlity test H0: =0 (the model is not useful) Ha: 0 (there is a useful linear relaUonship between the variables) Example Revisited (ex 20 on page 126) Perform a model uUlity test H0: =0 (the model is not useful) Ha: 0 (there is a useful linear relaUonship between the variables) SAS output b=0.09424 and sb=0.02215 Example Revisited (ex 20 on page 126) Perform a model uUlity test H0: =0 (the model is not useful) Ha: 0 (there is a useful linear relaUonship between the variables) SAS output b=0.09424 and sb=0.02215 Test staUsUc t=b/sb = 0.009424/0.002215=4.25 df=22, and pvalue=0; so reject H0 in favor of Ha. Conclusion: there is a useful linear relaUonship between the variables. ExampleHeight and Weight The following data set gives the average heights and weights for American women aged 3039 (source: The World Almanac and Book of Facts, 1975). Total observaUons 15. SAS Code proc reg data=example; model weight = height; plot weight*height; run; Note the `clb' opUon will produce a confidence interval for b in addiUon to the hypothesis test ExampleHeight and Weight SAS outputHeight and Weight Using the SAS output for inference Construct a 95% confidence interval for Test whether or not there is a significant linear relaUonship (H0: = 0) SAS Code proc reg data=example; model weight = height / alpha=0.05 clb; plot weight*height; run; Note the `clb' opUon will produce a confidence interval for b in addiUon to the hypothesis test SAS output Equivalent CorrelaUon test H0: = 0 H0: = 0 (the model is not useful) Ha: 0 Ha: 0 Test staUsUcs is: Where r is the sample correlaUon coefficient SUll based on t distribuUon with df = n 2 Example (ex 15 on page 508) The value of the sample correlaUon coefficient is 0.449 for the n=14 observaUons on x = hydorgen content and y = gas porosity. Carry out a test at significance level 0.05 to decide whether these two variables are linearly related in the populaUon from which the data was selected. Example (ex 15 on page 508) SoluUon (=0.05): H0: = 0 (the model is not useful) Ha: 0 Also, from the data that r=0.449 and n=14 df=12, pvalue=0.11 > 0.05 Conclusion: no evidence of linear relaUonship. => do not reject H0 SAS output Weight and Height Example (Revisit) Carry out a significant test to decide if weight and height are linearly related: H0: = 0 (no linear relaAonship) Ha: 0 =0.05 (default) Also, from the SAS output that R2=0.9910 and n=15 df=13, pvalue<0.001 => reject H0 Conclusion: strong evidence of linear relaUonship. ...
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 Spring '08
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 Statistics, Linear Regression

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