1. The Simple Linear Regression Model

2. Estimators in Simple Linear Regression and

3. Sampling distributions of the estimators

4. Recall that if y 1, y 2, y 3 …, y n are 1.Independent 2.Normally distributed with means  1,  2,  3 …,  n and standard deviations  1,  2,  3 …,  n Then L = c 1 y 1 + c 2 y 2 + c 3 y 3 + … + c n y n is normal with mean and standard deviation

5. Sampling distribution the slope

6. Note : Also

7. Thus Hence where and standard deviation is normal with mean

8. Thus since and

9. Also

10. and standard deviation Henceis normal with mean

11. Sampling distribution of the intercept

12. The sampling distribution intercept of the least squares line : It can be shown that has a normal distribution with mean and standard deviation

13. Proof: where Thus

15. Also now

16. Hence and

17. and standard deviation Summary is normal with mean is normal with mean and standard deviation 1. 2.

18. Sampling distribution of the estimate of variance

19. The sampling distribution of s 2 This estimate of  is said to be based on n – 2 degrees of freedom

20. The sampling distribution of s 2 Recall that y 1, y 2, …, y n are independent, normal with mean  +  x i and standard deviation  Let Then z 1, z 2, …, z n are independent, normal with mean 0 and standard deviation 1, and Has a  2 distribution with n degrees of freedom

21. If  and  are replaced by their estimators: then has a  2 distribution with n-2 degrees of freedom Note:

22. Thus This verifies the statement made earlier that s 2 is an unbiased estimator of  2. and

23. and standard deviation Summary is normal with mean is normal with mean and standard deviation 1. 2.

24. and standard deviation Recall is normal with mean Therefore has a standard normal distribution

25. has a t distribution with n – 2 degrees of freedom and

26. (1 –  )100% Confidence Limits for slope  : t  /2 critical value for the t-distribution with n – 2 degrees of freedom

27. and standard deviation Also is normal with mean Therefore has a standard Normal distribution

28. and has a t distribution with n – 2 degrees of freedom

29. (1 –  )100% Confidence Limits for intercept  : t  /2 critical value for the t-distribution with n – 2 degrees of freedom

30. The following data showed the per capita consumption of cigarettes per month (X) in various countries in 1930, and the death rates from lung cancer for men in 1950. TABLE : Per capita consumption of cigarettes per month (X i ) in n = 11 countries in 1930, and the death rates, Y i (per 100,000), from lung cancer for men in 1950. Country (i)X i Y i Australia4818 Canada5015 Denmark3817 Finland11035 Great Britain11046 Holland4924 Iceland236 Norway259 Sweden3011 Switzerland5125 USA13020

32. Fitting the Least Squares Line

33. First compute the following three quantities:

34. Computing Estimate of Slope and Intercept

35. 95% Confidence Limits for slope  : t.025 = 2.262 critical value for the t-distribution with 9 degrees of freedom 0.0706 to 0.3862

36. 95% Confidence Limits for intercept  : -4.34 to 17.85 t.025 = 2.262 critical value for the t-distribution with 9 degrees of freedom

37. (1 –  )100% Confidence Limits for a point on the regression line  +  x 0 : x y regression line  +  0 x 0 x0x0 y =  +  0 x

38. Let then and

39. Proof: where Note and Thus

42. Also now

43. Hence and

44. (1 –  )100% Confidence Limits for a point on the regression line intercept  +  x 0 : t  /2 critical value for the t-distribution with n - 2 degrees of freedom

45. Prediction In linear regression model

46. (1 –  )100% Prediction Limits for y when x = x 0 : t  /2 critical value for the t-distribution with n - 2 degrees of freedom