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Unformatted text preview: the Mean Consider the relationship between exam 2 and final exam scores … Least squares regression line (or estimated regression function): ˆ
y 21.67 + 1.046(x) also E Y = 21.67 + 1.046(x) We also have: s 8.24671 How would you predict the final exam score for Barb who scored 60 points on exam 2? ˆ
y 21.67 + 1.046(60) = 84.43 points. How would you estimate the mean final exam score for all students who scored 60 points on exam 2? E Y = 21.67 + 1.046(60) = 84.43 points. So our estimate for predicting a future observation and for estimating the mean response are found using the same least squares regression equation. What about their standard errors? (We would need the standard errors to be able to produce an interval estimate.) Idea: Consider a population of individuals and a population of means: n Population of individuals Population of means What is the standard deviation for a population of individuals? What is the standard deviation for a population of means? n Which standard deviation is larger? So a prediction interval for an individual response will be (wider or narrower) than a confidence interval for a mean response. 197 Here are the (somewhat messy) formulas: Confidence interval for a mean response: ˆ* y t s.e.(fit) s.e.(fit ) s where (x x)
1 n x i x 2 Sum that comes up a lot! 2 df = n – 2 the x that you are doing
Prediction interval for an individual response: the predicting/estimating at ˆ
y t * s.e.(pred) 2
s.e.(pred) s 2 s.e.(fit ) the sdf = n xtra variability
= e – 2 2 where that makes PI wider Try It! Exam 2 vs Final Construct a 95% confidence interval for the mean final exam score for all students who scored x = 60 points on exam 2. Recall: n = 6, x 51 , x x 2 S XX 940 , y 21.67 +1.046(x), and s = 8.24761. ˆ ˆ
y 21.67 + 1.046(60) = 84.43 cm. t* = 2.78 (with df = 4)
s.e.(fit ) s 1 n (x x) 2 x i x 2 8.24761 1 (60 51) 2 4.147
y t *s.e.(fit ) 84.43 (2.78)(4.147) 84.43 11.53 => ( 72.9, 95.96) Construct a 95% prediction interval for the final exam score for an in...
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