Ch Inference for Linear Regression

Transcription

1 Ch Inference for Linear Regression

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3 ACT = (GPA) For every increase of 1 in GPA, we predict the ACT score to increase by population regression line β (true slope) μ y = α + βx mean response α (true y-intercept)

4 GPA: Student #: ACT: ACT = + (GPA) sample regression line y = a + bx a b residuals 2 s n 2 std dev of the residuals. μ y = α + βx α β σ std dev of responses about the population regression line b On Exam formula sheet, y = b 0 + b 1 x is used. Just know coefficient of x is always slope.

5 Sampling Distribution of b N(β, SE b ) Approximately normal Centered at β = 5.17 (statistic b is an unbiased estimator) Measured by the standard error of the slope std dev of residuals = s x n 1 s s x n 1

6 Actual relationship between x & y is linear. μ y = α + βx To check, graph scatterplot to see that overall pattern is linear. Individual observations are independent of each other. When sampling without replacement, check 10% condition. For a fixed x, y varies according to a normal distribution. Make a boxplot, histogram, stemplot, or Normal probability plot of the residuals and check for clear skewness or major outliers. The std dev of y (σ), is the same for each value of x. Graph residual plot; scatter should be roughly the same for all x values. See if the data were produced by random sampling or a randomized experiment. For the normal condition, the AP Exam will most likely show either a histogram or normal probability plot for multiple choice questions. If you re given an FRQ, just graph a boxplot. Remember LINER

7 This is the regression model when the conditions for inference are met. The line is the population regression line.

8 Student #: GPA: ACT: Linear: Scatterplot of GPA vs ACT is fairly linear. Graph scatterplot Independent: Sampling without replacement, so check 10% cond. There are more than 10 9 = 90 students in the HS. Normal: Boxplot of residuals looks approximately normal. Find residuals Plot boxplot Equal Variance: Residual plot shows a similar amount of scatter about Graph residual the residual = 0 line. plot Random: random sample of 9

9 ACT = (GPA) State: We want to estimate the true slope, β, of the population regression line of GPA vs ACT at a 95% confidence level. Plan: t interval for the slope β Linear: Scatterplot of GPA vs ACT is fairly linear. Since we re using SE b (std error), we need to use a t dist always

10 Independent: There are more than 10 9 = 90 students in the HS. Normal: Boxplot of residuals looks approximately normal. Equal Variance: Residual plot shows a similar amount of scatter about the residual = 0 line for GPA values. Random: random sample of 9

11 Do: Estimate ± Margin of Error b ± t SE b df = n 2 = 9 2 = 7 Why is it n 2? Doesn t matter std dev of residuals s = std error of slope SE b = 7.51 ± ± , resid 2 n 2 s s x n 1 (1.29) 1-VarStats for residuals list and use x 2 = = = = 1.29 We now have to find s SE b = s x n 1 std dev of the x values s x = VarStats for GPA values On average, we will have an error of 3.25 when trying to predict an ACT score from a GPA. On average, we will have an error of 1.29 when trying to predict the true slope of the LSRL.

12 With calculator: STAT TESTS LinRegTInt (G) Xlist: L 1 Ylist: L 2 Freq: 1 C-Level: 0.95 RegEQ: leave blank Calculate Not all calculators have LinRegTInt I used LinRegTInt on my calculator and got a confidence interval , with df = 7. y = a + bx , b = df = 7 df s = a = r 2 = r = confidence interval sample slope std dev of residuals sample y-intercept coefficient of determination correlation

13 Conclude: We are 95% confident that the interval from 4.46 to captures the true slope, β, of the population regression line of GPA vs ACT scores.

14 State: H 0 : H a : positive relationship β = 0 β > 0 no linear relationship, slope = 0 β true slope of the population regression line of GPA vs ACT α = 0.05 Plan: t test for the slope β Conditions are the same as previous example

15 Do: Sampling Distribution of b b = 7.51 N 0, 1.29 df = n 2 = 9 2 = 7 SE b t = statistic parameter st. dev. of statistic = b β SE b = = 5.82 area = tcdf 5.82, 99999, 7 = p-value

16 With calculator: STAT TESTS LinRegTTest (F) Xlist: L 1 Ylist: L 2 Freq: 1 β & ρ: 0 < 0 > 0 RegEQ: leave blank Calculate Not all calculators have LinRegTTest I used LinRegTTest on my calculator and got a t-score of 5.82, a p-value of with df = 7. y = a + bx β > 0 and ρ > 0 t = p = E-4 df = 7 df a = b = s = r 2 = r = t-score p-value sample y-intercept sample slope std dev of residuals coefficient of determination correlation

17 Conclude: Assuming H 0 is true β = 0, there is a probability of obtaining a b value of 7.51 or more purely by chance. This provides strong evidence against H 0 and is statistically significant at α = 0.05 level ( <.05). Therefore, we reject H 0 and can conclude that there is a positive linear relationship between GPA and ACT scores.

18 y = a + bx will never use std dev of residuals a = b = SE a = SE b = r 2 coefficienct of determination t-score for α t-score for β 82.8% of the variation in y is explained or accounted for by the LSRL. 17.2% of the variation is explained by other factors besides GPA. never use p-value for α p-value for β for two-sided test so 2