1 / 24

Inference for  0 and  1

Inference for  0 and  1. Confidence intervals and hypothesis tests. Example 1: Relation between leg strength and punting distance?. PUNTER LEG DIST 1 170 162.50 2 140 144.00 3 180 147.50 4 160 163.50

tiger
Download Presentation

Inference for  0 and  1

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


  1. Inference for0 and 1 Confidence intervals and hypothesis tests

  2. Example 1: Relation between leg strength and punting distance? PUNTER LEG DIST 1 170 162.50 2 140 144.00 3 180 147.50 4 160 163.50 5 170 192.00 6 150 171.75 7 170 162.00 8 110 104.83 9 120 105.67 10 130 117.58 11 120 140.25 12 140 150.16 13 160 165.16 • 13 punters in American football • DIST = average length (feet) of 10 punts • LEG = strength of leg (pounds lifted)

  3. Example 2: Relation between state latitude and skin cancer mortality? • Mortality rate of white males due to malignant skin melanoma from 1950-1959. • LAT = degrees (north) latitude of center of state • MORT = mortality rate due to malignant skin melanoma per 10 million people • # State LAT MORT • 1 Alabama 33.0 219 • 2 Arizona 34.5 160 • 3 Arkansas 35.0 170 • 4 California 37.5 182 • 5 Colorado 39.0 149 •    49 Wyoming 43.0 134

  4. Example 3: Relation between use of amphetamines and food consumption? X = Amphetamine dose (mg/kg) 0 2.5 5.0 112.6 73.3 38.5 102.1 84.8 81.3 90.2 67.3 57.1 81.5 55.3 62.3 105.6 80.7 51.5 93.0 90.0 48.3 106.6 75.5 42.7 108.3 77.1 57.9 • 24 rats randomly allocated to dose of amphetamine (saline (0), 2.5, and 5.0 mg/kg) • Y = amount of food (grams of food consumed per kilogram of body weight) in following 3-hour period

  5. Point estimates b0 and b1 The b0 and b1 values vary. They depend on the particular (xi, yi) sample obtained.

  6. Assumptions about error terms i • E(i) = 0 • i and j are uncorrelated • Var(i) = 2 • (New!!) i are normally distributed NOTE: All results thus far (such as least squares estimates, Gauss-Markov Theorem, and mean square error) only depend on first three assumptions. Today’s results depend on normality of the error terms.

  7. b1 is normally distributed and variance Sampling distribution of b1 Providing error terms i are normally distributed: with mean 1

  8. Recall: Confidence interval for  using  Sample estimate ± margin of error

  9. Confidence interval for 1 using b1 Sample estimate ± margin of error

  10. Test statistic Recall: Hypothesis testing for  using  The null (H0:  = 0) versus the alternative (HA:  ≠ 0) P-value = How likely is it that we’d get a test statistic t* as extreme as we did if the null hypothesis is true? The P-value is determined by comparing the test statistic t* to a t-distribution with n-1 degrees of freedom.

  11. Test statistic Hypothesis testing for 1 using b1 The null (H0: 1 = ) versus the alternative (HA: 1 ≠ ) P-value = How likely is it that we’d get a test statistic t* as extreme as we did if the null hypothesis is true? The P-value is determined by comparing the test statistic t* to a t-distribution with n-2 degrees of freedom.

  12. b0 is normally distributed and variance Sampling distribution of b0 Providing error terms i are normally distributed: with mean 0

  13. Confidence interval for 0 using b0 Sample estimate ± margin of error

  14. Test statistic Hypothesis testing for 0 using b0 The null (H0: 0 =  ) versus the alternative (HA: 0 ≠ ) P-value = How likely is it that we’d get a test statistic t* as extreme as we did if the null hypothesis is true? The P-value is determined by comparing the test statistic t* to a t-distribution with n-2 degrees of freedom.

  15. Example 1: Inference The regression equation is punt = 14.9 + 0.903 leg Predictor Coef SE Coef T P Constant 14.91 31.37 0.48 0.644 leg 0.9027 0.2101 4.30 0.001 S = 16.58 R-Sq = 62.7% R-Sq(adj) = 59.3% Unusual Observations Obs leg punt Fit SE Fit Residual St Resid 3 180 147.50 177.39 8.20 -29.89 -2.07R R denotes an observation with a large standardized residual

  16. Example 2: Inference The regression equation is Mortality = 389 - 5.98 Latitude Predictor Coef SE Coef T P Constant 389.19 23.81 16.34 0.000 Latitude -5.9776 0.5984 -9.99 0.000 S = 19.12 R-Sq = 68.0% R-Sq(adj) = 67.3% Unusual Observations Obs Latitude Mortality Fit SE Fit Residual St Resid 7 39.0 200.00 156.06 2.75 43.94 2.32R 9 28.0 197.00 221.82 7.42 -24.82 -1.41 X 30 35.0 141.00 179.97 3.85 -38.97 -2.08R R denotes an observation with a large standardized residual X denotes an observation whose X value gives it large influence.

  17. Example 3: Inference The regression equation is consumption = 99.3 - 9.01 dose Predictor Coef SE Coef T P Constant 99.331 3.680 26.99 0.000 dose -9.008 1.140 -7.90 0.000 S = 11.40 R-Sq = 73.9% R-Sq(adj) = 72.8% Unusual Observations Obs dose consumpt Fit SE Fit Residual St Resid 18 5.00 81.30 54.29 3.68 27.01 2.50R R denotes an observation with a large standardized residual.

  18. Inference for 0 and 1 in Minitab • Select Stat. • Select Regression. • Select Regression … • Specify Response (y) and Predictor (x). • Click on OK.

More Related