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Department of Epidemiology and Public Health Unit of Biostatistics and Computational Sciences Ordinary linear regression PD Dr. C. Schindler Swiss Tropical and Public Health Institute University of Basel christian.schindler@unibas.ch Annual meeting of the Swiss Societies of Clinical Neurophysiology and of Neurology, Lugano, May 3 rd 2012

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Example: Association between blood volume and body weight in women

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Question: How does the mean of blood volume depend on body weight in women?

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The regression line y = 893 + 45.7 · x y = 893 + 45.7 · 70 = 4092

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In this example, the regression line describes the mean of blood volume of women as a function of weight. * syn. outcome variable ** syn. explanatory or predictor variable In general: The regression line describes the mean of the dependent variable Y* as a function of the independent variable X**.

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0 0 y = + · x = intercept = y-value of the line at x = 0 = slope of the line = change in y, if x increases by one unit x y = y / x x y Regression equation and regression parameters Regression parameters

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The values of the parameters must be determined from empirical data. They are estimates of the respective true parameter values at the population level. Therefore, they are referred to as parameter estimates.

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^ = estimated intercept = 893 ml: for a weight of 0 kg, a blood volume of 893 ml would be predicted. ^= estimated slope = 45.7 ml/kg: According to this model, the mean of blood volume in women is supposed to increase by 45.7 ml with each additional kg of weight. Of course, this interpretation does not make sense, since valid predictions can only be made for values of weight between 50 and 80 kg (range of observed values) Interpretation of parameter estimates Note: and denote the parameters of the true regression line at the population level.

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Residuals and predicted values Residual plot Residual = deviation of the observed value of the dependent variable (here: blood volume) from the value which the model predicts for the respective value of the independent variable (here: weight) (-> predicted value).

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Definition and properties of the regression line 2. The regression line always runs through the point (mean of X, mean of Y) i.e., for the mean of the independent variable, the regression line always predicts the mean of the dependent variable. 1.Among all possible lines, the regression line stands out as the one with the smallest possible variance of the residuals.

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Regression output of a statistics program (SPSS) The rightmost column (Sig) contains the p-values of the two parameter estimates. They refer to the deviation of these estimates from 0. The t-value (4. column) equals the ratio between the parameter estimate (B) and its standard error (Std. Error). The standardized coefficient equals Pearsons correlation coefficient. 1 parameter estimate, 2 intercept, 3 slope Coefficients a Model Unstandardized Coefficients Standardized Coefficients tSig. B 1 Std. ErrorBeta 1 (Constant) 2 893.253369.8272.415.020 Gewicht ) 3 45.6825.846.7777.815.000 a. Dependent Variable: blutvol

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With a p-value < 0.0001, the deviation of the estimated slope from 0 is highly significant. *If the true slope is 0 then two situations are possible: a) the mean of Y does not depend on X at all or b) the mean of Y depends on X in a specific non-linear way (see next slide) The hypothesis, that the slope of the true regression line be 0 can therefore be rejected at the usual significance level of 0.05 (in fact even at a significance level of 0.0001). Slope

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Here = 0. The mirror symmetriy of the curve with respect to the vertical axis at x = 0 forces the regression line to run horizontally. y = 0.1 · x 2 y x y = 0 · x + 8

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With a p-value < 0.05, the deviation of the estimated intercept from 0 is statistically significant as well at the usual level of 0.05. Therefore, the hypothesis that the true regression line pass through the origin of the coordinate system, can also be rejected at the usual level of 0.05. Intercept

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Approximate 95%-confidence interval of the slope (Parameter estimate ± 2 standard error) 45.7 ± 2 · 5.8 = (34.1, 57.3) It is thus quite certain that the true regression slope is higher than 30 and lower than 60 ml/kg. We can be 95% confident that the slope of the regression line at the population level lies between 34.1 and 57.3 ml/kg.

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Coefficients a Model Unstandardized Coefficients Standardized Coefficients tSig. BStd. ErrorBeta 1 (Constant) 893.253369.8272.415.020 Gewicht 45.6825.846.7777.815.000 a. Dependent Variable: blutvol Model Summary ModelRR Square Adjusted R Square Std. Error of the Estimate 1.777 a.604.594308.45008 a. Predictors: (Constant), Gewicht Standard deviation of residuals Proportion of variance of Y, which is explained by the model Other important parameters of a regression model (SPSS)

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Decomposition of total variance Total variance = variance of predicted values + variance of residuals Total variance = sum of squared deviations of the individual values of Y from their mean value. Variance of residuals = sum of squared residuals (residual sum of squares) Variance of predicted values = sum of squared deviations of the predicted values of Y from the sample mean of Y. explained variance unexplained variance

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R 2 -value (or measure of determination) of the model Note: R 2 = 1 The data are completely explained by the model, i.e., all the points lie on the regression line. R 2 = 0 slope of the regression line = 0. explained variance* total variance* = total variance* - unexplained variance* total variance* * of Y

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Regression line with 95%-confidence intervals Confidence intervals of predicted values become wider with increasing distance from the center.

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Power considerations X s s bSE 1-n )( residuals SE(b) is proportional to the standard deviation of residuals -> the residuals should be as small as possible SE(b) is inversely proportional to the square root of n-1 -> n should be sufficiently large SE(b) is inversely proportional to the standard deviation of X -> the range of X should be as large as possible (standard error of the slope)

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Conditions for the validity of a regression model

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The residual plot should display a horizontal point cloud (no banana or wave shape). -> validity of parameter estimates, confidence intervals and p-values) The (vertical) variability of the residuals should be more or less constant across the whole range of the independent variable (condition of homoscedasticity). -> validity of confidence intervals and p-values a) b)

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the distribution of residuals should be approximately normal (visual assess- ment by normal probability plot). -> validity of confidence intervals and p-values 1.Each observational unit should only occupy one row of the data table (i.e., each subject should contribute one observation to the analysis). 2.If the individual observational units can be grouped into clusters (families, hospitals, etc.) then the cluster means of residuals must not vary systemati- cally between the clusters (i.e., cluster means of residuals should differ from 0 only by chance*). -> validity of confidence intervals and p-values c) d) *If they dont, one should introduce the cluster variable as additional fixed or random factor into the regression model.

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Beware: Not all relations can be well described by a regression line. Very often, relation(s) between dependent and independent variable(s) are non-linear. Linear association Non-linear association y = -22.6 + 2.3 · x y = -1.6 + 4.26 · x – 0.039 · x 2

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Multiple regression models (illustration based on concrete example) Association between systolic blood pressure, gender, age and overweight

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Different purposes of regression models 1.Prediction models ex. Prediction of blood volume based on weight. Prediction of clinical outcome after t years. 2.Reference models ex. Growth curves, reference values for functional parameters as a function of sex, age, etc. 3. Explanatory models* describe the parallel influences of different predictor variables on a given outcome variable. e.g., Influence of sex, age and obesity on systolic blood pressure. * also serve to protect effect estimates against confounding.

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Aim 1: Reference model for adult systolic blood pressure (SBP) in Lugano as a function of sex and age. Sample used: SAPALDIA-subjects from Lugano with normal weight (i.e., BMI < 25 kg/m 2 )

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SAPALDIA study (Swiss Cohort Study on Air Pollution and Lung and Heart Diseases in Adults) 1st survey (1991): n = 9651 lung health (symptoms/lung function) + allergies 2nd survey (2002): n 6500 lung health + allergies + cardiovascular health (blood pressure, 24hr – ECG) 8 study areas (Basel, Geneva, Lugano, Aarau, Wald, Payerne, Davos, Montana) Study subjects were between 18 and 60 years old in 1991 and had to be resident in the respective area for at least 3 years.

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SAPALDIA: Study areas

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Statistical method Ordinary linear regression (quantitative outcome) Simple model: E[SBP | sex, age] = b 0 + b 1 · female + b 2 · age_50 female = binary variable with 1 in women and 0 in men. age_50 = age – 50 (age centered at 50 yrs) E[SBP | sex, age] = mean of SBP (as a function of sex and age) predicted value of Y ( ) expected value of Y ( )

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Source | SS df MS Number of obs = 480 -------------+------------------------------ F( 2, 477) = 69.24 Model | 37845.7847 2 18922.8923 Prob > F = 0.0000 Residual | 130361.613 477 273.294787 R-squared = 0.2250 -------------+------------------------------ Adj R-squared = 0.2217 Total | 168207.398 479 351.16367 Root MSE = 16.532 ------------------------------------------------------------------------------ bpsys | Coef. Std. Err. t P>|t| [95% Conf. Interval] -------------+---------------------------------------------------------------- female | -13.41972 1.593982 -8.42 0.000 -16.55181 -10.28762 age_50 |.5503499.0650948 8.45 0.000.4224418.678258 _cons | 128.5214 1.296175 99.15 0.000 125.9745 131.0683 ------------------------------------------------------------------------------ Result of regression model (program STATA) 1. The age-adjusted mean of systolic blood pressure was significantly lower among women (i.e., by 13.4 mm Hg). 3. The value of the intercept parameter, 128.5 mm Hg, is the estimated mean of SBP in 50 year old men (they have female = 0 and age_50 = 0). 2. The gender-adjusted mean of SBP showed a mean increase of 0.55 mm Hg per year.

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Point line is slightly curved -> distribution of residuals is slightly skewed Normal probability plot (QQ-plot)

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Residual plot (vertical) variability of residuals increases from left to right x-axis: predicted values y-axis: residuals

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If the distribution of residuals is left skewed and their (vertical) variability gets larger with increasing predicted values, then a logarithmic transformation of the data often helps. We will thus consider the new outcome variable Y = ln(SBP)

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Statistical method Ordinary linear regression (quantitative outcome) Alternative model: E[ln(SBP) | sex, age] = b 0 + b 1 · female + b 2 · age_50 E[ln(Y) | sex, age] = mean of ln(Y) as a function of sex and age exp{E[ln(Y) | sex, age]} = geometric mean of Y as a function of sex and age. median of Y as a function of sex and age (if residuals are symmetrically distributed) e E[ln(Y) | sex, age]

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Residual plot x-axis: predicted values y-axis: residuals Point line is almost linear -> distribution of residuals close to normal -.4 -.2 0.2.4 Residuals -.4-.20.2.4 Inverse Normal

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(vertical) variability of residuals increases less strongly from left to right x-axis: predicted values y-axis: residuals Residual plot

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Source | SS df MS Number of obs = 480 -------------+------------------------------ F( 2, 477) = 68.25 Model | 2.49178871 2 1.24589436 Prob > F = 0.0000 Residual | 8.70727941 477.018254255 R-squared = 0.2225 -------------+------------------------------ Adj R-squared = 0.2192 Total | 11.1990681 479.0233801 Root MSE =.13511 ------------------------------------------------------------------------------ lnbpsys | Coef. Std. Err. t P>|t| [95% Conf. Interval] -------------+---------------------------------------------------------------- female | -.1109775.0130272 -8.52 0.000 -.1365753 -.0853798 age_50 |.0043791.000532 8.23 0.000.0033337.0054244 _cons | 4.846471.0105933 457.50 0.000 4.825656 4.867287 ------------------------------------------------------------------------------ 1. The age-adjusted mean of ln(SBP) was lower by 0.11 in women. The geometric mean ratio of SBP between women and men was exp(-0.11) = 0.90. The geometric mean of SBP was lower in women by 10%. 2. On average, the geom. mean of SBP increased by a factor of exp(0.0043) = 1.0043, i.e., by 0.43% per year of age. 3. The estimated geometric mean of SBP in 50 year old men is exp(4.846) = 127.2.

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Source | SS df MS Number of obs = 480 -------------+------------------------------ F( 3, 476) = 45.77 Model | 2.5071076 3.835702532 Prob > F = 0.0000 Residual | 8.69196052 476.018260421 R-squared = 0.2239 -------------+------------------------------ Adj R-squared = 0.2190 Total | 11.1990681 479.0233801 Root MSE =.13513 ------------------------------------------------------------------------------- lnbpsys | Coef. Std. Err. t P>|t| [95% Conf. Interval] --------------+---------------------------------------------------------------- female | -.1103769.0130459 -8.46 0.000 -.1360115 -.0847423 age_50 |.0043318.0005346 8.10 0.000.0032814.0053823 age_50squared |.0000422.000046 0.92 0.360 -.0000483.0001326 _cons | 4.840393.0125019 387.17 0.000 4.815827 4.864959 ------------------------------------------------------------------------------- Is the relation between ln(SBP) and age linear? may be assessed by adding the square of age_50: age_50squared = age_50 2 The square term is clearly not significant with a p-value of 0.36.

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Source | SS df MS Number of obs = 480 -------------+------------------------------ F( 3, 476) = 47.74 Model | 2.59024374 3.86341458 Prob > F = 0.0000 Residual | 8.60882438 476.018085766 R-squared = 0.2313 -------------+------------------------------ Adj R-squared = 0.2264 Total | 11.1990681 479.0233801 Root MSE =.13448 ------------------------------------------------------------------------------- lnbpsys | Coef. Std. Err. t P>|t| [95% Conf. Interval] --------------+---------------------------------------------------------------- female | -.1139236.0130282 -8.74 0.000 -.1395236 -.0883236 age_50 |.0027492.0008766 3.14 0.002.0010267.0044716 female_age_50 |.0025665.0011 2.33 0.020.000405.0047279 _cons | 4.847934.0105629 458.96 0.000 4.827179 4.86869 ------------------------------------------------------------------------------- Is the relation between ln(SBP) and age independent of gender? may be assessed by adding the interaction term: female_age_50 = female*age_50 The interaction term is statistically significant with a p-value of 0.02. The slope between ln(SBP) and age is higher in women (i.e., 0.0027+0.0026 = 0.0053) than in men (i.e., 0.0027).

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Graphical representation of the model on the log-scale

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Graphical representation of the model on the original scale:

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Variable selection strategies in prediction / reference models 1. Between two models select the one which is more significant. 2. Between two models select the one with the lower AIC-value (AIC = Akaike information criterion). 3. Between two models select the one with the lower BIC-value (BIC = Bayesian information criterion). 2) and 3) are better than 1), because they estimate performance of the model in new data. They are strongly linked to cross-validation. 3) is stricter than 2) and is preferable if parsimony of the model is an important criterion.

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Aim 2: Assessment of the association between adult systolic blood pressure (SBP) in Lugano and overweight. We consider variable overweight with values: 0 in persons with BMI 25kg/m 2 1 in persons with BMI > 25 kg/m 2

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Source | SS df MS Number of obs = 924 -------------+------------------------------ F( 1, 922) = 98.63 Model | 2.14124857 1 2.14124857 Prob > F = 0.0000 Residual | 20.0169737 922.021710384 R-squared = 0.0966 -------------+------------------------------ Adj R-squared = 0.0957 Total | 22.1582222 923.024006741 Root MSE =.14734 ------------------------------------------------------------------------------ lnbpsys | Coef. Std. Err. t P>|t| [95% Conf. Interval] -------------+---------------------------------------------------------------- overweight |.0963513.0097019 9.93 0.000.0773109.1153917 _cons | 4.779094.0067253 710.61 0.000 4.765896 4.792293 ------------------------------------------------------------------------------ 1. The mean of ln(SBP) was higher by 0.096 in overweight persons compared to persons of normal weight. The geometric mean ratio of SBP between overweight and normal weight persons was exp(0.096) = 1.10. The geometric mean of SBP was higher by 10% in overweight persons. Regression model: ln(SBP) = b 0 + b 1 · overweight 2. The estimated geometric mean of SBP in normal weight persons is exp(4.779) = 119.0.

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Source | SS df MS Number of obs = 924 -------------+------------------------------ F( 4, 919) = 103.79 Model | 6.89527577 4 1.72381894 Prob > F = 0.0000 Residual | 15.2629465 919.016608212 R-squared = 0.3112 -------------+------------------------------ Adj R-squared = 0.3082 Total | 22.1582222 923.024006741 Root MSE =.12887 ------------------------------------------------------------------------------- lnbpsys | Coef. Std. Err. t P>|t| [95% Conf. Interval] --------------+---------------------------------------------------------------- female | -.103432.0091201 -11.34 0.000 -.1213307 -.0855334 age_50 |.0036905.0005486 6.73 0.000.0026138.0047672 female_age_50 |.0022885.000742 3.08 0.002.0008323.0037447 overweight |.054128.0088583 6.11 0.000.0367431.071513 _cons | 4.840025.0083731 578.05 0.000 4.823592 4.856457 ------------------------------------------------------------------------------- Adjustment for gender and age: The gender and age-adjusted mean of ln(SBP) was higher by 0.054 in overweight persons compared to persons of normal weight. The adjusted geometric mean ratio of SBP between overweight and normal weight persons was exp(0.054) = 1.055. The adjusted geometric mean of SBP was higher by 5.5% in overweight persons.

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Arithmetic of confounding OW SBP age + + association between OW and age = + association between SBP and age = + Confounding of association between SBP and OW by age = + + = +. association between OW and F = - association between SBP and F = - Confounding of association between SBP and OW by sex = - - = +. OW SBP female - - Both, age and sex are positive confounders of the association between SBP and OW. => If age and sex are included in the model, the slope between SBP and OW decreases.

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Adjustment for clustering of data Example: multi-center studies If clustering is ignored, then this may lead to a) a loss of power (RCTs with randomisation stratified by center) b) confounding (observational studies with different study areas) Remedy: Introduce study center as a fixed factor into the regression model or use mixed linear model with random effects for the different centers.

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Source | SS df MS Number of obs = 3243 -------------+------------------------------ F( 11, 3231) = 140.38 Model | 23.5304364 11 2.13913058 Prob > F = 0.0000 Residual | 49.2344884 3231.015238158 R-squared = 0.3234 -------------+------------------------------ Adj R-squared = 0.3211 Total | 72.7649248 3242.022444456 Root MSE =.12344 ------------------------------------------------------------------------------- lnbpsys | Coef. Std. Err. t P>|t| [95% Conf. Interval] --------------+---------------------------------------------------------------- female | -.0887425.0045394 -19.55 0.000 -.0976429 -.079842 age_50 |.0031324.0002716 11.53 0.000.0025999.0036649 female_age_50 |.0031068.000379 8.20 0.000.0023637.0038499 overweight |.0600691.0045947 13.07 0.000.0510602.0690779 _Iarea_161 | -.0076996.0078966 -0.98 0.330 -.0231825.0077832 _Iarea_162 |.0201633.0098355 2.05 0.040.0008788.0394477 _Iarea_163 | -.0084698.0084272 -1.01 0.315 -.0249929.0080534 _Iarea_164 | -.0076526.0093321 -0.82 0.412 -.02595.0106449 _Iarea_165 | -.0411928.0086567 -4.76 0.000 -.058166 -.0242196 _Iarea_166 | -.0126127.0083109 -1.52 0.129 -.0289078.0036825 _Iarea_167 | -.0260132.0095668 -2.72 0.007 -.0447708 -.0072556 _cons | 4.841831.0071065 681.32 0.000 4.827897 4.855765 SAPALDIA-example (fixed area effects) All but one study area gets a parameter estimate, expressing its difference to the one area which serves as the reference (here: area 160).

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Random-effects ML regression Number of obs = 3243 Group variable: area Number of groups = 8 Random effects u_i ~ Gaussian Obs per group: min = 259 avg = 405.4 max = 624 LR chi2(4) = 1227.18 Log likelihood = 2176.8602 Prob > chi2 = 0.0000 ------------------------------------------------------------------------------- lnbpsys | Coef. Std. Err. z P>|z| [95% Conf. Interval] --------------+---------------------------------------------------------------- female | -.0889006.0045361 -19.60 0.000 -.0977913 -.08001 age_50 |.0031431.0002713 11.58 0.000.0026113.003675 female_age_50 |.0031.0003787 8.19 0.000.0023578.0038422 overweight |.0597827.0045912 13.02 0.000.0507841.0687812 _cons | 4.831426.0068285 707.54 0.000 4.818043 4.84481 --------------+---------------------------------------------------------------- /sigma_u |.0150779.0045517.0083441.0272459 /sigma_e |.1233706.0015339.1204006.1264139 rho |.014717.0087657.0041602.0430383 SAPALDIA-example (mixed linear model with random area effects) Random area effects u are viewed as independent outcomes of a normal distribution with u = 0 and u = 0.015 (residual standard deviation within areas = 0.123).

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Thank you for your attention!

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