search for: gendermale

...quot;Male", "Female", "Male", "Female", "Male", "Female")) model <- lm(income ~ education * gender, data = data) # Test if the beta for "education" is significantly different between genders test <- linearHypothesis(model, "genderMale - genderFemale = 0") print(test) This, however, produces an error that I can't find a way to resolve. Can this test actually be done in this manner, or is this a case of AI run amok. Guidance would be appreciated. --John Sparks [[alternative HTML version deleted]]

Just a general question concerning the woolf test (package vcd), when we have stratified data (2x2 tables) and when the p.value of the woolf-test is below 0.05 then we assume that there is a heterogeneity and a common odds ratio cannot be computed? Does this mean that we have to try to add more stratification variables (stratify more) to make the woolf-test p.value insignificant? Also in the

...Fresh Female 6 677 Fresh Female 7 592 Rancid Male 8 538 Rancid Male 9 476 Rancid Male 10 508 Rancid Female 11 505 Rancid Female 12 539 Rancid Female > lm(fixed=Value~Gender,data=Food) Call: lm(data = Food, fixed = Value ~ Gender) Coefficients: (Intercept) LardRancid GenderMale 651.4 -142.8 35.5 Warning message: extra arguments fixed are just disregarded. in: lm.fit(x, y, offset = offset, singular.ok = singular.ok, ...) > lm(fixed=Value~Lard+Gender,data=Food) Call: lm(data = Food, fixed = Value ~ Lard + Gender) Coefficients: (Intercept) L...

...le", "Male", "Female", "Male", "Female")) >> model <- lm(income ~ education * gender, data = data) >> # Test if the beta for "education" is significantly different between genders >> test <- linearHypothesis(model, "genderMale - genderFemale = 0") That last line appears unlikely to be parsed correctly. In R a ?=? sign is interpreted as assignment whereas a ?==? (doubled equals) is a logical operator. Since I?ve only got a iPhone at hand I can?t test. In the future you should include full text of errors, preferably...

..., "Female", "Male", "Female", "Male", "Female")) > model <- lm(income ~ education * gender, data = data) > # Test if the beta for "education" is significantly different between genders > test <- linearHypothesis(model, "genderMale - genderFemale = 0") > print(test) > > This, however, produces an error that I can't find a way to resolve. > > Can this test actually be done in this manner, or is this a case of AI run amok. > > Guidance would be appreciated. > --John Sparks > > > &...

...Estimate Std. Error t value Pr(>|t|) (Intercept) -7.39879 1.97605 -3.744 0.000219 *** log(pes) 1.78020 0.40118 4.437 1.31e-05 *** originsite 0.06572 0.01935 3.397 0.000781 *** originwild 0.07655 0.03552 2.155 0.032011 * gendermale -9.32418 2.37476 -3.926 0.000109 *** log(pes):gendermale 1.90393 0.47933 3.972 9.06e-05 *** --- Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1 Residual standard error: 0.1433 on 281 degrees of freedom Multiple R-Squared: 0....

...0 Prob > chi2 = 0.0000 Here is the R glm.nb output: Deviance Residuals: Min 1Q Median 3Q Max -1.9785 -1.0627 -0.4147 0.2865 2.8193 Coefficients: Estimate Std. Error z value Pr(>|z|) (Intercept) 2.716069 0.234174 11.598 < 2e-16 *** gendermale -0.431185 0.139516 -3.091 0.00200 ** mathnce -0.001601 0.005300 -0.302 0.76259 langnce -0.014348 0.005372 -2.671 0.00756 ** --- Signif. codes: 0 `***' 0.001 `**' 0.01 `*' 0.05 `.' 0.1 ` ' 1 (Dispersion parameter for Negative Binomial(0.7762) family take...

...trying to put gender='Male' in newdata to create a expected survival curve for a pseudo cohort by using survfit based on Cox regression. My codes are shown below: fit<- coxph(Surv(end, status2)~gender, data=wlwsn1) Summary(fit) coef exp(coef) se(coef) z p genderMale 0.204 1.23 0.0912 2.23 0.025 temp<-data.frame(gender='Male) wlwsn1curve<-survfit(fit, newdata=temp) Then I got error message: Error in `contrasts<-`(`*tmp*`, value = "contr.treatment") : contrasts can be applied only to factors with 2 or more levels I do not know...

...9 0.7348 smokesever 0.0498764 0.0326254 1.529 0.1271 smokesnever 0.0394109 0.0349142 1.129 0.2597 diseasestate1 0.0018739 0.0176817 0.106 0.9157 diseasestate2 -0.0009858 0.0178651 -0.055 0.9560 age 0.0002841 0.0006290 0.452 0.6518 gendermale 0.1164889 0.0128748 9.048 <2e-16 *** --- Signif. codes: 0 ?***? 0.001 ?**? 0.01 ?*? 0.05 ?.? 0.1 ? ? 1 Residual standard error: 0.1257 on 397 degrees of freedom Multiple R-squared: 0.1933, Adjusted R-squared: 0.1791 F-statistic: 13.59 on 7 and 397 DF, p-value: 8.9...

...;binomial') #Reults Independent Sampling design (with replacement) svydesign(~0, weights = ~weight, data = mat.test) Call: svyglm(formula = trust ~ edu + prov + gender, design = test, family = "binomial") Coefficients: (Intercept) edusecondary provON provPQ genderMale -2.658e+01 -8.454e-04 5.317e+01 -1.408e-02 NA Degrees of Freedom: 599 Total (i.e. Null); 596 Residual Null Deviance: 759.6 Residual Deviance: 3.406e-09 AIC: 8 Warning messages: 1: In eval(expr, envir, enclos) : non-integer #successes in a binomial glm! 2: glm.f...

Hello, I have a question about glm: if i have a binary covariate (unit=1,0) the reference group would be 0? (prediction for unit=1) example: dat1<-data.frame(y,unit,x1,x2) log_u <- glm(y~.,family=binomial,data=dat1) summary(log_u) Estimate Std. Error z value Pr(>|z|) (Intercept) -0.54247 0.24658 -2.200 0.0278 * unit1 -0.13052 0.22861 -0.571 0.5680 aps

...ot;)) Deviance Residuals: Min 1Q Median 3Q Max -0.7660 -0.3053 -0.2462 -0.1984 3.2166 Coefficients: Estimate Std. Error z value Pr(>|z|) (Intercept) -2.439e+00 2.985e-02 -81.714 < 2e-16 *** age -6.436e-03 3.743e-04 -17.193 < 2e-16 *** genderMALE 1.785e-01 9.424e-03 18.945 < 2e-16 *** gemedu 2.128e-02 4.698e-03 4.528 5.95e-06 *** gemhinc 1.062e-01 6.185e-03 17.166 < 2e-16 *** es_gdppc 9.765e-06 1.400e-06 6.977 3.02e-12 *** imf_pop 4.131e-06 1.707e-05 0.242 0.809 estbbo_m 5.932e+00 1.088e-01...