data("Salaries")
data <- na.omit(Salaries)
data %>% group_by(sex,rank, discipline) %>% shapiro_test(salary)
## # A tibble: 12 × 6
## rank discipline sex variable statistic p
## <fct> <fct> <fct> <chr> <dbl> <dbl>
## 1 AsstProf A Female salary 0.870 0.226
## 2 AsstProf B Female salary 0.889 0.354
## 3 AssocProf A Female salary 0.863 0.269
## 4 AssocProf B Female salary 0.635 0.00117
## 5 Prof A Female salary 0.934 0.549
## 6 Prof B Female salary 0.974 0.923
## 7 AsstProf A Male salary 0.941 0.300
## 8 AsstProf B Male salary 0.941 0.0458
## 9 AssocProf A Male salary 0.878 0.0113
## 10 AssocProf B Male salary 0.967 0.416
## 11 Prof A Male salary 0.952 0.000259
## 12 Prof B Male salary 0.978 0.0435
ex1 <- anova(aov( salary ~ sex + rank + discipline, data))
ex1
## Analysis of Variance Table
##
## Response: salary
## Df Sum Sq Mean Sq F value Pr(>F)
## sex 1 6.9800e+09 6.9800e+09 13.617 0.0002559 ***
## rank 2 1.3709e+11 6.8546e+10 133.719 < 2.2e-16 ***
## discipline 1 1.8283e+10 1.8283e+10 35.666 5.254e-09 ***
## Residuals 392 2.0094e+11 5.1261e+08
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
ex12 <- anova(aov( salary ~ sex * rank * discipline, data))
ex12
## Analysis of Variance Table
##
## Response: salary
## Df Sum Sq Mean Sq F value Pr(>F)
## sex 1 6.9800e+09 6.9800e+09 13.4603 0.000278 ***
## rank 2 1.3709e+11 6.8546e+10 132.1851 < 2.2e-16 ***
## discipline 1 1.8283e+10 1.8283e+10 35.2574 6.457e-09 ***
## sex:rank 2 2.3520e+08 1.1760e+08 0.2268 0.797203
## sex:discipline 1 4.5581e+08 4.5581e+08 0.8790 0.349069
## rank:discipline 2 4.7483e+08 2.3742e+08 0.4578 0.632997
## sex:rank:discipline 2 1.3239e+08 6.6196e+07 0.1277 0.880195
## Residuals 385 1.9965e+11 5.1856e+08
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
ex2 <- lm(salary ~ sex * yrs.since.phd * yrs.service, data)
report(ex2)
## We fitted a linear model (estimated using OLS) to predict salary with sex,
## yrs.since.phd and yrs.service (formula: salary ~ sex * yrs.since.phd *
## yrs.service). The model explains a statistically significant and substantial
## proportion of variance (R2 = 0.33, F(7, 389) = 26.87, p < .001, adj. R2 =
## 0.31). The model's intercept, corresponding to sex = Female, yrs.since.phd = 0
## and yrs.service = 0, is at 66955.99 (95% CI [45666.11, 88245.87], t(389) =
## 6.18, p < .001). Within this model:
##
## - The effect of sex [Male] is statistically non-significant and positive (beta
## = 4328.42, 95% CI [-18177.17, 26834.02], t(389) = 0.38, p = 0.706; Std. beta =
## 0.35, 95% CI [-0.03, 0.74])
## - The effect of yrs since phd is statistically significant and positive (beta =
## 1855.48, 95% CI [61.49, 3649.47], t(389) = 2.03, p = 0.043; Std. beta = 0.47,
## 95% CI [-0.27, 1.20])
## - The effect of yrs service is statistically non-significant and positive (beta
## = 1274.56, 95% CI [-1389.64, 3938.77], t(389) = 0.94, p = 0.348; Std. beta =
## 0.13, 95% CI [-0.63, 0.89])
## - The effect of sex [Male] × yrs since phd is statistically non-significant and
## positive (beta = 348.94, 95% CI [-1515.52, 2213.39], t(389) = 0.37, p = 0.713;
## Std. beta = -0.02, 95% CI [-0.79, 0.75])
## - The effect of sex [Male] × yrs service is statistically non-significant and
## positive (beta = 421.97, 95% CI [-2340.98, 3184.92], t(389) = 0.30, p = 0.764;
## Std. beta = -0.03, 95% CI [-0.83, 0.76])
## - The effect of yrs since phd × yrs service is statistically non-significant
## and negative (beta = -43.08, 95% CI [-128.48, 42.32], t(389) = -0.99, p =
## 0.322; Std. beta = -0.24, 95% CI [-0.71, 0.23])
## - The effect of (sex [Male] × yrs since phd) × yrs service is statistically
## non-significant and negative (beta = -22.52, 95% CI [-109.28, 64.23], t(389) =
## -0.51, p = 0.610; Std. beta = -0.12, 95% CI [-0.60, 0.36])
##
## Standardized parameters were obtained by fitting the model on a standardized
## version of the dataset. 95% Confidence Intervals (CIs) and p-values were
## computed using a Wald t-distribution approximation.
#The effect of yrs since phd is statistically significant and positive
plot(ex2)
chisq <- aggregate(data[c("yrs.since.phd", "yrs.service")], by = list(data$rank), FUN = mean)
ex3 <- anova(aov((yrs.since.phd+yrs.service) ~ rank, data))
ex3
## Analysis of Variance Table
##
## Response: (yrs.since.phd + yrs.service)
## Df Sum Sq Mean Sq F value Pr(>F)
## rank 2 113886 56943 160.69 < 2.2e-16 ***
## Residuals 394 139618 354
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
#yes
#I already have 100% from laboratories, if I get any points from this report is
#there a possibility to transfer them to a different category as i lack 2.32%
#or 1.32 pts to get 5 instead of 4.5.