Basic boxplot of wages distribution

boxplot(wage1_region$wage, main = "General overiew on hourly earnings", ylab = "avarage hourly earnings")

median(wage1_region$wage)

## [1] 4.6

A general overview shows the wages of the cleaned dataset which median is equal to 4.6. As we can see, there are also many outliers of people that earns much more than avarage.

Frequency table of the wages

attach(wage1_region)

interval <- (max(wage)-min(wage))/10
wage_limits <- cut(wage, seq(min(wage), max(wage), by = interval))
wage_tab <- table(wage_limits)

wage_tab <- as.data.frame(wage_tab)
kable(wage_tab, format = "html", col.names=c("Wages range", "Frequency"))

Tabular Accurancy Index

wage_tai <- classIntervals(wage, n = 10, style="fixed", fixedBreaks=seq(min(wage), max(wage), by = interval) )
tai <- jenks.tests(wage_tai)
kable(tai, format = "html")

Wage & Marriage & Region

ggplot(wage1_region, aes(married, wage)) + geom_boxplot()

ggplot(wage1_region, aes(married, wage, fill=region)) + geom_boxplot()

Density plots

qplot(wage, data = wage1_region, geom = "density", color = female) + theme_bw() + facet_wrap(~region)

qplot(wage, data = wage1_region, geom = "histogram", color = female) + theme_bw() + facet_wrap(~region)

## `stat_bin()` using `bins = 30`. Pick better value with `binwidth`.

We are calclutaing the proportion of married men that lives in west and all married men.

Manual Solution

n <- nrow(married_male) 
k <- nrow(married_west_male) 
p_hat <- k/n
p_hat

## [1] 0.1946309

se_prop <- sqrt(p_hat*(1-p_hat)/n)
se_prop

## [1] 0.03243472

E = qnorm(.975)*se_prop


p_hat + c(-E, E)

## [1] 0.1310600 0.2582018

standard_error <- sd(married_male$wage)/sqrt(n)
standard_error

## [1] 0.3375064

The proportion (p hat) of married men that lives in west and all married men is equal to 0.19

The estimated confidence interval of the proportion of married men that lives in west and all married men is between 0.13 and 0.25.

Test check

p_hat_interval <- prop.test(k,n)
p_hat_interval

## 
##  1-sample proportions test with continuity correction
## 
## data:  k out of n, null probability 0.5
## X-squared = 54.362, df = 1, p-value = 1.667e-13
## alternative hypothesis: true p is not equal to 0.5
## 95 percent confidence interval:
##  0.1362006 0.2692053
## sample estimates:
##         p 
## 0.1946309

A test confirms the results of calculations

We are estiamting the mean of the population based on the sample of married men.

Manual Solution

attach(married_male)

## Następujące obiekty zostały zakryte z wage1_region:
## 
##     clerocc, construc, educ, exper, expersq, female, lwage, married,
##     ndurman, nonwhite, northcen, numdep, profocc, profserv, region,
##     services, servocc, smsa, south, tenure, tenursq, trade, trcommpu,
##     wage, west

n <- length(wage)
xbar <- mean(wage)
s <- sd(wage)


margin <- qt(0.975,df=n-1)*s/sqrt(n)


low <- xbar - margin
high <- xbar + margin


low

## [1] 6.919556

high

## [1] 8.253464

Based on calculations, a mean is estimated (with 95% confidence) to lie in the interval between 6,92 and 8,25

Test check

t_test <- t.test(wage)
t_test

## 
##  One Sample t-test
## 
## data:  wage
## t = 22.478, df = 148, p-value < 2.2e-16
## alternative hypothesis: true mean is not equal to 0
## 95 percent confidence interval:
##  6.919556 8.253464
## sample estimates:
## mean of x 
##   7.58651

A test confirms the results of calculations

Plots

summarySE <- function(data=NULL, measurevar, groupvars=NULL, na.rm=FALSE,
                      conf.interval=.95, .drop=TRUE) {

    # New version of length which can handle NA's: if na.rm==T, don't count them
    length2 <- function (x, na.rm=FALSE) {
        if (na.rm) sum(!is.na(x))
        else       length(x)
    }

    # This does the summary. For each group's data frame, return a vector with
    # N, mean, and sd
    datac <- ddply(data, groupvars, .drop=.drop,
      .fun = function(xx, col) {
        c(N    = length2(xx[[col]], na.rm=na.rm),
          mean = mean   (xx[[col]], na.rm=na.rm),
          sd   = sd     (xx[[col]], na.rm=na.rm)
        )
      },
      measurevar
    )

    # Rename the "mean" column    
    datac <- rename(datac, c("mean" = measurevar))

    datac$se <- datac$sd / sqrt(datac$N)  # Calculate standard error of the mean

    # Confidence interval multiplier for standard error
    # Calculate t-statistic for confidence interval: 
    # e.g., if conf.interval is .95, use .975 (above/below), and use df=N-1
    ciMult <- qt(conf.interval/2 + .5, datac$N-1)
    datac$ci <- datac$se * ciMult

    return(datac)
}



x1 <- summarySE(married_male, measurevar="wage", groupvars=c("region"))
x1

##     region  N     wage       sd        se        ci
## 1 northcen 44 7.687955 4.108166 0.6193294 1.2489967
## 2    south 76 6.615132 3.405525 0.3906406 0.7781959
## 3     west 29 9.978276 4.922120 0.9140147 1.8722743

# Error bars represent standard error of the mean
ggplot(x1, aes(x=region, y=wage, fill = region)) + 
    geom_bar(position=position_dodge(), stat="identity") +
    geom_errorbar(aes(ymin=wage-se, ymax=wage+se),
                  width=.2,                    # Width of the error bars
                  position=position_dodge(.9))

# Use 95% confidence intervals instead of SEM
ggplot(x1, aes(x=region, y=wage, fill = region)) + 
    geom_bar(position=position_dodge(), stat="identity") +
    geom_errorbar(aes(ymin=wage-ci, ymax=wage+ci),
                  width=.2,                    # Width of the error bars
                  position=position_dodge(.9))