data$priemer <- rowMeans(data[, 5:29])
data$suma <- rowSums(data[, 5:29])
#Vypocet uspesnosti v ramci kategorii
data <- mutate(data, internet = (Task_1 + Task_2 + Task_3 + Task_4 + Task_5)*100/5)
data <- mutate(data, bezpecnost = (Task_6 + Task_7 + Task_8 + Task_9 + Task_10)*100/5)
data <- mutate(data, soc_siete = (Task_11 + Task_12 + Task_13 + Task_14 + Task_15)*100/5)
data <- mutate(data, kanc_nastroje = (Task_16 + Task_17 + Task_18 + Task_19 + Task_20)*100/5)
data <- mutate(data, komplex_ulohy = (Task_21 + Task_22 + Task_23 + Task_24 + Task_25)*100/5)
mean = uspesnost, sd = smerodajna odychylka
Internet
with(data[data$rok == 2017,], describe(internet, na.rm = TRUE, skew = F, fast = T, ranges = F))
## vars n mean sd se
## X1 1 212 63.4 21.06 1.45
Bezpecnost
with(data[data$rok == 2017,], describe(bezpecnost, na.rm = TRUE, skew = F, fast = T, ranges = F))
## vars n mean sd se
## X1 1 212 38.87 27.38 1.88
Soc siete
with(data[data$rok == 2017,], describe(soc_siete, na.rm = TRUE, skew = F, fast = T, ranges = F))
## vars n mean sd se
## X1 1 212 46.51 25.18 1.73
Kanc. nastroje
with(data[data$rok == 2017,], describe(kanc_nastroje, na.rm = TRUE, skew = F, fast = T, ranges = F))
## vars n mean sd se
## X1 1 212 54.62 28.74 1.97
Komplexne ulohy
with(data[data$rok == 2017,], describe(komplex_ulohy, na.rm = TRUE, skew = F, fast = T, ranges = F))
## vars n mean sd se
## X1 1 212 45.09 28.49 1.96
Welshov t-test rozdielu dvoch nezávislých priemerov. Treba reportovat nasledujucim sposobom: Napr, celkové skóre v roku 2017 (M = 0,50; SD = …) bolo nižšie ako v roku 2016 (M = 0,58; SD = …), t(448) = 5,45; p = 8.2e-8.
x = 2016, y = 2017
t.test(data$priemer[data$rok == 2016], data$priemer[data$rok == 2017], var.equal = FALSE)
##
## Welch Two Sample t-test
##
## data: data$priemer[data$rok == 2016] and data$priemer[data$rok == 2017]
## t = 5.4511, df = 447.8, p-value = 8.283e-08
## alternative hypothesis: true difference in means is not equal to 0
## 95 percent confidence interval:
## 0.05465913 0.11629089
## sample estimates:
## mean of x mean of y
## 0.5824561 0.4969811
x = 2016, y = 2017
t.test(data$internet[data$rok == 2016], data$internet[data$rok == 2017], var.equal = FALSE)
##
## Welch Two Sample t-test
##
## data: data$internet[data$rok == 2016] and data$internet[data$rok == 2017]
## t = 10.147, df = 375.01, p-value < 2.2e-16
## alternative hypothesis: true difference in means is not equal to 0
## 95 percent confidence interval:
## 13.84087 20.49450
## sample estimates:
## mean of x mean of y
## 80.56391 63.39623
x = 2016, y = 2017
t.test(data$bezpecnost[data$rok == 2016], data$bezpecnost[data$rok == 2017], var.equal = FALSE)
##
## Welch Two Sample t-test
##
## data: data$bezpecnost[data$rok == 2016] and data$bezpecnost[data$rok == 2017]
## t = 3.0249, df = 434.53, p-value = 0.002635
## alternative hypothesis: true difference in means is not equal to 0
## 95 percent confidence interval:
## 2.445072 11.517193
## sample estimates:
## mean of x mean of y
## 45.84906 38.86792
x = 2016, y = 2017
t.test(data$kanc_nastroje[data$rok == 2016], data$kanc_nastroje[data$rok == 2017], var.equal = FALSE)
##
## Welch Two Sample t-test
##
## data: data$kanc_nastroje[data$rok == 2016] and data$kanc_nastroje[data$rok == 2017]
## t = 3.1582, df = 373, p-value = 0.001717
## alternative hypothesis: true difference in means is not equal to 0
## 95 percent confidence interval:
## 2.748093 11.816147
## sample estimates:
## mean of x mean of y
## 61.90476 54.62264
x = 2016, y = 2017
t.test(data$komplex_ulohy[data$rok == 2016], data$komplex_ulohy[data$rok == 2017], var.equal = FALSE)
##
## Welch Two Sample t-test
##
## data: data$komplex_ulohy[data$rok == 2016] and data$komplex_ulohy[data$rok == 2017]
## t = 1.6044, df = 407.54, p-value = 0.1094
## alternative hypothesis: true difference in means is not equal to 0
## 95 percent confidence interval:
## -0.8500317 8.3964772
## sample estimates:
## mean of x mean of y
## 48.86756 45.09434
Socialne siete
x = 2016, y = 2017