Lab3

library(ggplot2)
library(tidyverse)

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library(moments)  
library(effsize)   

ex1 <- read.csv("Exercise3_1.csv")
ex2 <- read.csv("Exercise3_2.csv")

ggplot(ex1, aes(x = mathScore)) +
  geom_histogram(bins = 6, color = "black")

The histogram looks balanced overall, not showing any outliers so the data seems normal. ## R Markdown

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summary(cars)

##      speed           dist       
##  Min.   : 4.0   Min.   :  2.00  
##  1st Qu.:12.0   1st Qu.: 26.00  
##  Median :15.0   Median : 36.00  
##  Mean   :15.4   Mean   : 42.98  
##  3rd Qu.:19.0   3rd Qu.: 56.00  
##  Max.   :25.0   Max.   :120.00

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shapiro.test(ex1$mathScore)

## 
##  Shapiro-Wilk normality test
## 
## data:  ex1$mathScore
## W = 0.96297, p-value = 0.4767

t.test(ex1$mathScore, mu = 50)

## 
##  One Sample t-test
## 
## data:  ex1$mathScore
## t = 2.5336, df = 24, p-value = 0.01823
## alternative hypothesis: true mean is not equal to 50
## 95 percent confidence interval:
##  51.00849 59.87151
## sample estimates:
## mean of x 
##     55.44

Since the p value was below o.o5, we reject the null hypothesis. This shows the students average math score is significantly different from the national average.

cohen.d(ex1$mathScore, rep(50, length(ex1$mathScore)),  na.rm = TRUE)

## 
## Cohen's d
## 
## d estimate: 0.7166067 (medium)
## 95 percent confidence interval:
##     lower     upper 
## 0.1299448 1.3032686

The effect was medium (d=0.72). This means the difference between the students’ average math score and national average is not very large.

A one sample t test showed that the students average matyh score (M=55.44) was significantly different from the national average of 50,T(24)=2.53, p=.018

---

## Problem 2 (Exercise3_2)

ex2 <- read.csv("Exercise3_2.csv")
head(ex2)

##   score
## 1     2
## 2     4
## 3     2
## 4     3
## 5     3
## 6     5

ggplot(ex2, aes(x = factor(score))) +
  geom_bar(color = "black") +
  xlab("score")

shapiro.test(ex2$score)

## 
##  Shapiro-Wilk normality test
## 
## data:  ex2$score
## W = 0.89306, p-value = 0.0002857

Since the p-value is below 0.05, we reject normality assumption. So the scores do not look normally distributed.

t.test(ex2$score, mu = 3)

## 
##  One Sample t-test
## 
## data:  ex2$score
## t = 1.0706, df = 49, p-value = 0.2896
## alternative hypothesis: true mean is not equal to 3
## 95 percent confidence interval:
##  2.859673 3.460327
## sample estimates:
## mean of x 
##      3.16

Since the p-value (0.2896) is above 0.05, we fail to reject the null. That means the average response is not significantly different from 3 (neutral)

d <- (mean(ex2$score) - 3) / sd(ex2$score)
d

## [1] 0.1514067

The effect sixcze (Cohens d) is about 0.15, so its small ***