Chris Timke s3868828
25/10/2020
## chr [1:250] ".509" ".395" ".384" "0.5" ".386" ".346" "" ".69" ".39" ".761" ...
students$Reaction_time <- as.numeric(students$Reaction_time)
studentsTest <- students %>% select(Reaction_time)
# number of NAs
sum(is.na(studentsTest))## [1] 22
## [1] 0
## [1] 13 17 31 65 175 184 213
reaction_time_outliers <- studentsTest %>% filter(Reaction_time >1)
studentsTest2 <- studentsTest %>% filter(Reaction_time <0.9 & Reaction_time >0.10)
boxplot(studentsTest2$Reaction_time) - While the distribution is skewed, we know the sampling distribution of a population mean with a sample size greater than 30, as is here n = 216, will be approximately normally distributed with a mean equal to the population mean.
studentsTest2 %>% summarise(Min = min(Reaction_time,na.rm = TRUE),
Q1 = quantile(Reaction_time,probs = .25,na.rm = TRUE),
Median = median(Reaction_time, na.rm = TRUE),
Q3 = quantile(Reaction_time,probs = .75,na.rm = TRUE),
Max = max(Reaction_time,na.rm = TRUE),
Mean = mean(Reaction_time, na.rm = TRUE),
SD = sd(Reaction_time, na.rm = TRUE),
n = n(),
Missing = sum(is.na(Reaction_time))) -> table
knitr::kable(table)| Min | Q1 | Median | Q3 | Max | Mean | SD | n | Missing |
|---|---|---|---|---|---|---|---|---|
| 0.129 | 0.35475 | 0.4305 | 0.58525 | 0.891 | 0.4729306 | 0.1677707 | 216 | 0 |
The Null hypothesis for this question is \[H_0: \mu = 0.2\sec\]
The Alternate hypothesis for this question is \[H_A: \mu \ne 0.2\sec\]
##
## One Sample t-test
##
## data: studentsTest2$Reaction_time
## t = 23.909, df = 215, p-value < 2.2e-16
## alternative hypothesis: true mean is not equal to 0.2
## 95 percent confidence interval:
## 0.4504302 0.4954309
## sample estimates:
## mean of x
## 0.4729306
## [1] 1.971059