library(s20x)
sleep.df = read.table(file.choose(), header = TRUE)
head(sleep.df)

#Put in sleep hours as a working variable
SleepHrs = sleep.df$SleepHrs

#Generate a vector of whether the student reported to be sleep deprived or not
reported = rep(c("Yes", "No"), c(50, 35))

#Rewrite, treating sleep hours as a factor
reported = factor(reported)

boxplot(SleepHrs ~ reported)

Exploratory Analysis: The data from the box plot shows that the data is not skewed and have an even spread. The centre for the students who reported to not be sleep deprived is 6 hours and for the ones who reported sleep deprivation is about 7 hours.

onewayPlot(SleepHrs ~ reported)

There are more data collected for the students reported to be sleep deprived compared to the data for those reported not being sleep deprived. The data for the students who reported Yes is more spread out to the 2 extremes being 10 hours of sleep and 3 hours of sleep. This can also due to the larger data set for the reported, hence, more diversity, spread.

summaryStats(SleepHrs ~ reported)
    Sample Size     Mean Median  Std Dev Midspread
No           35 6.228571      6 0.910259      1.00
Yes          50 6.640000      7 1.351643      1.75

The reported answers yes and no are not in a dataframe, however, the column vectors are in their own right so we can refer directly to them.

normcheck(lm(SleepHrs ~ reported))

The Q-Q check for the data set has a normal distribution as most data points fit into the linear line and the graph fits into the normal distribution bell shape graph. Some of the data points at the extremes are skewed from the data. But due to Central Limit Theoren, it can be assumed that the data has a normal distribution.

eovcheck(SleepHrs ~ reported)

The data is spread equally for the most part. The column on the right (Students who reported to be sleep deprived) have a bigger spread from -3 to 3 from the mean compared to the other group. As mentioned before, this might be due to the larger data set for the sleep deprived record. So the equality of variance for this data is valid.

t.test(SleepHrs ~ reported, var.equal = TRUE)

    Two Sample t-test

data:  SleepHrs by reported
t = -1.5677, df = 83, p-value = 0.1207
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
 -0.9334037  0.1105466
sample estimates:
 mean in group No mean in group Yes 
         6.228571          6.640000

The mean in the reported to be sleep deprived group is higher than that of the not reported to be sleep deprived group.

Methods and Assumption check The measurements and data were made on two distinct groups of students who reported to be sleep deprived and those who did not report to be sleep deprived, hence, a two sample t-test was conducted. It is assumed that the students who reported are independent of one another. The equality of variance and normality check seems satisfactory which means the standard two sample t-test is valid to be used. The model fitted is 〖SleepHours〗_ij= μ+ α_i+ ε_ij where α_i is the effect of whether the student reported to be sleep deprived or not and ε_ij □(iid/) N(0,σ^2)

Executive Summary The data collected was to assess whether there is a difference between students who reported to be sleep deprived to students who did not reported to be sleep deprived with respect to the average hours of sleep they reported. It was observed that there is a difference in the sleep hours between these two groups. The students who reported to not be sleep deprived have 1 hours less of average sleep hours in comparison to those who reported yes. Even though there is a larger group of students who reported to be sleep deprived, there are equal amount of data at the 2 extremes to the sample, hence, we can’t conclude that there is an influence on the data due to more data.

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