Respuesta :
Answer:
Following are the responses to the given question:
Step-by-step explanation:
For point a:
In R-Studio, we first insert the data set,
Please notice that perhaps the blue colored lines are input and the green lines are R-Studio results.
[tex]Class \ 1 = c(72,73,74,75,76,79,82,83,84,86,91,92,93,93,94,95,95,97,98,98)\\\\Class\ 2 = c(59,65,68,68,69,72,73,78,80,82,82,83,83,85,88,88,89,94,96,97,98)[/tex]
We will get the smallest observation, first, mid, third, and largest quartile for both classes and use a summary of 5 numbers,
[tex]\to five\ num(Class\ 1) \\\\\[1\] \ 72.0 77.5 88.5 94.5 98.0 \\\\\to five\ num(Class\ 2) \\\\\[1\]\ 59 72 82 88 98[/tex]
The table can be defined as follows:
[tex]Class\ 1\ \ \ \ \ \ \ \ \ \ \ \ Class\ 2 \\[/tex]
[tex]Smallest \ value\ \ \ \ \ \ \ 72\ \ \ \ \ \ \ 59\\First \ quartile \ Q1\ \ \ \ \ \ \ 77.5\ \ \ \ \ \ \ 72\\Median\ Q2 \ \ \ \ \ \ \ 88.5\ \ \ \ \ \ \ 82\\Third \ quartile \ Q3 \ \ \ \ \ \ \ 94.5\ \ \ \ \ \ \ 88\\Largest \ value \ \ \ \ \ \ \ 98\ \ \ \ \ \ \ 98\\[/tex]
The parallel boxplots in R-Studio as,
[tex]\text{boxplot(Class1, Class2, xlab = "Class", ylab = "midterm grades", main = "Boxplots")}[/tex]Please find the graph file.
For point b:
Its performance overall of Class 1 is better, while the median of class 1 is greater than class 2, as well as the value (grades) of class 1, is less dispersed in relation to class 2.
For point c:
The stated 90 percent confidence interval for a significant difference is (0.09299, 11.413) Users now calculate the difference among Class 1 and Class 2 plan presented of mean value as:
[tex]mean \ (Class \ 1) \\\\\[1\]\ 86.5\\\\mean\ (Class\ 2)\\\\[/tex]
[tex]\[ 1 \] \ 80.80952\\\\Difference = 86.5 - 80.80952 = 5.69048[/tex]
Its discrepancy among two estimations is between confidence interval of 90 percent (0.09299, 11.413). Its mean population of grades of two classes therefore differs significantly.
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