Mr. Turner has two Algebra 1 classes. With one class, he lectured and the students took notes. In the other class, the students worked in small groups to solve math problems. After the first test, Mr. Turner recorded the student grades to determine if his different styles of teaching might have impacted student learning.

Class 1: 80, 81, 81, 75, 70, 72, 74, 76, 77, 77, 77, 79, 84, 88, 90, 86, 80, 80, 78, 82

Class 2: 70, 90, 88, 89, 86, 86, 86, 86, 84, 82, 77, 79, 84, 84, 84, 86, 87, 88, 88, 88

1. Analyze his student grades by filling in the table below. Which class do you think was the lecture and which was the
small group? Why?

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ewomazinoade ewomazinoade · Answer 1 · 2022-04-22T15:59:16+02:00

Q1 Q2 IQR Mean Median MAD

Class 1 76.25 81.75 5.5 79.35 79.50 3.12

Class 2 84 88 4 84.60 86 3.85

The first step is to arrange the grades in the classes in ascending order.

Class 1: 70, 72, 74, 75, 76, 77, 77,77, 78, 79, 80, 80, 80, 81, 81, 82, 84, 86, 88, 90

Class 2: 70, 77, 79, 82, 84, 84, 84, 84, 86, 86, 86, 86, 86, 87, 88, 88, 88, 88, 89, 90

Q1 for class 1= 1/4(n + 1) = 21/4 = 5.25 = 76.25

Q2 for class 2 = 1/4(n + 1) = 5.25 = 84

Q3 for class 1= 3/4(n + 1) = 15.75 = 81.75

Q3 for class 2 = 3/4(n + 1) = 15.75 = 88

IQR for class 1 = Q3 - Q1 = 81.75 - 76.25 = 5.50

IQR for class 2 = Q3 - Q1 = 88 - 84 = 4

Mean for class 1 = sum of grades / total number of grades = 1587 / 20 = 79.35

Mean for class 2 = sum of grades / total number of grades= 1692 / 20 = 84.6

Median for class 1 = (n + 1) / 2 = 21/2 = 10.5 = 79.50

Median for class 1 = (n + 1) / 2 = 21/2 = 10.5 = 86

MAD = [tex]\frac{1}{n}[/tex]∑ l x - m(x) l

Where: n = number of observations

x = number in the data set

m = mean

Mean absolute deviation for class 1 = 62. 3/ 20 = 3.12

Mean absolute deviation for class 2. = 77/ 20 = 3.85

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