Math Problem Statement

By using this formula, Calculate the correlation of coefficient (r) for above data. Interpret the results.

By using this formula, calculate the correlation coefficient (r) for the above data. Interpret the results.

Solution

To calculate the correlation coefficient rr for the data provided, I'll use the formula shown in the second image. Here’s a step-by-step breakdown of the calculation:

Step 1: Set Up the Data

Let's define:

  • xx as the number of absences.
  • yy as the final grades.

Data for each student:

Studentxx (Absences)yy (Final Grade)
A682
B286
C1543
D974
E1258
F590
G878

Step 2: Calculate x\sum x, y\sum y, x2\sum x^2, y2\sum y^2, and xy\sum xy

  1. Calculate x\sum x and y\sum y: x=6+2+15+9+12+5+8=57\sum x = 6 + 2 + 15 + 9 + 12 + 5 + 8 = 57 y=82+86+43+74+58+90+78=511\sum y = 82 + 86 + 43 + 74 + 58 + 90 + 78 = 511

  2. Calculate x2\sum x^2 and y2\sum y^2: x2=62+22+152+92+122+52+82=36+4+225+81+144+25+64=579\sum x^2 = 6^2 + 2^2 + 15^2 + 9^2 + 12^2 + 5^2 + 8^2 = 36 + 4 + 225 + 81 + 144 + 25 + 64 = 579 y2=822+862+432+742+582+902+782=6724+7396+1849+5476+3364+8100+6084=38993\sum y^2 = 82^2 + 86^2 + 43^2 + 74^2 + 58^2 + 90^2 + 78^2 = 6724 + 7396 + 1849 + 5476 + 3364 + 8100 + 6084 = 38993

  3. Calculate xy\sum xy: xy=(6×82)+(2×86)+(15×43)+(9×74)+(12×58)+(5×90)+(8×78)\sum xy = (6 \times 82) + (2 \times 86) + (15 \times 43) + (9 \times 74) + (12 \times 58) + (5 \times 90) + (8 \times 78) =492+172+645+666+696+450+624=3745= 492 + 172 + 645 + 666 + 696 + 450 + 624 = 3745

Step 3: Calculate SS(x), SS(y), and SS(xy)

Using the formulas:

  1. SS(x): SS(x)=x2(x)2nSS(x) = \sum x^2 - \frac{(\sum x)^2}{n} SS(x)=5795727=57932497=579464.14=114.86SS(x) = 579 - \frac{57^2}{7} = 579 - \frac{3249}{7} = 579 - 464.14 = 114.86

  2. SS(y): SS(y)=y2(y)2nSS(y) = \sum y^2 - \frac{(\sum y)^2}{n} SS(y)=3899351127=389932611217=3899337231.57=1761.43SS(y) = 38993 - \frac{511^2}{7} = 38993 - \frac{261121}{7} = 38993 - 37231.57 = 1761.43

  3. SS(xy): SS(xy)=xyxynSS(xy) = \sum xy - \frac{\sum x \cdot \sum y}{n} SS(xy)=374557×5117=3745291277=37454153.86=408.86SS(xy) = 3745 - \frac{57 \times 511}{7} = 3745 - \frac{29127}{7} = 3745 - 4153.86 = -408.86

Step 4: Calculate rr

The correlation coefficient rr is given by: r=SS(xy)SS(x)SS(y)r = \frac{SS(xy)}{\sqrt{SS(x) \cdot SS(y)}} r=408.86114.86×1761.43r = \frac{-408.86}{\sqrt{114.86 \times 1761.43}} Calculating the denominator: 114.86×1761.43=202162.37=449.63\sqrt{114.86 \times 1761.43} = \sqrt{202162.37} = 449.63 Thus, r=408.86449.63=0.91r = \frac{-408.86}{449.63} = -0.91

Interpretation

The correlation coefficient r0.91r \approx -0.91 indicates a strong negative correlation between the number of absences and the final grade. This suggests that as the number of absences increases, the final grade tends to decrease significantly.

Would you like any further details or clarification on this calculation?


  1. What is the effect of outliers on the correlation coefficient?
  2. How does the interpretation of a negative correlation differ from a positive one?
  3. What does a correlation of r=0r = 0 signify about the relationship between two variables?
  4. How could we test the statistical significance of this correlation coefficient?
  5. What other measures can describe the strength of the relationship between two variables besides the correlation coefficient?

Tip: Remember that correlation does not imply causation; it only shows a relationship between two variables.

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Math Problem Analysis

Mathematical Concepts

Statistics
Correlation Coefficient
Data Analysis

Formulas

Correlation Coefficient formula: r = SS(xy) / √(SS(x) * SS(y))

Theorems

Pearson's Correlation Coefficient

Suitable Grade Level

Grades 10-12