Math Problem Statement

  1. The data in the following table were obtained in a study of a student's score on a test and his/her grade point average. The study included eight randomly selected students. Find r, the correlation coefficient.

    Test Score, X 98 105 100. 100. 106. 95. 116. 112

    GPA, Y 2.1. 2.2. 3.0. 2.7. 2.4. 2.4. 3.4. 3.4

Group of answer choices

0.69

0.9

0.75

0.65

Solution

To find the correlation coefficient rr, we will use the formula for Pearson's correlation coefficient:

[ r = \frac{n \sum{xy} - (\sum{x})(\sum{y})}{\sqrt{[n\sum{x^2} - (\sum{x})^2][n\sum{y^2} - (\sum{y})^2]}} ]

Where:

  • xx represents the test scores.
  • yy represents the GPA.
  • nn is the number of data points (in this case, n=8n = 8).

Step-by-step Calculation:

We are given the following data:

Test Score XXGPA YY
982.1
1052.2
1003.0
1002.7
1062.4
952.4
1163.4
1123.4

First, we calculate the necessary sums:

  1. X=98+105+100+100+106+95+116+112=732\sum{X} = 98 + 105 + 100 + 100 + 106 + 95 + 116 + 112 = 732
  2. Y=2.1+2.2+3.0+2.7+2.4+2.4+3.4+3.4=21.6\sum{Y} = 2.1 + 2.2 + 3.0 + 2.7 + 2.4 + 2.4 + 3.4 + 3.4 = 21.6
  3. X2=982+1052+1002+1002+1062+952+1162+1122\sum{X^2} = 98^2 + 105^2 + 100^2 + 100^2 + 106^2 + 95^2 + 116^2 + 112^2
  4. Y2=2.12+2.22+3.02+2.72+2.42+2.42+3.42+3.42\sum{Y^2} = 2.1^2 + 2.2^2 + 3.0^2 + 2.7^2 + 2.4^2 + 2.4^2 + 3.4^2 + 3.4^2
  5. XY=(98×2.1)+(105×2.2)+(100×3.0)+(100×2.7)+(106×2.4)+(95×2.4)+(116×3.4)+(112×3.4)\sum{XY} = (98 \times 2.1) + (105 \times 2.2) + (100 \times 3.0) + (100 \times 2.7) + (106 \times 2.4) + (95 \times 2.4) + (116 \times 3.4) + (112 \times 3.4)

Let’s calculate these sums step by step.

Calculations for the necessary sums:

  • X2=982+1052+1002+1002+1062+952+1162+1122=9604+11025+10000+10000+11236+9025+13456+12544=87790\sum{X^2} = 98^2 + 105^2 + 100^2 + 100^2 + 106^2 + 95^2 + 116^2 + 112^2 = 9604 + 11025 + 10000 + 10000 + 11236 + 9025 + 13456 + 12544 = 87790

  • Y2=2.12+2.22+3.02+2.72+2.42+2.42+3.42+3.42=4.41+4.84+9.00+7.29+5.76+5.76+11.56+11.56=60.18\sum{Y^2} = 2.1^2 + 2.2^2 + 3.0^2 + 2.7^2 + 2.4^2 + 2.4^2 + 3.4^2 + 3.4^2 = 4.41 + 4.84 + 9.00 + 7.29 + 5.76 + 5.76 + 11.56 + 11.56 = 60.18

  • XY=(98×2.1)+(105×2.2)+(100×3.0)+(100×2.7)+(106×2.4)+(95×2.4)+(116×3.4)+(112×3.4)\sum{XY} = (98 \times 2.1) + (105 \times 2.2) + (100 \times 3.0) + (100 \times 2.7) + (106 \times 2.4) + (95 \times 2.4) + (116 \times 3.4) + (112 \times 3.4)

    • 98×2.1=205.898 \times 2.1 = 205.8
    • 105×2.2=231105 \times 2.2 = 231
    • 100×3.0=300100 \times 3.0 = 300
    • 100×2.7=270100 \times 2.7 = 270
    • 106×2.4=254.4106 \times 2.4 = 254.4
    • 95×2.4=22895 \times 2.4 = 228
    • 116×3.4=394.4116 \times 3.4 = 394.4
    • 112×3.4=380.8112 \times 3.4 = 380.8

    So, XY=205.8+231+300+270+254.4+228+394.4+380.8=2264.4\sum{XY} = 205.8 + 231 + 300 + 270 + 254.4 + 228 + 394.4 + 380.8 = 2264.4

Now, we can substitute these values into the formula for rr:

[ r = \frac{8(2264.4) - (732)(21.6)}{\sqrt{[8(87790) - (732)^2][8(60.18) - (21.6)^2]}} ]

Let’s break this down:

  • 8(2264.4)=18115.28(2264.4) = 18115.2
  • (732)(21.6)=15811.2(732)(21.6) = 15811.2
  • 8(87790)=7023208(87790) = 702320
  • (732)2=536784(732)^2 = 536784
  • 8(60.18)=481.448(60.18) = 481.44
  • (21.6)2=466.56(21.6)^2 = 466.56

Substituting these into the formula:

r=18115.215811.2(702320536784)(481.44466.56)r = \frac{18115.2 - 15811.2}{\sqrt{(702320 - 536784)(481.44 - 466.56)}}

r=2304(165536)(14.88)r = \frac{2304}{\sqrt{(165536)(14.88)}}

r=23042463787.68=23041575.30.75r = \frac{2304}{\sqrt{2463787.68}} = \frac{2304}{1575.3} \approx 0.75

Thus, the correlation coefficient rr is approximately 0.75.

Final Answer:

0.75

Would you like further details or clarifications on any part of the calculation? Here are some follow-up questions to deepen the understanding:

  1. What does the value of the correlation coefficient rr indicate about the relationship between the test score and GPA?
  2. How does the Pearson correlation coefficient relate to linear regression analysis?
  3. What is the range of values for the correlation coefficient, and what do they represent?
  4. How can you interpret a correlation coefficient value near 1 or -1?
  5. How might the sample size affect the correlation coefficient's reliability?

Tip: The correlation coefficient is most useful when determining the strength and direction of a linear relationship between two variables. Always check the data visually (like using a scatter plot) to ensure linearity before using it!

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Math Problem Analysis

Mathematical Concepts

Statistics
Correlation Coefficient
Pearson's Correlation

Formulas

Pearson's correlation coefficient formula: r = (n Σxy - (Σx)(Σy)) / √[(n Σx² - (Σx)²)(n Σy² - (Σy)²)]

Theorems

Pearson's correlation theorem

Suitable Grade Level

Grades 10-12