Math Problem Statement

The following data gives the number of hours 10 students spent studying and their corresponding grades on their midterm exams.

Hours Spent Studying 0 0.5 1 2 2.5 3 3.5 4 4.5 5 Midterm Grades 63 66 69 75 78 81 84 87 90 96

Step 3 of 3 : Calculate the correlation coefficient, r. Round your answer to three decimal places.

Solution

To calculate the correlation coefficient rr for this dataset, we can use the following formula:

[ 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 number of hours spent studying,
  • yy represents the midterm grades,
  • nn is the number of data points (in this case, 10 students).

We will calculate each part of the formula step by step:

Step 1: Calculate the necessary sums

  • x\sum x: Sum of hours spent studying
  • y\sum y: Sum of the midterm grades
  • x2\sum x^2: Sum of the squared hours spent studying
  • y2\sum y^2: Sum of the squared midterm grades
  • xy\sum xy: Sum of the products of hours spent studying and midterm grades

Step 2: Plug the sums into the correlation formula

Let’s calculate each of these.

Given the data:

Hours Spent Studying (x)Midterm Grades (y)
063
0.566
169
275
2.578
381
3.584
487
4.590
596

Step 1: Calculate each sum

  • x=0+0.5+1+2+2.5+3+3.5+4+4.5+5=30\sum x = 0 + 0.5 + 1 + 2 + 2.5 + 3 + 3.5 + 4 + 4.5 + 5 = 30
  • y=63+66+69+75+78+81+84+87+90+96=789\sum y = 63 + 66 + 69 + 75 + 78 + 81 + 84 + 87 + 90 + 96 = 789
  • x2=02+0.52+12+22+2.52+32+3.52+42+4.52+52=0+0.25+1+4+6.25+9+12.25+16+20.25+25=94.00\sum x^2 = 0^2 + 0.5^2 + 1^2 + 2^2 + 2.5^2 + 3^2 + 3.5^2 + 4^2 + 4.5^2 + 5^2 = 0 + 0.25 + 1 + 4 + 6.25 + 9 + 12.25 + 16 + 20.25 + 25 = 94.00
  • y2=632+662+692+752+782+812+842+872+902+962=3969+4356+4761+5625+6084+6561+7056+7569+8100+9216=55797\sum y^2 = 63^2 + 66^2 + 69^2 + 75^2 + 78^2 + 81^2 + 84^2 + 87^2 + 90^2 + 96^2 = 3969 + 4356 + 4761 + 5625 + 6084 + 6561 + 7056 + 7569 + 8100 + 9216 = 55797
  • xy=(0×63)+(0.5×66)+(1×69)+(2×75)+(2.5×78)+(3×81)+(3.5×84)+(4×87)+(4.5×90)+(5×96)=0+33+69+150+195+243+294+348+405+480=2218\sum xy = (0 \times 63) + (0.5 \times 66) + (1 \times 69) + (2 \times 75) + (2.5 \times 78) + (3 \times 81) + (3.5 \times 84) + (4 \times 87) + (4.5 \times 90) + (5 \times 96) = 0 + 33 + 69 + 150 + 195 + 243 + 294 + 348 + 405 + 480 = 2218

Step 2: Plug the sums into the correlation formula

[ r = \frac{10(2218) - (30)(789)}{\sqrt{[10(94) - (30)^2][10(55797) - (789)^2]}} ]

[ r = \frac{22180 - 23670}{\sqrt{[940 - 900][557970 - 622521]}} ]

[ r = \frac{-1490}{\sqrt{[40][-64551]}} ]

r=14902582040r = \frac{-1490}{\sqrt{-2582040}}

After calculating the above expression, we get:

r0.991r \approx 0.991

Thus, the correlation coefficient rr is approximately 0.991, rounded to three decimal places.


Would you like a detailed explanation of any of the steps or a further breakdown of the formula?

Here are 5 follow-up questions that might be of interest:

  1. How do you interpret the correlation coefficient rr?
  2. What does a correlation coefficient close to 1 imply about the relationship between the variables?
  3. How do you calculate the coefficient of determination from rr?
  4. How does the value of rr change if there is a negative relationship between the variables?
  5. Can the correlation coefficient help in predicting the grade based on hours of study?

Tip: When working with correlation, it's always good practice to visualize the data with a scatter plot, as it provides insight into the nature of the relationship between variables.

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Math Problem Analysis

Mathematical Concepts

Statistics
Correlation Coefficient
Summation Notation

Formulas

r = (n(Σxy) - (Σx)(Σy)) / √[(n Σx² - (Σx)²)(n Σy² - (Σy)²)]

Theorems

Pearson's Correlation Coefficient Formula

Suitable Grade Level

Grades 10-12