Math Problem Statement

A study of bone density on 5 random women at a hospital produced the following results.

Age 37 49 57 61 65 Bone Density 360 355 330 320 310

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

Solution

To calculate the correlation coefficient rr, we use the 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:

  • nn is the number of data points (in this case, n=5n = 5),
  • xx represents the values for age,
  • yy represents the values for bone density,
  • x\sum x is the sum of the ages,
  • y\sum y is the sum of the bone densities,
  • xy\sum xy is the sum of the products of age and bone density for each data pair,
  • x2\sum x^2 is the sum of the squares of the ages,
  • y2\sum y^2 is the sum of the squares of the bone densities.

Given data:

  • Age (xx): 37, 49, 57, 61, 65
  • Bone Density (yy): 360, 355, 330, 320, 310

Let’s break down the necessary calculations.

Step-by-step calculations:

  1. Sum of xx (ages): x=37+49+57+61+65=269\sum x = 37 + 49 + 57 + 61 + 65 = 269

  2. Sum of yy (bone densities): y=360+355+330+320+310=1675\sum y = 360 + 355 + 330 + 320 + 310 = 1675

  3. Sum of x2x^2 (squares of ages): x2=372+492+572+612+652=1369+2401+3249+3721+4225=11965\sum x^2 = 37^2 + 49^2 + 57^2 + 61^2 + 65^2 = 1369 + 2401 + 3249 + 3721 + 4225 = 11965

  4. Sum of y2y^2 (squares of bone densities): y2=3602+3552+3302+3202+3102=129600+126025+108900+102400+96100=563025\sum y^2 = 360^2 + 355^2 + 330^2 + 320^2 + 310^2 = 129600 + 126025 + 108900 + 102400 + 96100 = 563025

  5. Sum of xyxy (product of age and bone density): xy=(37×360)+(49×355)+(57×330)+(61×320)+(65×310)\sum xy = (37 \times 360) + (49 \times 355) + (57 \times 330) + (61 \times 320) + (65 \times 310) xy=13320+17395+18810+19520+20150=89295\sum xy = 13320 + 17395 + 18810 + 19520 + 20150 = 89295

Now, let's plug these into the formula:

[ r = \frac{5 \times 89295 - 269 \times 1675}{\sqrt{[5 \times 11965 - (269)^2][5 \times 563025 - (1675)^2]}} ]

Simplifying the numerator:

5×89295=4464755 \times 89295 = 446475 269×1675=450575269 \times 1675 = 450575 446475450575=4100446475 - 450575 = -4100

Simplifying the denominator:

  1. 5×11965=598255 \times 11965 = 59825 (269)2=72361(269)^2 = 72361 5982572361=1253659825 - 72361 = -12536

  2. 5×563025=28151255 \times 563025 = 2815125 (1675)2=2805625(1675)^2 = 2805625 28151252805625=95002815125 - 2805625 = 9500

Now, the denominator is:

(12536)×9500=11959200010935.15\sqrt{(-12536) \times 9500} = \sqrt{119592000} \approx 10935.15

Finally, calculating rr:

r=410010935.150.375r = \frac{-4100}{10935.15} \approx -0.375

Final answer:

The correlation coefficient rr is approximately -0.375.

Would you like further details or clarifications? Here are some related questions you might find interesting:

  1. How do the values of rr indicate the strength and direction of a relationship between two variables?
  2. What are some other methods to measure the relationship between variables?
  3. How does the calculation of correlation change with more data points?
  4. Can the correlation coefficient be used for nonlinear relationships?
  5. What are the limitations of the correlation coefficient in statistical analysis?

Tip: The correlation coefficient ranges from -1 to 1. A value closer to 1 indicates a strong positive relationship, closer to -1 indicates a strong negative relationship, and around 0 means no linear relationship.

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Math Problem Analysis

Mathematical Concepts

Statistics
Correlation
Data Analysis

Formulas

r = (nΣxy - ΣxΣy) / sqrt[(nΣx² - (Σx)²)(nΣy² - (Σy)²)]

Theorems

Pearson correlation coefficient

Suitable Grade Level

Grades 11-12