Math Problem Statement

The table below gives the age and bone density for five randomly selected women. Using this data, consider the equation of the regression line, yˆ=b0+b1x , for predicting a woman's bone density based on her age. Keep in mind, the correlation coefficient may or may not be statistically significant for the data given. Remember, in practice, it would not be appropriate to use the regression line to make a prediction if the correlation coefficient is not statistically significant.

Age 40 45 53 57 63 Bone Density 350 348 341 327 324 Table

Step 2 of 6 : Find the estimated y-intercept. Round your answer to three decimal places.

Solution

To calculate the estimated y-intercept b0b_0 for the regression line equation y^=b0+b1x\hat{y} = b_0 + b_1 x, we need to use the following formulas for linear regression:

1. The formula for b1b_1 (the slope) is:

b1=nxyxynx2(x)2b_1 = \frac{n \sum x y - \sum x \sum y}{n \sum x^2 - (\sum x)^2}

2. The formula for b0b_0 (the y-intercept) is:

b0=yb1xnb_0 = \frac{\sum y - b_1 \sum x}{n}

Where:

  • xx represents age
  • yy represents bone density
  • nn is the number of data points (in this case, n=5n = 5)

Let's first calculate the necessary sums.

Data:

  • Age (x): 40, 45, 53, 57, 63
  • Bone Density (y): 350, 348, 341, 327, 324

Step 1: Calculate the sums

  • x=40+45+53+57+63=258\sum x = 40 + 45 + 53 + 57 + 63 = 258
  • y=350+348+341+327+324=1690\sum y = 350 + 348 + 341 + 327 + 324 = 1690
  • x2=402+452+532+572+632=1600+2025+2809+3249+3969=14652\sum x^2 = 40^2 + 45^2 + 53^2 + 57^2 + 63^2 = 1600 + 2025 + 2809 + 3249 + 3969 = 14652
  • xy=(40×350)+(45×348)+(53×341)+(57×327)+(63×324)=14000+15660+18073+18639+20412=86784\sum x y = (40 \times 350) + (45 \times 348) + (53 \times 341) + (57 \times 327) + (63 \times 324) = 14000 + 15660 + 18073 + 18639 + 20412 = 86784

Step 2: Calculate the slope b1b_1

b1=5×86784(258×1690)5×14652(258)2b_1 = \frac{5 \times 86784 - (258 \times 1690)}{5 \times 14652 - (258)^2} b1=4339204357207326066564b_1 = \frac{433920 - 435720}{73260 - 66564} b1=18006696b_1 = \frac{-1800}{6696} b10.269b_1 \approx -0.269

Step 3: Calculate the y-intercept b0b_0

b0=1690(0.269×258)5b_0 = \frac{1690 - (-0.269 \times 258)}{5} b0=1690+69.4625b_0 = \frac{1690 + 69.462}{5} b0=1759.4625b_0 = \frac{1759.462}{5} b0351.892b_0 \approx 351.892

Final Answer:

The estimated y-intercept b0b_0 is approximately 351.892.

Would you like to go over the next steps, or have any questions regarding this calculation?


Related Questions:

  1. How do we interpret the meaning of the slope b1b_1 in the context of this data?
  2. What is the significance of the y-intercept b0b_0 in the regression line equation?
  3. How do we calculate the correlation coefficient for this data, and what does it tell us?
  4. What would be the appropriate steps to check if the regression line is statistically significant?
  5. How can we use the regression line to make predictions for women of different ages?

Tip:

When calculating regression coefficients by hand, it's easy to make mistakes in summing up values. Double-check your sums to avoid errors in the final result!

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Math Problem Analysis

Mathematical Concepts

Linear Regression
Statistics
Slope and Y-intercept

Formulas

b1 = (n Σxy - Σx Σy) / (n Σx^2 - (Σx)^2)
b0 = (Σy - b1 Σx) / n

Theorems

Least Squares Method
Linear Regression Theorem

Suitable Grade Level

Grades 10-12