Math Problem Statement

The following data describes the number of times I sneezed (y) each year (x). a) Determine the line of best fit using a linear regression. b) Using the line of best fit, estimate how many times I sneezed in 2017. c) Using the line of best fit, estimate what year I sneezed 200 times.

Solution

Let's work through the given problem step-by-step:

Part (a): Determine the Line of Best Fit Using Linear Regression

We have the following data points for the years and the number of sneezes:

Year (x)Sneezes (y)
2018234
2019243
2020255
2021264

We will use linear regression to find the line of best fit, which has the general equation: y=mx+by = mx + b Where:

  • mm is the slope of the line.
  • bb is the y-intercept.

To find mm and bb, we need to compute the following sums:

  • x\sum x: Sum of x-values (years).
  • y\sum y: Sum of y-values (sneezes).
  • xy\sum xy: Sum of the product of corresponding x and y values.
  • x2\sum x^2: Sum of the square of x-values.

After calculating these, we can use the formulas: m=Nxy(x)(y)Nx2(x)2m = \frac{N\sum xy - (\sum x)(\sum y)}{N\sum x^2 - (\sum x)^2} b=ymxNb = \frac{\sum y - m\sum x}{N} Where NN is the number of data points (in this case, 4).

Part (b): Estimate Sneezes in 2017 Using the Line of Best Fit

Once we have the equation of the line from part (a), substitute x=2017x = 2017 into the equation to estimate the number of sneezes.

Part (c): Estimate the Year with 200 Sneezes Using the Line of Best Fit

Set y=200y = 200 in the line equation and solve for xx (year).


I'll compute these values for you.### Solutions: (a) Line of Best Fit:
The equation of the line of best fit using linear regression is: y=10.20x20349.90y = 10.20x - 20349.90

(b) Estimate Sneezes in 2017:
Using the line of best fit, the estimated number of times sneezed in 2017 is: y(2017)=10.20(2017)20349.90=223.5y(2017) = 10.20(2017) - 20349.90 = 223.5

(c) Estimate the Year with 200 Sneezes:
Setting y=200y = 200 in the line equation, the estimated year when 200 sneezes occurred is: x=200+20349.9010.202014.7x = \frac{200 + 20349.90}{10.20} \approx 2014.7 This corresponds to approximately the year 2015.

Let me know if you have questions or need more detailed explanations.

Related Questions:

  1. How do we determine the goodness of fit for a linear regression model?
  2. What other models could be used for fitting if the data is not linear?
  3. How can we interpret the slope and intercept in a real-world context?
  4. What would happen if we used more data points to determine the line of best fit?
  5. How do outliers affect the linear regression line?

Tip: When using linear regression, always check if the model assumptions hold true, such as linearity, independence, and homoscedasticity.

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Math Problem Analysis

Mathematical Concepts

Linear Regression
Algebra
Statistics

Formulas

y = mx + b
m = (NΣxy - ΣxΣy) / (NΣx^2 - (Σx)^2)
b = (Σy - mΣx) / N

Theorems

Least Squares Regression

Suitable Grade Level

High School (Grades 10-12)