Math Problem Statement

1. Suppose working with a set of 5 ordered pairs, we gathered the following: Σx = 20, Σx² = 90, Σy = 44, Σy² = 430, Σxy = 195. (a) Find the value of the linear correlation coefficient. (b) Find the value of the coefficient of determination. (c) Find the equation of the regression line. (d) Predict y value if x = 4.5 assuming the linear correlation is not significant.

Solution

Let's break down the problem and solve each part step by step. The given data is:

x=20,x2=90,y=44,y2=430,xy=195,n=5\sum x = 20, \, \sum x^2 = 90, \, \sum y = 44, \, \sum y^2 = 430, \, \sum xy = 195, \, n = 5

(a) Find the value of the linear correlation coefficient rr.

The formula for the linear correlation coefficient rr is:

r=nxyxy(nx2(x)2)(ny2(y)2)r = \frac{n \sum xy - \sum x \sum y}{\sqrt{\left(n \sum x^2 - (\sum x)^2\right)\left(n \sum y^2 - (\sum y)^2\right)}}

Plugging in the values:

r=5(195)(20)(44)(5)(90)(20)2(5)(430)(44)2r = \frac{5(195) - (20)(44)}{\sqrt{(5)(90) - (20)^2} \sqrt{(5)(430) - (44)^2}}

Let me compute that for you.The linear correlation coefficient rr is approximately 0.9180.918, rounded to three decimal places.

(b) Find the value of the coefficient of determination.

The coefficient of determination r2r^2 represents the proportion of the variance in the dependent variable that is predictable from the independent variable. It's calculated as:

r2=(r)2r^2 = (r)^2

Let me calculate that and round to the nearest whole percent.The coefficient of determination r2r^2 is approximately 84%, meaning that 84% of the variance in yy can be explained by the variance in xx.

(c) Find the equation of the regression line.

The equation of the regression line is given by:

y=mx+by = mx + b

where mm (the slope) and bb (the y-intercept) are calculated as follows:

m=nxyxynx2(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}

I'll compute the values of mm and bb for the equation.The equation of the regression line is:

y=1.9x+1.2y = 1.9x + 1.2

(d) Predict the yy-value if x=4.5x = 4.5 assuming the linear correlation is not significant.

Using the regression equation y=1.9x+1.2y = 1.9x + 1.2, we substitute x=4.5x = 4.5:

y=1.9(4.5)+1.2y = 1.9(4.5) + 1.2

Let me calculate that for you.It seems like I can’t do more advanced data analysis right now. Please try again later.

However, you can substitute x=4.5x = 4.5 into the regression equation y=1.9x+1.2y = 1.9x + 1.2 to get the predicted value manually:

y=1.9(4.5)+1.2=8.55+1.2=9.75y = 1.9(4.5) + 1.2 = 8.55 + 1.2 = 9.75

Thus, the predicted yy-value is approximately 9.759.75.


Let me know if you'd like further details or have any questions! Here are five related problems you can explore:

  1. What does a correlation coefficient of 0.918 indicate about the relationship between xx and yy?
  2. How would the coefficient of determination change if the correlation were weaker?
  3. What happens to the regression line if the data points show no correlation?
  4. How does increasing the number of data points affect the accuracy of the regression equation?
  5. If the slope of the line was negative, what would that imply about the relationship?

Tip: When interpreting regression lines, always consider the slope and correlation coefficient to understand the strength and direction of the relationship between variables.

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Math Problem Analysis

Mathematical Concepts

Statistics
Linear Regression
Correlation

Formulas

Linear correlation coefficient formula: r = (nΣxy - ΣxΣy) / sqrt[(nΣx² - (Σx)²)(nΣy² - (Σy)²)]
Coefficient of determination: r²
Equation of regression line: y = mx + b

Theorems

Linear regression theory
Pearson correlation coefficient

Suitable Grade Level

Grades 10-12