Question

a mathematics teacher wanted to see the correlation between test scores and homework. the homework grade (x) and test grade (y) are given in the accompanying table. write the linear regression equation that represents this set of data, rounding all coefficients to the nearest tenth. using this equation, estimate the homework grade, to the nearest integer, for a student with a test grade of 31.

homework grade (x) | test grade (y)
58 | 52
75 | 66
66 | 68
85 | 74
52 | 38
74 | 65
78 | 76

linear regression equation:
final answer:

the annual profits for a company are given in the following table, where x represents the number of years since 1998, and y represents the profit in thousands of dollars. write the linear regression equation that represents this set of data, rounding all coefficients to the nearest hundredth. using this equation, find the projected profit (in thousands of dollars) for 2006, rounded to the nearest thousand dollars.

years since 1998 (x) | profits (y) (in thousands of dollars)
0 | 72
1 | 84
2 | 104
3 | 156

linear regression equation:
final answer:

s-id.6a--summarize, represent, and interpret data on two categorical and quantitative variables. represent data on two quantitative variables on a scatter plot and describe how the variables are related. fit a least squares regression line to linear data using technology. use the fitted function to solve problems.

Explanation:

First Problem (Homework and Test Grades)

Step 1: Calculate necessary sums

We have data points \((x,y)\): \((58,52)\), \((75,66)\), \((66,68)\), \((85,74)\), \((52,38)\), \((74,65)\), \((78,76)\)
\(n = 7\)
\(\sum x = 58 + 75 + 66 + 85 + 52 + 74 + 78 = 488\)
\(\sum y = 52 + 66 + 68 + 74 + 38 + 65 + 76 = 439\)
\(\sum xy = 58\times52 + 75\times66 + 66\times68 + 85\times74 + 52\times38 + 74\times65 + 78\times76\)
\(= 3016 + 4950 + 4488 + 6290 + 1976 + 4810 + 5928 = 31458\)
\(\sum x^2 = 58^2 + 75^2 + 66^2 + 85^2 + 52^2 + 74^2 + 78^2\)
\(= 3364 + 5625 + 4356 + 7225 + 2704 + 5476 + 6084 = 34834\)

Step 2: Calculate slope \(m\) and intercept \(b\)

The formula for the slope \(m\) of the linear regression line \(y = mx + b\) is:
\(m=\frac{n\sum xy-\sum x\sum y}{n\sum x^2 - (\sum x)^2}\)
Substitute the values:
\(m=\frac{7\times31458 - 488\times439}{7\times34834 - 488^2}\)
\(7\times31458 = 220206\), \(488\times439 = 214232\), \(7\times34834 = 243838\), \(488^2 = 238144\)
\(m=\frac{220206 - 214232}{243838 - 238144}=\frac{5974}{5694}\approx1.05\) (rounded to nearest tenth)

The formula for the intercept \(b\) is:
\(b=\frac{\sum y - m\sum x}{n}\)
\(b=\frac{439 - 1.05\times488}{7}=\frac{439 - 512.4}{7}=\frac{-73.4}{7}\approx - 10.5\) (rounded to nearest tenth)

So the linear regression equation is \(y = 1.05x - 10.5\) (Wait, actually, we need to solve for \(x\) when \(y = 31\). Wait, the problem says "estimate the homework grade (x) for a test grade (y) of 31". So we need to rearrange the equation \(y=mx + b\) to \(x=\frac{y - b}{m}\)

Wait, maybe I mixed up \(x\) and \(y\). The homework grade is \(x\), test grade is \(y\). So we want to find \(x\) when \(y = 31\). So first, find the regression equation of \(x\) on \(y\) (since we are predicting \(x\) from \(y\)).

Let's re - define: let the independent variable be \(y\) and dependent be \(x\).

\(n = 7\), \(\sum y=439\), \(\sum x = 488\), \(\sum yx=\sum xy = 31458\), \(\sum y^2=52^2 + 66^2 + 68^2 + 74^2 + 38^2 + 65^2 + 76^2\)
\(= 2704+4356 + 4624+5476+1444+4225+5776 = 28605\)

Slope \(m_{x - y}=\frac{n\sum yx-\sum y\sum x}{n\sum y^2-(\sum y)^2}\)
\(m_{x - y}=\frac{7\times31458-439\times488}{7\times28605 - 439^2}\)
\(7\times31458 = 220206\), \(439\times488 = 214232\), \(7\times28605 = 200235\), \(439^2 = 192721\)
\(m_{x - y}=\frac{220206 - 214232}{200235 - 192721}=\frac{5974}{7514}\approx0.8\) (rounded to nearest tenth)

Intercept \(b_{x - y}=\frac{\sum x - m_{x - y}\sum y}{n}=\frac{488-0.8\times439}{7}=\frac{488 - 351.2}{7}=\frac{136.8}{7}\approx19.5\)

So the equation is \(x = 0.8y+19.5\)

Now, substitute \(y = 31\):
\(x=0.8\times31 + 19.5=24.8+19.5 = 44.3\approx44\)

Second Problem (Company Profits)

Step 1: Calculate necessary sums

We have data points \((x,y)\): \((0,72)\), \((1,84)\), \((2,104)\), \((3,156)\)
\(n = 4\)
\(\sum x=0 + 1+2 + 3=6\)
\(\sum y=72 + 84+104 + 156=416\)
\(\sum xy=0\times72+1\times84 + 2\times104+3\times156=0 + 84+208 + 468=760\)
\(\sum x^2=0^2 + 1^2+2^2 + 3^2=0 + 1+4 + 9 = 14\)

Step 2: Calculate slope \(m\) and intercept \(b\)

The formula for the slope \(m\) of the linear regression line \(y=mx + b\) is:
\(m=\frac{n\sum xy-\sum x\sum y}{n\sum x^2-(\sum x)^2}\)
Substitute the values:
\(m=\frac{4\times760-6\times416}{4\times14 - 6^2}=\frac{3040 - 2496}{56 - 36}=\frac{544}{20}=27.2\) (rounded to nearest hundredth)

The formula for the intercept \(b\) is:
\(b=\frac{\sum y - m\sum x}{n}=\frac{416-27.2\times6}{4}=\frac{416 - 163.2}{4}=\frac{252.8}{4}=63.2\)

So the linear regression equation is \(y = 27.2x+63.2\)

For the year 2006, the number of years since 1998 is \(x = 2006 - 1998=8\)

Substitute \(x = 8\) into the equation:
\(y=27.2\times8 + 63.2=217.6+63.2 = 280.8\approx281\) (in thousands of dollars)

Answer:

(First Problem - Homework Grade Estimation):
Linear Regression Equation (for \(x\) in terms of \(y\)): \(x = 0.8y+19.5\) (or if we consider \(y\) in terms of \(x\) first and then solve for \(x\), the process is as above). The estimated homework grade for a test grade of 31 is \(\boxed{44}\)