Consider the following data for a dependent variable y and two independent variables, x1 and x2. x1: 30, 47, 24, 51, 40, 51, 75, 36, 59, 77; x2: 13, 11, 18, 17, 5, 19, 8, 12, 14, 17; y: 95, 109, 112, 179, 94, 175, 170, 118, 143, 211 Round your all answers to two decimal places. Enter negative values as negative numbers, if necessary. Develop an estimated regression equation relating y to x2.
The Correct Answer and Explanation is:
To develop the estimated regression equation relating yyy to x2x_2x2, we can perform a simple linear regression using x2x_2x2 as the independent variable and yyy as the dependent variable. The regression equation has the general form:y=β0+β1x2y = \beta_0 + \beta_1 x_2y=β0+β1x2
Where:
- yyy is the dependent variable,
- x2x_2x2 is the independent variable,
- β0\beta_0β0 is the intercept,
- β1\beta_1β1 is the coefficient (slope) of x2x_2x2.
Step 1: Organize the Data
Given values for x2x_2x2 and yyy are:
- x2=[13,11,18,17,5,19,8,12,14,17]x_2 = [13, 11, 18, 17, 5, 19, 8, 12, 14, 17]x2=[13,11,18,17,5,19,8,12,14,17]
- y=[95,109,112,179,94,175,170,118,143,211]y = [95, 109, 112, 179, 94, 175, 170, 118, 143, 211]y=[95,109,112,179,94,175,170,118,143,211]
Step 2: Calculate the Required Sums
- Sum of x2x_2x2:
∑x2=13+11+18+17+5+19+8+12+14+17=124\sum x_2 = 13 + 11 + 18 + 17 + 5 + 19 + 8 + 12 + 14 + 17 = 124∑x2=13+11+18+17+5+19+8+12+14+17=124
- Sum of yyy:
∑y=95+109+112+179+94+175+170+118+143+211=1406\sum y = 95 + 109 + 112 + 179 + 94 + 175 + 170 + 118 + 143 + 211 = 1406∑y=95+109+112+179+94+175+170+118+143+211=1406
- Sum of x22x_2^2×22 (the square of each value in x2x_2x2):
∑x22=132+112+182+172+52+192+82+122+142+172=169+121+324+289+25+361+64+144+196+289=1962\sum x_2^2 = 13^2 + 11^2 + 18^2 + 17^2 + 5^2 + 19^2 + 8^2 + 12^2 + 14^2 + 17^2 = 169 + 121 + 324 + 289 + 25 + 361 + 64 + 144 + 196 + 289 = 1962∑x22=132+112+182+172+52+192+82+122+142+172=169+121+324+289+25+361+64+144+196+289=1962
- Sum of x2⋅yx_2 \cdot yx2⋅y (the product of each corresponding pair of x2x_2x2 and yyy):
∑x2⋅y=(13×95)+(11×109)+(18×112)+(17×179)+(5×94)+(19×175)+(8×170)+(12×118)+(14×143)+(17×211)=1235+1199+2016+3043+470+3325+1360+1416+2002+3587=17053\sum x_2 \cdot y = (13 \times 95) + (11 \times 109) + (18 \times 112) + (17 \times 179) + (5 \times 94) + (19 \times 175) + (8 \times 170) + (12 \times 118) + (14 \times 143) + (17 \times 211) = 1235 + 1199 + 2016 + 3043 + 470 + 3325 + 1360 + 1416 + 2002 + 3587 = 17053∑x2⋅y=(13×95)+(11×109)+(18×112)+(17×179)+(5×94)+(19×175)+(8×170)+(12×118)+(14×143)+(17×211)=1235+1199+2016+3043+470+3325+1360+1416+2002+3587=17053
Step 3: Compute the Slope (β1\beta_1β1) and Intercept (β0\beta_0β0)
Using the formulas for the slope and intercept in simple linear regression:β1=n∑x2y−∑x2∑yn∑x22−(∑x2)2\beta_1 = \frac{n \sum x_2 y – \sum x_2 \sum y}{n \sum x_2^2 – (\sum x_2)^2}β1=n∑x22−(∑x2)2n∑x2y−∑x2∑yβ0=∑y−β1∑x2n\beta_0 = \frac{\sum y – \beta_1 \sum x_2}{n}β0=n∑y−β1∑x2
Where nnn is the number of data points (in this case, n=10n = 10n=10).
- Calculate β1\beta_1β1:
β1=10(17053)−(124)(1406)10(1962)−(124)2=170530−17412419620−15376=−35944244≈−0.85\beta_1 = \frac{10(17053) – (124)(1406)}{10(1962) – (124)^2} = \frac{170530 – 174124}{19620 – 15376} = \frac{-3594}{4244} \approx -0.85β1=10(1962)−(124)210(17053)−(124)(1406)=19620−15376170530−174124=4244−3594≈−0.85
- Calculate β0\beta_0β0:
β0=1406−(−0.85)(124)10=1406+105.410=1511.410≈151.14\beta_0 = \frac{1406 – (-0.85)(124)}{10} = \frac{1406 + 105.4}{10} = \frac{1511.4}{10} \approx 151.14β0=101406−(−0.85)(124)=101406+105.4=101511.4≈151.14
Step 4: Form the Regression Equation
Now that we have the values for β0\beta_0β0 and β1\beta_1β1, we can write the estimated regression equation:y=151.14−0.85x2y = 151.14 – 0.85 x_2y=151.14−0.85×2
Conclusion
The estimated regression equation that relates yyy to x2x_2x2 is:y=151.14−0.85x2y = 151.14 – 0.85 x_2y=151.14−0.85×2
This equation can now be used to predict yyy for any given value of x2x_2x2.
