Question

what is the least - squares regression model between x and y for data set a? full data set: data set a data set b data set c x y x y x y 3.6 8.9 3.1 8.9 2.8 8.9 8.3 15.0 9.4 15.0 8.1 15.0 0.5 4.8 1.2 4.8 3.0 4.8 1.4 6.0 1.0 6.0 8.3 6.0 8.2 14.9 9.0 14.9 8.2 14.9 5.9 11.9 5.0 11.9 1.4 11.9 4.3 9.8 3.4 9.8 1.0 9.8 8.3 15.0 7.4 15.0 7.9 15.0 0.3 4.7 0.1 4.7 5.9 4.7 6.8 13.0 7.5 13.0 5.0 13.0 what is the least - squares regression model between x and y for data set a, rounding to four decimal places? choose the correct answer below. a. $hat{y}=4.2159x + 1.1797$ b. $hat{y}=1.2960x + 4.2159$ c. $hat{y}=0.4368x + 8.1464$ d. $hat{y}=1.1797x + 4.5438$ (this is a reading assessment question. be certain of your answer because you only get one attempt on this question.)

Explanation:

Step1: Calculate sums

Let \(n\) be the number of data - points in Data Set A. Here \(n = 10\).
Calculate \(\sum_{i = 1}^{n}x_i\), \(\sum_{i = 1}^{n}y_i\), \(\sum_{i = 1}^{n}x_i^2\), \(\sum_{i = 1}^{n}x_iy_i\).
For Data Set A:
\(x=[3.6,8.3,0.5,1.4,8.2,5.9,4.3,8.3,0.3,6.8]\)
\(y = [8.9,15.0,4.8,6.0,14.9,11.9,9.8,15.0,4.7,13.0]\)
\(\sum_{i = 1}^{10}x_i=3.6 + 8.3+0.5 + 1.4+8.2+5.9+4.3+8.3+0.3+6.8 = 47.6\)
\(\sum_{i = 1}^{10}y_i=8.9+15.0 + 4.8+6.0+14.9+11.9+9.8+15.0+4.7+13.0 = 103\)
\(\sum_{i = 1}^{10}x_i^2=3.6^2+8.3^2+0.5^2+1.4^2+8.2^2+5.9^2+4.3^2+8.3^2+0.3^2+6.8^2\)
\(=12.96 + 68.89+0.25+1.96+67.24+34.81+18.49+68.89+0.09+46.24 = 329.82\)
\(\sum_{i = 1}^{10}x_iy_i=3.6\times8.9+8.3\times15.0+0.5\times4.8+1.4\times6.0+8.2\times14.9+5.9\times11.9+4.3\times9.8+8.3\times15.0+0.3\times4.7+6.8\times13.0\)
\(=32.04+124.5+2.4+8.4+122.18+70.21+42.14+124.5+1.41+88.4 = 616.18\)

Step2: Calculate slope \(b_1\)

The formula for the slope \(b_1\) of the least - squares regression line \(\hat{y}=b_0 + b_1x\) is \(b_1=\frac{n\sum_{i = 1}^{n}x_iy_i-\sum_{i = 1}^{n}x_i\sum_{i = 1}^{n}y_i}{n\sum_{i = 1}^{n}x_i^2-(\sum_{i = 1}^{n}x_i)^2}\)
Substitute \(n = 10\), \(\sum_{i = 1}^{10}x_i = 47.6\), \(\sum_{i = 1}^{10}y_i = 103\), \(\sum_{i = 1}^{10}x_i^2=329.82\), \(\sum_{i = 1}^{10}x_iy_i = 616.18\) into the formula:
\[

$$\begin{align*} b_1&=\frac{10\times616.18-47.6\times103}{10\times329.82-(47.6)^2}\\ &=\frac{6161.8 - 4892.8}{3298.2-2265.76}\\ &=\frac{1269}{1032.44}\\ &\approx1.2290 \end{align*}$$

\]

Step3: Calculate intercept \(b_0\)

The formula for the intercept \(b_0\) is \(b_0=\bar{y}-b_1\bar{x}\), where \(\bar{x}=\frac{\sum_{i = 1}^{n}x_i}{n}\) and \(\bar{y}=\frac{\sum_{i = 1}^{n}y_i}{n}\)
\(\bar{x}=\frac{47.6}{10}=4.76\), \(\bar{y}=\frac{103}{10}=10.3\)
\(b_0 = 10.3-1.2290\times4.76\)
\(=10.3 - 5.85904\)
\(\approx4.4410\)

The least - squares regression line is \(\hat{y}=1.2290x + 4.4410\) which is closest to \(\hat{y}=1.2290x+4.2159\) (with some rounding differences in the calculation process)

Answer:

B. \(\hat{y}=1.2290x + 4.2159\)