Question

find the equation of the regression line for the given data. then construct a scatter plot of the data and draw the regression line. (the pair of variables have a significant correlation.) then use the regression equation to predict the value of y for each of the given x - values, if meaningful. the number of hours 6 students spent for a test and their scores on that test are shown below.
hours spent studying, x: 0 1 2 3 5 5
test score, y: 38 44 51 47 64 68
(a) x = 4 hours
(b) x = 3.5 hours
(c) x = 14 hours
(d) x = 4.5 hours
find the regression equation
\\(\hat{y}=\square x+\square\\)
(round the slope to three decimal places as needed. round the y - intercept to two decimal places as needed.)

Explanation:

Step1: Calculate the means of x and y

Let \(x = [0,1,2,3,5,5]\) and \(y=[38,44,51,47,64,68]\)
\(\bar{x}=\frac{0 + 1+2+3+5+5}{6}=\frac{16}{6}\approx2.667\)
\(\bar{y}=\frac{38 + 44+51+47+64+68}{6}=\frac{312}{6} = 52\)

Step2: Calculate the numerator and denominator for the slope (b1)

\[

$$\begin{align*} \sum_{i = 1}^{n}(x_i-\bar{x})(y_i - \bar{y})&=(0 - 2.667)(38-52)+(1 - 2.667)(44 - 52)+(2-2.667)(51 - 52)+(3-2.667)(47 - 52)+(5-2.667)(64 - 52)+(5-2.667)(68 - 52)\\ &=(- 2.667)\times(-14)+(-1.667)\times(-8)+(-0.667)\times(-1)+(0.333)\times(-5)+(2.333)\times12+(2.333)\times16\\ &=37.338+13.336 + 0.667-1.665+27.996+37.328\\ &=115.99 \end{align*}$$

\]
\[

$$\begin{align*} \sum_{i=1}^{n}(x_i-\bar{x})^2&=(0 - 2.667)^2+(1 - 2.667)^2+(2-2.667)^2+(3-2.667)^2+(5-2.667)^2+(5-2.667)^2\\ &=7.112889+2.778889 + 0.444889+0.111889+5.444889+5.444889\\ &=21.338334 \end{align*}$$

\]
The slope \(b_1=\frac{\sum_{i = 1}^{n}(x_i-\bar{x})(y_i - \bar{y})}{\sum_{i=1}^{n}(x_i-\bar{x})^2}=\frac{115.99}{21.338334}\approx5.435\)

Step3: Calculate the y - intercept (b0)

\(b_0=\bar{y}-b_1\bar{x}=52-5.435\times2.667=52 - 14.505145\approx37.49\)

The regression equation is \(\hat{y}=5.435x + 37.49\)

Step4: Predict the values of y

(a) When \(x = 4\)

\(\hat{y}=5.435\times4+37.49=21.74+37.49 = 59.23\)

(b) When \(x = 3.5\)

\(\hat{y}=5.435\times3.5+37.49=19.0225+37.49=56.5125\approx56.51\)

(c) When \(x = 14\)

\(\hat{y}=5.435\times14+37.49=76.09+37.49 = 113.58\) (This may not be meaningful as it may be outside the range of the data used to fit the model)

(d) When \(x = 4.5\)

\(\hat{y}=5.435\times4.5+37.49=24.4575+37.49=61.9475\approx61.95\)

Answer:

The regression equation is \(\hat{y}=5.435x + 37.49\)
(a) 59.23
(b) 56.51
(c) 113.58
(d) 61.95