The problem provides data of mid-semester test scores ($x$) and end-of-year examination scores ($y$) for 10 students. The goal is to: a) Find the values of $a$ and $b$ for the regression line equation $y = a + bx$ using a calculator. b) Draw a scatter plot of the data and the regression line. c) Show that the regression line passes through the point formed by the mean of $x$ and the mean of $y$.

1. Problem Description

The problem provides data of mid-semester test scores ($x$) and end-of-year examination scores ($y$) for 10 students. The goal is to:

a) Find the values of $a$ and $b$ for the regression line equation $y = a + bx$ using a calculator.

b) Draw a scatter plot of the data and the regression line.

c) Show that the regression line passes through the point formed by the mean of $x$ and the mean of $y$.

2. Solution Steps

a) Finding the regression line equation $y = a + bx$.

First, let's list the data points (x, y):

(15, 63), (12, 65), (19, 82), (16, 75), (8, 52), (12, 60), (6, 48), (14, 64), (11, 58), (13, 68)

n = 10

We need to calculate the following sums:

$\sum x_i$, $\sum y_i$, $\sum x_i y_i$, $\sum x_i^2$

$\sum x_i = 15+12+19+16+8+12+6+14+11+13 = 126$

$\sum y_i = 63+65+82+75+52+60+48+64+58+68 = 635$

$\sum x_i y_i = (15*63)+(12*65)+(19*82)+(16*75)+(8*52)+(12*60)+(6*48)+(14*64)+(11*58)+(13*68) = 945+780+1558+1200+416+720+288+896+638+884 = 8325$

$\sum x_i^2 = 15^2+12^2+19^2+16^2+8^2+12^2+6^2+14^2+11^2+13^2 = 225+144+361+256+64+144+36+196+121+169 = 1716$

Now, calculate the means:

$\bar{x} = \frac{\sum x_i}{n} = \frac{126}{10} = 12.6$

$\bar{y} = \frac{\sum y_i}{n} = \frac{635}{10} = 63.5$

The formula for $b$ is:

$b = \frac{n(\sum x_i y_i) - (\sum x_i)(\sum y_i)}{n(\sum x_i^2) - (\sum x_i)^2}$

$b = \frac{10(8325) - (126)(635)}{10(1716) - (126)^2} = \frac{83250 - 80010}{17160 - 15876} = \frac{3240}{1284} = 2.523364486 \approx 2.523$

The formula for $a$ is:

$a = \bar{y} - b\bar{x}$

$a = 63.5 - (2.523364486)(12.6) = 63.5 - 31.79436709 \approx 31.706$

So, the regression line equation is approximately $y = 31.706 + 2.523x$

b) Drawing a scatter plot and the regression line.

This part requires a graph, which I cannot generate. However, you can plot the points (x, y) from the table and then draw the line $y = 31.706 + 2.523x$ on the same graph.

c) Showing the regression line passes through $(\bar{x}, \bar{y})$.

We already calculated $\bar{x} = 12.6$ and $\bar{y} = 63.5$.

Now, we substitute $\bar{x}$ into the regression line equation and check if we get $\bar{y}$.

$y = 31.706 + 2.523(12.6) = 31.706 + 31.790 = 63.496 \approx 63.5$

Since the calculated value is approximately equal to $\bar{y}$, the regression line passes through the point $(\bar{x}, \bar{y})$.

3. Final Answer

a) $a \approx 31.706$, $b \approx 2.523$. The regression line equation is approximately $y = 31.706 + 2.523x$.

b) Scatter plot should be drawn using the data points, and the regression line plotted according to the equation $y = 31.706 + 2.523x$.

c) The regression line passes through the point $(\bar{x}, \bar{y})$ because when $x=\bar{x}=12.6$ is substituted into the regression line equation, the resulting $y$ is approximately equal to $\bar{y}=63.5$.