Question

1. the table below gives the number of hours spent studying for a science exam (x) and the final exam grade (y).

x	2	5	1	0	4	2	3

predict the exam grade of a student who studied for 6 hours.
equation_______________________ r = __ grade expected____
describe the correlation _________________________________________

Explanation:

Step1: Calculate necessary sums

First, we need to find the sums of \( x \), \( y \), \( xy \), and \( x^2 \).
Given \( x = [2, 5, 1, 0, 4, 2, 3] \) and \( y = [77, 92, 70, 63, 90, 75, 84] \)
\( n = 7 \)
\( \sum x = 2 + 5 + 1 + 0 + 4 + 2 + 3 = 17 \)
\( \sum y = 77 + 92 + 70 + 63 + 90 + 75 + 84 = 551 \)
\( \sum xy = (2\times77)+(5\times92)+(1\times70)+(0\times63)+(4\times90)+(2\times75)+(3\times84) \)
\( = 154 + 460 + 70 + 0 + 360 + 150 + 252 = 1446 \)
\( \sum x^2 = 2^2 + 5^2 + 1^2 + 0^2 + 4^2 + 2^2 + 3^2 = 4 + 25 + 1 + 0 + 16 + 4 + 9 = 59 \)

Step2: Calculate slope (\( m \)) and intercept (\( b \)) of regression line

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:
\( m=\frac{7\times1446 - 17\times551}{7\times59 - 17^2} \)
\( 7\times1446 = 10122 \), \( 17\times551 = 9367 \), \( 7\times59 = 413 \), \( 17^2 = 289 \)
\( m=\frac{10122 - 9367}{413 - 289}=\frac{755}{124}\approx6.089 \)
The formula for the intercept \( b \) is:
\( b=\frac{\sum y - m\sum x}{n} \)
\( b=\frac{551 - 6.089\times17}{7} \)
\( 6.089\times17\approx103.513 \)
\( b=\frac{551 - 103.513}{7}=\frac{447.487}{7}\approx63.927 \)
So the regression equation is \( y = 6.089x + 63.927 \) (we can round to reasonable decimals, e.g., \( y\approx6.09x + 63.93 \))

Step3: Calculate correlation coefficient (\( r \))

The formula for the correlation coefficient \( r \) is:
\( r=\frac{n\sum xy - \sum x\sum y}{\sqrt{[n\sum x^2 - (\sum x)^2][n\sum y^2 - (\sum y)^2]}} \)
First, calculate \( \sum y^2 = 77^2 + 92^2 + 70^2 + 63^2 + 90^2 + 75^2 + 84^2 \)
\( = 5929 + 8464 + 4900 + 3969 + 8100 + 5625 + 7056 = 44043 \)
\( n\sum y^2 - (\sum y)^2 = 7\times44043 - 551^2 \)
\( 7\times44043 = 308301 \), \( 551^2 = 303601 \)
\( 308301 - 303601 = 4700 \)
\( n\sum x^2 - (\sum x)^2 = 124 \) (from step 2)
\( n\sum xy - \sum x\sum y = 755 \) (from step 2)
\( r=\frac{755}{\sqrt{124\times4700}} \)
\( \sqrt{124\times4700}=\sqrt{582800}\approx763.41 \)
\( r=\frac{755}{763.41}\approx0.989 \)

Step4: Predict the grade for \( x = 6 \)

Substitute \( x = 6 \) into the regression equation \( y = 6.089x + 63.927 \)
\( y = 6.089\times6 + 63.927 \)
\( 6.089\times6\approx36.534 \)
\( y\approx36.534 + 63.927\approx100.46 \) (we can also use the rounded slope and intercept, or more precise calculation. Alternatively, with more precise \( m \) and \( b \), the result may vary slightly, but approximately 100 or 100.5)

Answer:

Equation: \( y \approx 6.09x + 63.93 \) (or more precise \( y=\frac{755}{124}x+\frac{551 - \frac{755}{124}\times17}{7} \))
\( r \approx 0.99 \) (strong positive correlation)
Grade expected: Approximately 100 (or 100.5)
Correlation description: Strong positive correlation (since \( r \) is close to 1)