Question

independent practice

1. the amount of hours spent doing homework or studying throughout the week can have a posi

impact on your weekly quiz grades. data for 7 students showing hours studied vs weekly quiz
grades is shown below.

hours studied	0	2	3	5	6	8	10

a. what is the linear regression equation for the data modeled above?

y =

b. what is the correlation coefficient for the data above? describe it. (circle choices)

r = positive or negative strong or weak

c. based on the linear regression equation above, each hour studied is estimated to raise your
grade by how many points?

d. based on the linear regression equation above, what would be the estimated score for
someone who studies 0 hours?

e. using the linear regression equation, estimate the score of someone who studies for 13 hours

Explanation:

Step 1: Define Variables

Let \( x \) be the hours studied and \( y \) be the weekly quiz grade. The data points are: \((0, 45)\), \((2, 51)\), \((3, 55)\), \((5, 60)\), \((6, 63)\), \((8, 70)\), \((10, 73)\).

Step 2: Calculate \( \sum x \), \( \sum y \), \( \sum xy \), \( \sum x^2 \)

• \( \sum x = 0 + 2 + 3 + 5 + 6 + 8 + 10 = 34 \)
• \( \sum y = 45 + 51 + 55 + 60 + 63 + 70 + 73 = 417 \)
• \( \sum xy = (0\times45)+(2\times51)+(3\times55)+(5\times60)+(6\times63)+(8\times70)+(10\times73) = 0 + 102 + 165 + 300 + 378 + 560 + 730 = 2235 \)
• \( \sum x^2 = 0^2 + 2^2 + 3^2 + 5^2 + 6^2 + 8^2 + 10^2 = 0 + 4 + 9 + 25 + 36 + 64 + 100 = 238 \)

Step 3: Calculate Slope \( m \)

The formula for the slope \( m \) of the linear regression line \( y = mx + b \) is:

$$ m=\frac{n\sum xy-\sum x\sum y}{n\sum x^2 - (\sum x)^2} $$

where \( n = 7 \) (number of data points).
Substitute the values:

$$ m=\frac{7\times2235 - 34\times417}{7\times238 - 34^2}=\frac{15645 - 14178}{1666 - 1156}=\frac{1467}{510}\approx2.876 $$

Step 4: Calculate Intercept \( b \)

The formula for the intercept \( b \) is:

$$ b=\frac{\sum y - m\sum x}{n} $$

Substitute the values:

$$ b=\frac{417 - 2.876\times34}{7}=\frac{417 - 97.784}{7}=\frac{319.216}{7}\approx45.602 $$

Step 5: Linear Regression Equation

So the linear regression equation is \( y \approx 2.88x + 45.60 \) (rounded to two decimal places).

Step 6: 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 \( \sum y^2 = 45^2 + 51^2 + 55^2 + 60^2 + 63^2 + 70^2 + 73^2 = 2025 + 2601 + 3025 + 3600 + 3969 + 4900 + 5329 = 25449 \)
Substitute the values:

$$ r=\frac{7\times2235 - 34\times417}{\sqrt{[7\times238 - 34^2][7\times25449 - 417^2]}}=\frac{1467}{\sqrt{510\times(178143 - 173889)}}=\frac{1467}{\sqrt{510\times4254}}=\frac{1467}{\sqrt{2170540}}\approx\frac{1467}{1473.27}\approx0.996 $$

Since \( r \approx 0.996 \), it is positive (because \( r>0 \)) and strong (because \( |r| \) is close to 1).

Step 7: Part (c) - Points per Hour Studied

The slope \( m \approx 2.88 \) represents the estimated increase in grade per hour studied. So each hour studied is estimated to raise the grade by approximately \( 2.88 \) points.

Step 8: Part (d) - Estimated Score for 0 Hours Studied

Substitute \( x = 0 \) into the regression equation \( y = 2.88x + 45.60 \). When \( x = 0 \), \( y \approx 45.60 \), which is close to the actual value of 45 (the small difference is due to rounding).

Step 9: Estimated Score for 13 Hours Studied

Substitute \( x = 13 \) into the regression equation:

$$ y = 2.88\times13 + 45.60 = 37.44 + 45.60 = 83.04 $$

Answer:

s:
a. \( y \approx \boldsymbol{2.88x + 45.60} \) (or more precise \( y \approx 2.88x + 45.60 \))
b. \( r \approx \boldsymbol{0.996} \), Positive, Strong
c. Approximately \( \boldsymbol{2.88} \) points (or the more precise value from the slope)
d. Approximately \( \boldsymbol{45.60} \) (or close to 45)
e. Approximately \( \boldsymbol{83.04} \) (or more precise calculation)