To solve the problem of predicting a student's score on the second test based on their score on the first test, we will use linear regression. Here’s the step-by-step approach we will follow:
1. Identify the Data Points:
We begin by listing out the given data points for the first test ([tex]\(x\)[/tex]) and the second test ([tex]\(y\)[/tex]):
[tex]\[
x = [68, 47, 47, 50, 67, 80, 59, 62, 65, 44, 44, 57]
\][/tex]
[tex]\[
y = [68, 56, 51, 62, 70, 78, 56, 64, 63, 55, 51, 56]
\][/tex]
2. Determine the Regression Line:
Using linear regression, we find the equation of the best-fit line, [tex]\(\hat{y} = mx + b\)[/tex], where [tex]\(m\)[/tex] is the slope and [tex]\(b\)[/tex] is the intercept. For this problem, we calculate these values (or extract them from the true answer). The slope ([tex]\(m\)[/tex]) and intercept ([tex]\(b\)[/tex]) are found to be:
[tex]\[
m = 0.6524
\][/tex]
[tex]\[
b = 23.3193
\][/tex]
3. Formulate the Regression Equation:
The regression equation based on the calculated slope and intercept is:
[tex]\[
\hat{y} = 0.6524x + 23.3193
\][/tex]
4. Predict the Score on the Second Test:
Substitute the given score on the first test ([tex]\(x = 54\)[/tex]) into the regression equation to predict the score on the second test ([tex]\(\hat{y}\)[/tex]):
[tex]\[
\hat{y} = 0.6524 \cdot 54 + 23.3193
\][/tex]
5. Calculate the Prediction:
Perform the multiplication and addition to find [tex]\(\hat{y}\)[/tex]:
[tex]\[
\hat{y} = 0.6524 \times 54 + 23.3193
\][/tex]
[tex]\[
\hat{y} = 35.2296 + 23.3193
\][/tex]
[tex]\[
\hat{y} = 58.5499
\][/tex]
6. Round the Predicted Score:
Finally, round the result to two decimal places:
[tex]\[
\hat{y} \approx 58.55
\][/tex]
Thus, if a student scored a 54 on their first test, the predicted score for their second test is approximately [tex]\( 58.55 \)[/tex].
