Estimated Grade Prediction Using Regression Analysis

Math Problem Statement

The table below gives the number of hours spent unsupervised each day as well as the overall grade averages for five randomly selected middle school students. Using this data, consider the equation of the regression line, yˆ=b0+b1x , for predicting the overall grade average for a middle school student based on the number of hours spent unsupervised each day. Keep in mind, the correlation coefficient may or may not be statistically significant for the data given. Remember, in practice, it would not be appropriate to use the regression line to make a prediction if the correlation coefficient is not statistically significant.

Hours Unsupervised 1 2 3 4 5 Overall Grades 93 88 80 72 68 Table

Step 5 of 6: Find the estimated value of y when x=1 . Round your answer to three decimal places.

Solution

To find the estimated value of $y$ (the overall grade average) when $x = 1$ (when a student spends 1 hour unsupervised), we first need to calculate the equation of the regression line $\hat{y} = b_0 + b_1 x$ , where:

$b_0$ is the y-intercept
$b_1$ is the slope
$x$ is the number of hours unsupervised

To calculate $b_0$ and $b_1$ , we need to use the following formulas:

Formula for $b_1$ (slope):

$b_1 = \frac{n \sum{xy} - \sum{x} \sum{y}}{n \sum{x^2} - (\sum{x})^2}$

Formula for $b_0$ (intercept):

$b_0 = \frac{\sum{y} - b_1 \sum{x}}{n}$

Where:

$x$ is the number of hours unsupervised
$y$ is the overall grade average
$n$ is the number of data points (5 in this case)

We are given the following data:

\hline \text{Hours Unsupervised (x)} & \text{Overall Grades (y)} \\ \hline 1 & 93 \\ 2 & 88 \\ 3 & 80 \\ 4 & 72 \\ 5 & 68 \\ \hline \end{array}$$ Let's now compute the required values. ### Step 1: Compute the necessary sums We need the sums for $$x$$, $$y$$, $$xy$$, and $$x^2$$: $$\sum{x} = 1 + 2 + 3 + 4 + 5 = 15$$ $$\sum{y} = 93 + 88 + 80 + 72 + 68 = 401$$ $$\sum{xy} = (1)(93) + (2)(88) + (3)(80) + (4)(72) + (5)(68) = 93 + 176 + 240 + 288 + 340 = 1137$$ $$\sum{x^2} = (1)^2 + (2)^2 + (3)^2 + (4)^2 + (5)^2 = 1 + 4 + 9 + 16 + 25 = 55$$ ### Step 2: Compute $$b_1$$ and $$b_0$$ Now we can compute the slope $$b_1$$: $$b_1 = \frac{5(1137) - (15)(401)}{5(55) - (15)^2} = \frac{5685 - 6015}{275 - 225} = \frac{-330}{50} = -6.6$$ Next, we calculate the y-intercept $$b_0$$: $$b_0 = \frac{401 - (-6.6)(15)}{5} = \frac{401 + 99}{5} = \frac{500}{5} = 100$$ ### Step 3: Write the regression equation The equation of the regression line is: $$\hat{y} = 100 - 6.6x$$ ### Step 4: Find the estimated value of $$y$$ when $$x = 1$$ Now that we have the regression equation, we can substitute $$x = 1$$ into the equation: $$\hat{y} = 100 - 6.6(1) = 100 - 6.6 = 93.4$$ So, the estimated overall grade average when a student spends 1 hour unsupervised is **93.4**. ### Final Answer: The estimated value of $$y$$ when $$x = 1$$ is **93.400**. --- Do you need further explanation on any of the steps or calculations? Let me know if you have any more questions. --- Here are some related questions to consider: 1. How would you compute the correlation coefficient for this data? 2. What would happen to the regression line if the correlation coefficient was not statistically significant? 3. How do you interpret the slope and intercept of the regression line in the context of this data? 4. What are the limitations of using linear regression for predicting grades based on hours unsupervised? 5. How could you use this regression model to predict grades for other students? Tip: Always check if the data fits a linear model before applying regression—if the data shows a non-linear pattern, a different type of model might be more appropriate.

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Math Problem Analysis

Mathematical Concepts

Regression Analysis
Linear Equations
Statistical Prediction

Formulas

Regression equation: yˆ = b0 + b1x
Slope formula: b1 = (nΣxy - ΣxΣy) / (nΣx² - (Σx)²)
Intercept formula: b0 = (Σy - b1Σx) / n

Theorems

Least Squares Regression
Correlation Coefficient

Suitable Grade Level

Grades 9-12

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