Math Problem Statement

IThe table below gives the number of hours spent unsupervised each day as well as the overall grade averages for seven randomly selected middle school students. Using this data, consider the equation of the regression line, y bob₁x, 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

2

2.5

3.5

4

5.5

Overall Grades

1

99 81 73 72 67 65

6

63

Step 2 of 6: Find the estimated intercept. Round your answer to three decimal

places.

Solution

To calculate the estimated intercept (b0b_0) of the regression line (y=b0+b1xy = b_0 + b_1x), you need the following steps:

  1. Obtain the slope (b1b_1):
    b1=(xixˉ)(yiyˉ)(xixˉ)2b_1 = \frac{\sum (x_i - \bar{x})(y_i - \bar{y})}{\sum (x_i - \bar{x})^2}

  2. Find the intercept (b0b_0):
    After finding b1b_1, the intercept b0b_0 can be computed as: b0=yˉb1xˉb_0 = \bar{y} - b_1 \bar{x} where:

    • xˉ\bar{x} is the mean of the xx values (Hours Unsupervised).
    • yˉ\bar{y} is the mean of the yy values (Overall Grades).

Step-by-Step Solution:

Given data:

  • xx (Hours Unsupervised): 2,2.5,3.5,4,5.52, 2.5, 3.5, 4, 5.5
  • yy (Overall Grades): 99,81,73,67,6399, 81, 73, 67, 63

1. Compute the Means (xˉ\bar{x} and yˉ\bar{y}):

xˉ=xin,yˉ=yin\bar{x} = \frac{\sum x_i}{n}, \quad \bar{y} = \frac{\sum y_i}{n}

  • n=5n = 5
  • xˉ=2+2.5+3.5+4+5.55=3.5\bar{x} = \frac{2 + 2.5 + 3.5 + 4 + 5.5}{5} = 3.5
  • yˉ=99+81+73+67+635=76.6\bar{y} = \frac{99 + 81 + 73 + 67 + 63}{5} = 76.6

2. Compute b1b_1 (Slope):

b1=(xixˉ)(yiyˉ)(xixˉ)2b_1 = \frac{\sum (x_i - \bar{x})(y_i - \bar{y})}{\sum (x_i - \bar{x})^2} First, calculate xixˉx_i - \bar{x} and yiyˉy_i - \bar{y}, then multiply and square:

xix_iyiy_ixixˉx_i - \bar{x}yiyˉy_i - \bar{y}(xixˉ)(yiyˉ)(x_i - \bar{x})(y_i - \bar{y})(xixˉ)2(x_i - \bar{x})^2
2991.5-1.522.422.433.6-33.62.252.25
2.5811.0-1.04.44.44.4-4.41.01.0
3.5730.00.03.6-3.60.00.00.00.0
4670.50.59.6-9.64.8-4.80.250.25
5.5632.02.013.6-13.627.2-27.24.04.0
  • (xixˉ)(yiyˉ)=33.64.4+0.04.827.2=70.0\sum (x_i - \bar{x})(y_i - \bar{y}) = -33.6 - 4.4 + 0.0 - 4.8 - 27.2 = -70.0
  • (xixˉ)2=2.25+1.0+0.0+0.25+4.0=7.5\sum (x_i - \bar{x})^2 = 2.25 + 1.0 + 0.0 + 0.25 + 4.0 = 7.5

b1=70.07.5=9.333b_1 = \frac{-70.0}{7.5} = -9.333

3. Compute b0b_0 (Intercept):

b0=yˉb1xˉb_0 = \bar{y} - b_1 \bar{x}

  • yˉ=76.6\bar{y} = 76.6, xˉ=3.5\bar{x} = 3.5, b1=9.333b_1 = -9.333 b0=76.6(9.333)(3.5)=76.6+32.666=109.266b_0 = 76.6 - (-9.333)(3.5) = 76.6 + 32.666 = 109.266

Final Answer:

The estimated intercept is: b0=109.266b_0 = 109.266


Would you like further details or have any questions?

Related Questions:

  1. How do you interpret the slope (b1b_1) in the context of this data?
  2. What is the significance of the intercept (b0b_0) for these grades?
  3. How can we determine if the correlation coefficient is statistically significant?
  4. What are the limitations of using this regression line for predictions?
  5. How would the calculations change if another data point was added?

Tip:

Always visualize your data with a scatterplot before performing regression to check for linearity!

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Math Problem Analysis

Mathematical Concepts

Statistics
Linear Regression
Correlation Coefficient

Formulas

b₁ = Σ((xᵢ - x̄)(yᵢ - ȳ)) / Σ((xᵢ - x̄)²)
b₀ = ȳ - b₁x̄

Theorems

Least Squares Method

Suitable Grade Level

Grades 10-12