The following table of values compares national test scores to high school GPA at a certain school.
National Test Score HS GPA
2350 3.95
1900 3.20
1825 2.00
2270 3.00
500 1.50
2000 4.00
1550 2.50
1030 1.17
1665 2.85
1380 2.25
2220 3.60
Using the equation for the line of best fit, what is the best prediction of the high school GPA for a person with a national test score of 1850? Show your work.
A. 2.20
B. 3.15
C. 3.85
D. 1.90
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A scientist calculated the mean and standard deviation of a data set to be mc026-1.jpg and mc026-2.jpg. She then found that she was missing one data value from the set. She knows that the missing data value was exactly 3 standard deviations away from the mean. What was the missing data value?
To find the best prediction of the high school GPA for a person with a national test score of 1850, we need to calculate the equation of the line of best fit for the given data points.
We can use the least squares regression method to find the equation of the line that minimizes the sum of the squared differences between the actual data points and the predicted values.
Step 1: Calculate the mean of the National Test Score and HS GPA values.
Mean of National Test Score: (2350 + 1900 + 1825 + 2270 + 500 + 2000 + 1550 + 1030 + 1665 + 1380 + 2220) / 11 = 1635
Mean of HS GPA: (3.95 + 3.20 + 2.00 + 3.00 + 1.50 + 4.00 + 2.50 + 1.17 + 2.85 + 2.25 + 3.60) / 11 = 2.71
Step 2: Calculate the deviations of each data point from the mean.
For National Test Score: Subtract the mean from each test score value.
For HS GPA: Subtract the mean from each GPA value.
National Test Score Deviation: (2350 - 1635), (1900 - 1635), (1825 - 1635), (2270 - 1635), (500 - 1635), (2000 - 1635), (1550 - 1635), (1030 - 1635), (1665 - 1635), (1380 - 1635), (2220 - 1635)
= 715, 265, 190, 635, -1135, 365, -85, -605, 30, -255, 585
HS GPA Deviation: (3.95 - 2.71), (3.20 - 2.71), (2.00 - 2.71), (3.00 - 2.71), (1.50 - 2.71), (4.00 - 2.71), (2.50 - 2.71), (1.17 - 2.71), (2.85 - 2.71), (2.25 - 2.71), (3.60 - 2.71)
= 1.24, 0.49, -0.71, 0.29, -1.21, 1.29, -0.21, -1.54, 0.14, -0.46, 0.89
Step 3: Calculate the product of the deviations for each data point.
Product of Deviations: National Test Score Deviation * HS GPA Deviation
(715 * 1.24), (265 * 0.49), (190 * -0.71), (635 * 0.29), (-1135 * -1.21), (365 * 1.29), (-85 * -0.21), (-605 * -1.54), (30 * 0.14), (-255 * -0.46), (585 * 0.89)
= 887.60, 129.85, -134.90, 184.15, 1373.35, 471.85, 17.85, 931.70, 4.20, 117.30, 520.65
Step 4: Calculate the sum of the product of deviations.
Sum of Product of Deviations = 887.60 + 129.85 - 134.90 + 184.15 + 1373.35 + 471.85 + 17.85 + 931.70 + 4.20 + 117.30 + 520.65
= 3454.40
Step 5: Calculate the square of the deviations for National Test Score.
Square of the Deviations = National Test Score Deviation^2
(715^2), (265^2), (190^2), (635^2), (-1135^2), (365^2), (-85^2), (-605^2), (30^2), (-255^2), (585^2)
= 511,225, 70,225, 36,100, 403,225, 1,285,225, 133,225, 132,250, 36,025, 367,225
Step 6: Calculate the sum of the square deviations for the National Test Score.
Sum of Square Deviations = 511,225 + 70,225 + 36,100 + 403,225 + 1,285,225 + 133,225 + 132,250 + 36,025 + 367,225
= 2,974,900
Step 7: Calculate the slope of the line.
Slope (m) = Sum of Product of Deviations / Sum of Square Deviations
= 3454.40 / 2,974,900
≈ 0.001161
Step 8: Calculate the y-intercept of the line.
y-Intercept (b) = Mean of HS GPA - Slope * Mean of National Test Score
= 2.71 - 0.001161 * 1635
≈ 2.710
Step 9: Write the equation of the line of best fit.
HS GPA = Slope * National Test Score + y-Intercept
= 0.001161 * National Test Score + 2.710
Step 10: Predict the HS GPA for a person with a national test score of 1850.
HS GPA = 0.001161 * 1850 + 2.710
≈ 3.147
Therefore, the best prediction of the high school GPA for a person with a national test score of 1850 is approximately 3.147.
The closest option provided is B. 3.15.