an educator wants to see how the number of absences for a student in her class affects the students final…

Explanation:

Step1: Calculate necessary sums

Let (n = 6). Calculate (\sum x=10 + 12+2+0+8+5=37), (\sum y=70 + 65+96+94+75+82 = 482), (\sum x^{2}=10^{2}+12^{2}+2^{2}+0^{2}+8^{2}+5^{2}=100 + 144+4+0+64+25 = 337), (\sum y^{2}=70^{2}+65^{2}+96^{2}+94^{2}+75^{2}+82^{2}=4900+4225+9216+8836+5625+6724 = 39526), (\sum xy=(10\times70)+(12\times65)+(2\times96)+(0\times94)+(8\times75)+(5\times82)=700+780 + 192+0+600+410=2682).

Step2: Calculate the correlation - coefficient (r)

The formula for (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}]}}) [ \begin{align*} n\sum xy-(\sum x)(\sum y)&=6\times2682-37\times482\ &=16092-17834\ &=- 1742\ n\sum x^{2}-(\sum x)^{2}&=6\times337 - 37^{2}\ &=2022-1369\ &=653\ n\sum y^{2}-(\sum y)^{2}&=6\times39526-482^{2}\ &=237156 - 232324\ &=4832\ r&=\frac{-1742}{\sqrt{653\times4832}}\ &=\frac{-1742}{\sqrt{3155296}}\ &=\frac{-1742}{1776.319}\ &\approx - 0.9807 \end{align*} ]

Step3: Characterize (r)

Since (r\approx - 0.9807), the value of (r) is close to (-1). This indicates a strong negative linear relationship between the number of absences and the final grade. That is, as the number of absences increases, the final grade tends to decrease.

Step4: Calculate (r^{2})

(r^{2}=(-0.9807)^{2}\approx0.9618)

Step5: Interpret (r^{2})

The coefficient of determination (r^{2}\approx0.9618) means that approximately (96.18%) of the variation in the final - grade can be explained by the linear relationship with the number of absences.

Step6: Find the least - square regression equation

The slope (b) of the least - square regression line is given by (b=\frac{n\sum xy-(\sum x)(\sum y)}{n\sum x^{2}-(\sum x)^{2}}) and the (y) - intercept (a) is given by (a=\overline{y}-b\overline{x}), where (\overline{x}=\frac{\sum x}{n}=\frac{37}{6}\approx6.16667) and (\overline{y}=\frac{\sum y}{n}=\frac{482}{6}\approx80.33333) We already know that (b=\frac{-1742}{653}\approx - 2.6677) (a = 80.33333-(-2.6677)\times6.16667) (a = 80.33333 + 16.4577) (a\approx96.79103) The least - square regression equation is (\hat{y}=96.79-2.67x) (rounded to 2 decimal places)

Step7: Calculate the standard error of estimate (S_{e})

First, we find the predicted values (\hat{y}) for each (x) value using (\hat{y}=96.79-2.67x). For (x = 10), (\hat{y}=96.79-2.67\times10=70.09), residual (e_1=70 - 70.09=-0.09) For (x = 12), (\hat{y}=96.79-2.67\times12=64.85), residual (e_2=65 - 64.85 = 0.15) For (x = 2), (\hat{y}=96.79-2.67\times2=91.45), residual (e_3=96 - 91.45 = 4.55) For (x = 0), (\hat{y}=96.79-2.67\times0=96.79), residual (e_4=94 - 96.79=-2.79) For (x = 8), (\hat{y}=96.79-2.67\times8=75.43), residual (e_5=75 - 75.43=-0.43) For (x = 5), (\hat{y}=96.79-2.67\times5=83.44), residual (e_6=82 - 83.44=-1.44) (\sum e^{2}=(-0.09)^{2}+(0.15)^{2}+(4.55)^{2}+(-2.79)^{2}+(-0.43)^{2}+(-1.44)^{2}) (\sum e^{2}=0.0081 + 0.0225+20.7025+7.7841+0.1849+2.0736=30.7757) The standard error of estimate (S_{e}=\sqrt{\frac{\sum e^{2}}{n - 2}}=\sqrt{\frac{30.7757}{4}}\approx2.77) The standard error of estimate (S_{e}\approx2.77) represents the average amount that the observed values of (y) deviate from the predicted values (\hat{y}) in the regression model.

Step8: Predict the final grade for (x = 7)

(\hat{y}=96.79-2.67\times7=96.79 - 18.69 = 78.1)

Step9: Predict the final grade for (x = 15)

(\hat{y}=96.79-2.67\times15=96.79-40.05 = 56.74)

Answer:

1. (r\approx - 0.9807)
2. There is a strong negative linear relationship between the number of absences and the final grade.
3. (r^{2}\approx0.9618)
4. Approximately (96.18%) of the variation in the final - grade can be explained by the linear relationship with the number of absences.
5. (\hat{y}=96.79-2.67x)
6. (S_{e}\approx2.77), which represents the average amount that the observed values of (y) deviate from the predicted values (\hat{y}) in the regression model.
7. (78.1)
8. (56.74)