Question

an educator wants to see how the number of absences for a student in her class affects the students final grade. the data obtained from a sample are shown.

no. of absences x	10	12	2	0	8	5
final grade y	70	65	96	94	75	82

based on the above data, answer the following questions. use complete sentences to answer questions. show your work typed using tables where appropriate.

1. find r (round 4 decimal places)
2. characterize r
3. find r²(round 4 decimal places)
4. interpret r²
5. find the least - square regression equation. (go 5 decimal places first before rounding final answer to 2 decimal places)
6. find the standard error of estimate and describe what the number says. (round the residuals squared to 2 decimal places)
7. find the prediction for the final grade when a student has 7 absences.
8. find the prediction for the final grade when a student has 15 absences.

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. \(56.74\)