3. the table below gives the number of hours spent studying for a science exam (x) and the final exam grade (y). predict the exam grade of a student who studied for 6 hours. type a response

Question

Accepted Answer

# Explanation:
## Step1: Calculate the mean of $x$ and $y$
$$
\begin{align*}
\bar{x}&=\frac{2 + 5+1+0+4+2+3}{7}=\frac{17}{7}\approx2.43\
\bar{y}&=\frac{77 + 92+70+63+90+75+84}{7}=\frac{551}{7}\approx78.71
\end{align*}
$$
## Step2: Calculate the numerator and denominator for the slope $b$
$$
\begin{align*}
\text{Numerator}&=\sum_{i = 1}^{n}(x_{i}-\bar{x})(y_{i}-\bar{y})\
&=(2 - 2.43)(77-78.71)+(5 - 2.43)(92 - 78.71)+(1 - 2.43)(70 - 78.71)+(0 - 2.43)(63 - 78.71)+(4 - 2.43)(90 - 78.71)+(2 - 2.43)(75 - 78.71)+(3 - 2.43)(84 - 78.71)\
&=(- 0.43)(-1.71)+(2.57)(13.29)+(-1.43)(-8.71)+(-2.43)(-15.71)+(1.57)(11.29)+(-0.43)(-3.71)+(0.57)(5.29)\
&=0.7353+34.1553 + 12.4553+38.1753+17.7253+1.5953+3.0153\
&=107.86
\end{align*}
$$
$$
\begin{align*}
\text{Denominator}&=\sum_{i = 1}^{n}(x_{i}-\bar{x})^{2}\
&=(2 - 2.43)^{2}+(5 - 2.43)^{2}+(1 - 2.43)^{2}+(0 - 2.43)^{2}+(4 - 2.43)^{2}+(2 - 2.43)^{2}+(3 - 2.43)^{2}\
&=(-0.43)^{2}+(2.57)^{2}+(-1.43)^{2}+(-2.43)^{2}+(1.57)^{2}+(-0.43)^{2}+(0.57)^{2}\
&=0.1849+6.6049+2.0449+5.9049+2.4649+0.1849+0.3249\
&=17.71
\end{align*}
$$
$$b=\frac{107.86}{17.71}\approx6.09$$
## Step3: Calculate the intercept $a$
$$a=\bar{y}-b\bar{x}=78.71-6.09\times2.43=78.71 - 14.80=63.91$$
## Step4: Form the regression equation and predict
The regression equation is $y = 63.91+6.09x$. When $x = 6$, $y=63.91+6.09\times6=63.91 + 36.54=100.45\approx100$

# Answer:
100

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QUESTION IMAGE

Question

Explanation:

Step1: Calculate the mean of \(x\) and \(y\)

Step2: Calculate the numerator and denominator for the slope \(b\)

Step3: Calculate the intercept \(a\)

Step4: Form the regression equation and predict

Answer: