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Complete parts (a) through (d) using the given statistics.

\begin{itemize}
\item LinReg: [tex]$y = a + bx$[/tex]
\item [tex]$b = 1.894335513$[/tex]
\end{itemize}

[tex]\[
\begin{tabular}{|c|c|c|}
\hline
& Mean & Standard Deviation \\
\hline
Height, $x$ & 63.86 & 3.82 \\
\hline
Armspan, $y$ & 159.77 & 7.78 \\
\hline
\end{tabular}
\][/tex]

\begin{itemize}
\item [tex]$s_x = 3.82$[/tex] (Round to two decimal places as needed.)
\item [tex]$b = 1.90$[/tex] (Round to two decimal places as needed.)
\end{itemize}

c. Determine the corresponding values and verify the [tex]$y$[/tex]-intercept by using the formula:
[tex]\[a = \bar{y} - b \bar{x}\][/tex]

[tex]\[
\begin{aligned}
&\bar{y} = 159.77 \text{ (Round to two decimal places as needed.)} \\
&\bar{x} = 63.86 \text{ (Round to two decimal places as needed.)} \\
&a = 38.44 \text{ (Round to two decimal places as needed.)}
\end{aligned}
\][/tex]

d. Using the regression equation from part (a), predict the armspan (in cm) for someone 61 inches tall.

[tex]\[
\text{Predicted Armspan} = \square \text{ cm} \quad \text{(Round to one decimal place as needed.)}
\][/tex]


Sagot :

Certainly! Let's break down the solution step-by-step.

### Part c: Determine the y-intercept [tex]\( a \)[/tex]

We are given the following information:
- Mean height ([tex]\( \bar{x} \)[/tex]): 63.86
- Mean armspan ([tex]\( \bar{y} \)[/tex]): 159.77
- Slope ([tex]\( b \)[/tex]): 1.90

To find the y-intercept ([tex]\( a \)[/tex]), we use the formula for the y-intercept in a linear regression line:
[tex]\[ a = \bar{y} - b \cdot \bar{x}\][/tex]

Substituting the given values:
[tex]\[ a = 159.77 - 1.90 \cdot 63.86 \][/tex]

Performing the calculations, we get:
[tex]\[ a = 159.77 - 121.334 \][/tex]
[tex]\[ a = 38.436 \][/tex]

Rounding to two decimal places, we obtain:
[tex]\[ a = 38.44 \][/tex]

So, the verified y-intercept is:
[tex]\[ a = 38.44 \][/tex]

### Part d: Predict the armspan for someone 61 inches tall

Now that we have our linear regression equation [tex]\( y = a + bx \)[/tex], where:
- [tex]\( a = 38.44 \)[/tex] (y-intercept)
- [tex]\( b = 1.90 \)[/tex] (slope)
- [tex]\( x \)[/tex] is the height for prediction (61 inches)

We substitute [tex]\( x = 61 \)[/tex] into the regression equation to predict the armspan:

[tex]\[ y = 38.44 + 1.90 \cdot 61 \][/tex]

Performing the calculations, we get:
[tex]\[ y = 38.44 + 115.9 \][/tex]
[tex]\[ y = 154.34 \][/tex]

Rounding to one decimal place, we obtain:
[tex]\[ y = 154.3 \][/tex]

Therefore, the predicted armspan for someone 61 inches tall is:
[tex]\[ \text{Predicted Armspan} = 154.3 \, \text{cm} \][/tex]
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