Question

1. the table below gives the height and shoe sizes of six randomly selected men.

height	67	70	73.5	75	78	66

if a man has a shoe size of 10.5, what would be his predicted height?
equation________ r = __ height expected____
describe the correlation __________

Explanation:

Step1: Define Variables

Let \( x \) be shoe size and \( y \) be height. We have data points: \((8.5, 67)\), \((9.5, 70)\), \((11, 73.5)\), \((12, 75)\), \((13, 78)\), \((8, 66)\).

Step2: Calculate Mean of \( x \) and \( y \)

Mean of \( x \) (\(\bar{x}\)):

$$ \bar{x}=\frac{8.5 + 9.5 + 11 + 12 + 13 + 8}{6}=\frac{62}{6}\approx10.333 $$

Mean of \( y \) (\(\bar{y}\)):

$$ \bar{y}=\frac{67 + 70 + 73.5 + 75 + 78 + 66}{6}=\frac{429.5}{6}\approx71.583 $$

Step3: Calculate Slope (\( m \)) of Regression Line

Slope formula: \( m=\frac{\sum (x_i - \bar{x})(y_i - \bar{y})}{\sum (x_i - \bar{x})^2} \)

First, compute \( (x_i - \bar{x})(y_i - \bar{y}) \) and \( (x_i - \bar{x})^2 \) for each point:

• For \( (8.5, 67) \):

\( (8.5 - 10.333)(67 - 71.583)\approx(-1.833)(-4.583)\approx8.403 \)
\( (8.5 - 10.333)^2\approx(-1.833)^2\approx3.360 \)

• For \( (9.5, 70) \):

\( (9.5 - 10.333)(70 - 71.583)\approx(-0.833)(-1.583)\approx1.318 \)
\( (9.5 - 10.333)^2\approx(-0.833)^2\approx0.694 \)

• For \( (11, 73.5) \):

\( (11 - 10.333)(73.5 - 71.583)\approx(0.667)(1.917)\approx1.279 \)
\( (11 - 10.333)^2\approx(0.667)^2\approx0.445 \)

• For \( (12, 75) \):

\( (12 - 10.333)(75 - 71.583)\approx(1.667)(3.417)\approx5.694 \)
\( (12 - 10.333)^2\approx(1.667)^2\approx2.778 \)

• For \( (13, 78) \):

\( (13 - 10.333)(78 - 71.583)\approx(2.667)(6.417)\approx17.129 \)
\( (13 - 10.333)^2\approx(2.667)^2\approx7.111 \)

• For \( (8, 66) \):

\( (8 - 10.333)(66 - 71.583)\approx(-2.333)(-5.583)\approx12.926 \)
\( (8 - 10.333)^2\approx(-2.333)^2\approx5.443 \)

Sum of \( (x_i - \bar{x})(y_i - \bar{y}) \):
\( 8.403 + 1.318 + 1.279 + 5.694 + 17.129 + 12.926\approx46.75 \)

Sum of \( (x_i - \bar{x})^2 \):
\( 3.360 + 0.694 + 0.445 + 2.778 + 7.111 + 5.443\approx19.831 \)

Slope:

$$ m=\frac{46.75}{19.831}\approx2.357 $$

Step4: Calculate Intercept (\( b \)) of Regression Line

Intercept formula: \( b = \bar{y}-m\bar{x} \)

$$ b = 71.583 - 2.357(10.333)\approx71.583 - 24.357\approx47.226 $$

Step5: Regression Equation

The linear regression equation is \( \hat{y}=mx + b \), so:

$$ \hat{y}=2.357x + 47.226 $$

Step6: Calculate Correlation Coefficient (\( r \))

Correlation formula: \( r=\frac{\sum (x_i - \bar{x})(y_i - \bar{y})}{\sqrt{\sum (x_i - \bar{x})^2\sum (y_i - \bar{y})^2}} \)

First, compute \( \sum (y_i - \bar{y})^2 \):

• For \( (8.5, 67) \): \( (67 - 71.583)^2\approx20.904 \)
• For \( (9.5, 70) \): \( (70 - 71.583)^2\approx2.506 \)
• For \( (11, 73.5) \): \( (73.5 - 71.583)^2\approx3.675 \)
• For \( (12, 75) \): \( (75 - 71.583)^2\approx11.660 \)
• For \( (13, 78) \): \( (78 - 71.583)^2\approx41.170 \)
• For \( (8, 66) \): \( (66 - 71.583)^2\approx31.170 \)

Sum: \( 20.904 + 2.506 + 3.675 + 11.660 + 41.170 + 31.170\approx111.085 \)

Now, \( r \):

$$ r=\frac{46.75}{\sqrt{19.831\times111.085}}\approx\frac{46.75}{\sqrt{2193.0}}\approx\frac{46.75}{46.83}\approx0.998 $$

Step7: Predict Height for Shoe Size 10.5

Substitute \( x = 10.5 \) into the regression equation:

$$ \hat{y}=2.357(10.5) + 47.226\approx24.7485 + 47.226\approx71.9745\approx72 $$

Answer:

• Equation: \(\boldsymbol{\hat{y} \approx 2.36x + 47.23}\) (or more precise \( \hat{y}=2.357x + 47.226 \))
• \( r \approx \boldsymbol{0.998} \) (strong positive correlation)
• Height expected: \(\boldsymbol{\approx 72}\) (or more precisely ~72.0)
• Correlation: Strong positive correlation (since \( r \approx 0.998 \), close to 1)