Question

police sometimes measure shoe prints at crime scenes so that they can learn something about criminals. listed below are shoe print lengths, foot lengths, and heights of males. construct a scatterplot, find the value of the linear correlation coefficient ( r ), and find the ( p )-value of ( r ). determine whether there is sufficient evidence to support a claim of linear correlation between the two variables. based on these results, does it appear that police can use a shoe print length to estimate the height of a male? use a significance level of ( alpha = 0.01 ).

shoe print (cm)	foot length (cm)	height (cm)
30.4	25.4	181.5
31.1	27.5	182.3
31.5	26.3	181.7
27.2	26.1	168.6

construct a scatterplot. choose the correct graph below.

a. scatterplot a

b. scatterplot b

c. scatterplot c

d. scatterplot d

the linear correlation coefficient is ( r = square ).
(round to three decimal places as needed.)

Explanation:

Step1: Identify Variables

Let \( x \) be Foot Length (cm) and \( y \) be Height (cm). The data points are: \((25.2, 172.6)\), \((25.4, 181.5)\), \((27.5, 182.3)\), \((26.3, 181.7)\), \((26.1, 168.6)\).

Step2: Calculate Means

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

$$ \bar{x} = \frac{25.2 + 25.4 + 27.5 + 26.3 + 26.1}{5} = \frac{130.5}{5} = 26.1 $$

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

$$ \bar{y} = \frac{172.6 + 181.5 + 182.3 + 181.7 + 168.6}{5} = \frac{886.7}{5} = 177.34 $$

Step3: Calculate Deviations

For each data point, compute \( (x_i - \bar{x}) \), \( (y_i - \bar{y}) \), \( (x_i - \bar{x})(y_i - \bar{y}) \), and \( (x_i - \bar{x})^2 \), \( (y_i - \bar{y})^2 \).

• For \((25.2, 172.6)\):

\( x - \bar{x} = -0.9 \), \( y - \bar{y} = -4.74 \),
\( (x - \bar{x})(y - \bar{y}) = 4.266 \),
\( (x - \bar{x})^2 = 0.81 \), \( (y - \bar{y})^2 = 22.4676 \)

• For \((25.4, 181.5)\):

\( x - \bar{x} = -0.7 \), \( y - \bar{y} = 4.16 \),
\( (x - \bar{x})(y - \bar{y}) = -2.912 \),
\( (x - \bar{x})^2 = 0.49 \), \( (y - \bar{y})^2 = 17.3056 \)

• For \((27.5, 182.3)\):

\( x - \bar{x} = 1.4 \), \( y - \bar{y} = 4.96 \),
\( (x - \bar{x})(y - \bar{y}) = 6.944 \),
\( (x - \bar{x})^2 = 1.96 \), \( (y - \bar{y})^2 = 24.6016 \)

• For \((26.3, 181.7)\):

\( x - \bar{x} = 0.2 \), \( y - \bar{y} = 4.36 \),
\( (x - \bar{x})(y - \bar{y}) = 0.872 \),
\( (x - \bar{x})^2 = 0.04 \), \( (y - \bar{y})^2 = 19.0096 \)

• For \((26.1, 168.6)\):

\( x - \bar{x} = 0 \), \( y - \bar{y} = -8.74 \),
\( (x - \bar{x})(y - \bar{y}) = 0 \),
\( (x - \bar{x})^2 = 0 \), \( (y - \bar{y})^2 = 76.3876 \)

Step4: Sum Deviations

Sum of \( (x_i - \bar{x})(y_i - \bar{y}) \) (\( SS_{xy} \)):

$$ 4.266 - 2.912 + 6.944 + 0.872 + 0 = 9.17 $$

Sum of \( (x_i - \bar{x})^2 \) (\( SS_{xx} \)):

$$ 0.81 + 0.49 + 1.96 + 0.04 + 0 = 3.3 $$

Sum of \( (y_i - \bar{y})^2 \) (\( SS_{yy} \)):

$$ 22.4676 + 17.3056 + 24.6016 + 19.0096 + 76.3876 = 159.772 $$

Step5: Calculate Correlation Coefficient

Using the formula \( r = \frac{SS_{xy}}{\sqrt{SS_{xx} \cdot SS_{yy}}} \):

$$ r = \frac{9.17}{\sqrt{3.3 \times 159.772}} = \frac{9.17}{\sqrt{527.2476}} \approx \frac{9.17}{22.962} \approx 0.400 $$

Answer:

\( 0.400 \) (rounded to three decimal places)