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 α=0.01

Answer 1

It does not appear that police can use a shoe print length to estimate the height of a​ male.

The given parameters are:

$\begin{array}{cccccc}{Shoe\ Print} & {28.6} & {29.4} & {32.2} & {32.4} & {27.3} \ \\ Height (cm) & {172.5} & {176.7} & {188.4} & {170.1} & {179.2} \ \end{array}$

Rewrite as:

$\begin{array}{cccccc}{x} & {28.6} & {29.4} & {32.2} & {32.4} & {27.3} \ \\ y & {172.5} & {176.7} & {188.4} & {170.1} & {179.2} \ \end{array}$

See attachment for scatter plot

To determine the correlation coefficient, we extend the table as follows:

$\begin{array}{cccccc}{x} & {28.6} & {29.4} & {32.2} & {32.4} & {27.3} & y & {172.5} & {176.7} & {188.4} & {170.1} & {179.2} & x^2 & {817.96} & {864.36} & {1036.84} & {1049.76} & {745.29} & y^2 & {29756.25} & {31222.89} & {35494.56} & {28934.01} & {32112.64} & x \times y & {4933.5} & {5194.98} & {6066.48} & {5511.24} & {4892.16} \ \end{array}$

The correlation coefficient (r) is:

$r = \frac{\sum(x - \bar x)(y - \bar y)}{\sqrt{SS_x * SS_y}}$

We have:

$n =5$

$\sum xy =4933.5+5194.98+6066.48+5511.24+4892.16 =26598.36$

$\sum x =28.6+29.4+32.2+32.4+27.3=149.9$

$\sum y =172.5+176.7+188.4+170.1+179.2=886.9$

$\sum x^2 =817.96+864.36+1036.84+1049.76+745.29=4514.21$

$\sum y^2 =29756.25+31222.89+35494.56+28934.01+32112.64=157520.35$

Calculate mean of x and y

$\bar x = \frac{\sum x}{n} = \frac{149.9}{5} = 29.98$

$\bar y = \frac{\sum y}{n} = \frac{886.9}{5} = 177.38$

Calculate SSx and SSy

$SS_x = \sum (x - \bar x)^2 =(28.6-29.98)^2 + (29.4-29.98)^2 + (32.2-29.98)^2 + (32.4-29.98)^2 + (27.3-29.98)^2 =20.208$

$SS_y = \sum (y - \bar x)^2 =(172.5-177.38)^2 + (176.7-177.38)^2 + (188.4-177.38)^2 + (170.1-177.38)^2 + (179.2-177.38)^2 =202.028$

Calculate $\sum(x - \bar x)(y - \bar y)$

$\sum(x - \bar x)(y - \bar y) = (28.6-29.98)*(172.5-177.38) + (29.4-29.98)*(176.7-177.38) + (32.2-29.98)*(188.4-177.38) + (32.4-29.98)*(170.1-177.38) + (27.3-29.98) *(179.2-177.38) =9.098$

So:

$r = \frac{\sum(x - \bar x)(y - \bar y)}{\sqrt{SS_x * SS_y}}$

$r = \frac{9.098}{\sqrt{20.208 * 202.028}}$

$r = \frac{9.098}{\sqrt{4082.581824}}$

$r = \frac{9.098}{63.90}$

$r = 0.142$

Calculate test statistic:

$t = \frac{r}{\sqrt{\frac{1 - r^2}{n-2}}}$

$t = \frac{0.142}{\sqrt{\frac{1 - 0.142^2}{5-2}}}$

$t = \frac{0.142}{\sqrt{\frac{0.979836}{3}}}$

$t = \frac{0.142}{\sqrt{0.326612}}$

$t = \frac{0.142}{0.5715}$

$t = 0.248$

Calculate the degrees of freedom

$df = n - 2 = 5 - 2 = 3$

The $t_{\alpha/2}$ value at:

$df =3$

$t = 0.248$

$\alpha = 0.01$

The value is:

$t_{0.01/2} = \±5.841$

This means that we reject the null hypothesis if the t value is not between -5.841 and 5.841

We calculate the t value as:

$t = 0.248$

$-5.841 < 0.248 < 5.841$

Hence, we do not reject the null hypothesis because they do not appear to have any correlation.

Read more about regression at:

https://brainly.com/question/18405415