Question

Biologist Theodore Garland, Jr. studied the relationship between running speeds and morphology of 49 species of cursorial mammals (mammals adapted to or specialized for running). One of the relationships he investigated was maximal sprint speed in kilometers per hour and the ratio of metatarsal-to-femur length. A least-squares regression on the data he collected produces the equation ^ y = 37.67 + 33.18 x where x is metatarsal-to-femur ratio and ^ y is predicted maximal sprint speed in kilometers per hour. The standard error of the intercept is 5.69 and the standard error of the slope is 7.94. Construct a 96% confidence interval for the slope of the population regression line. Give your answers precise to at least two decimal places.

Answer 1

Answer:

Confidence interval for the intercept:

$37.67-2.1123(5.69) \leq \beta_0 \leq 37.67+2.1123(5.69)\\25.651 \leq \beta_0 \leq 49.6889$

Confidence interval for the slope:

$33.18-2.1123(7.94) \leq \beta_1 \leq 33.18+2.1123(7.94)\\16.4083 \leq \beta_1 \leq 49.9517$

Step-by-step explanation:

We start defining our equation's terms, starting from the linear regression model $\hat{y} =\hat{\beta}_{0} + \hat{\beta}_{1}x$

In this model ${\beta}_{0}$ is the intercept estimator and ${\beta}_{1}$ is the slope estimator.

in the equation y = 37.67 + 33.18x]

${\beta}_{0} = 37.67$ and ${\beta}_{1} = 33.81$

Then we have the standard errors (se) for each estimator:

$se(\hat{\beta_{0}}}) = 5.69$ and $se(\hat{\beta_{1}}}) = 7.94$

The sample number is 49 species (here we assume that all the individuals of a the same species are summarised with a central tendency measure, i.e. mean, median or mode, if each species contained more than one individual).

The formula for the 96% confidence interval for the intercept ${\beta}_{0}$ we have:

$\hat{\beta_0} - t_{\alpha/2,n-2} se(\beta_0) \leq \beta _0 \leq \hat{\beta_0} + t_{\alpha/2,n-2}$

Where  $t_{\alpha/2,n-2}$ represents the p-value in a t distribution when α=0.04 (so that we have a confidence interval of 96%, or 0.96), two-tailed, and n-2 degrees of freedom. In this example, n-2 = 47, and the t-value (47 degrees of freedom, 0.04 significance level, two tails) is ± 2.1123.

We input these values into our formula:

$37.67-2.1123(5.69) \leq \beta_0 \leq 37.67+2.1123(5.69)\\25.651 \leq \beta_0 \leq 49.6889$

Similarly, the 96% confidence interval for the slope  ${\beta}_{1}$ is:

$\hat{\beta_1} - t_{\alpha/2,n-2} se(\beta_1) \leq \beta _1 \leq \hat{\beta_1} + t_{\alpha/2,n-2}$

Where $t_{\alpha/2,n-2} =± 2.1123$

And into the formula:

$33.18-2.1123(7.94) \leq \beta_1 \leq 33.18+2.1123(7.94)\\16.4083 \leq \beta_1 \leq 49.9517$

The confidence interval does not include 0, so there is enough evidence saying that there is enough correlation between the metatarsal-to-femur ratio and maximal sprint speed in kilometers per hour. This study shows that measuring the lengths of metatarsal 3 and femur in mammals is a reliable predictor of maximal speed for cursorial mammals.