Answer:
Confidence interval for the intercept:
[tex]37.67-2.1123(5.69) \leq \beta_0 \leq 37.67+2.1123(5.69)\\25.651 \leq \beta_0 \leq 49.6889[/tex]
Confidence interval for the slope:
[tex]33.18-2.1123(7.94) \leq \beta_1 \leq 33.18+2.1123(7.94)\\16.4083 \leq \beta_1 \leq 49.9517[/tex]
Step-by-step explanation:
We start defining our equation's terms, starting from the linear regression model [tex]\hat{y} =\hat{\beta}_{0} + \hat{\beta}_{1}x[/tex]
In this model [tex]{\beta}_{0} [/tex] is the intercept estimator and [tex]{\beta}_{1} [/tex] is the slope estimator.
in the equation y = 37.67 + 33.18x]
[tex]{\beta}_{0} = 37.67 [/tex] and [tex]{\beta}_{1} = 33.81[/tex]
Then we have the standard errors (se) for each estimator:
[tex]se(\hat{\beta_{0}}}) = Â 5.69[/tex] and [tex]se(\hat{\beta_{1}}}) = Â 7.94[/tex]
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 [tex]{\beta}_{0} [/tex] we have:
[tex]\hat{\beta_0} - t_{\alpha/2,n-2} se(\beta_0) \leq  \beta _0 \leq \hat{\beta_0} + t_{\alpha/2,n-2}[/tex]
Where  [tex]t_{\alpha/2,n-2}[/tex] 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:
[tex]37.67-2.1123(5.69) \leq \beta_0 \leq 37.67+2.1123(5.69)\\25.651 \leq \beta_0 \leq 49.6889[/tex]
Similarly, the 96% confidence interval for the slope  [tex]{\beta}_{1} [/tex] is:
[tex]\hat{\beta_1} - t_{\alpha/2,n-2} se(\beta_1) \leq  \beta _1 \leq \hat{\beta_1} + t_{\alpha/2,n-2}[/tex]
Where [tex]t_{\alpha/2,n-2} =± 2.1123[/tex]
And into the formula:
[tex]33.18-2.1123(7.94) \leq \beta_1 \leq 33.18+2.1123(7.94)\\16.4083 \leq \beta_1 \leq 49.9517[/tex]
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.
