Question

Refer to Exercise $10.86$and find a $99\%$ confidence interval for the difference between the mean wing lengths of the two subspecies.

Short answer

The interval is $-4.488$to $-1.111$.

Step-by-step solution

Given information  

Given in the question that, We need to find a $99\%$confidence interval for the difference between the mean wing lengths of the two subspecies.

Explanation 

Consider the table

Migratory population1	Non migratory population2
${\overline{x}}_1=82.1$	${\overline{x}}_2=84.9$
$s_1=1.501$	$s_2=1.698$
$n_1=14$	$n_2=14$

For $99\%$ confidence interval,

$\mathrm{df}=\frac{{\left[\left(\frac{s_1^2}{n_1}\right)+\left(\frac{s_2^2}{n_2}\right)\right]}^2}{\frac{{\left(\frac{s_1^2}{n_1}\right)}^2}{n_1-1}+\frac{{\left(\frac{s_2^2}{n_2}\right)}^2}{n_2-1}}$

$=\frac{\left[\left(\frac{1.{501}^2}{14}\right)+{\left(\frac{1.{698}^2}{14}\right)}^2\right.}{{\left.\frac{{\left(\frac{1.{501}^2}{14}\right)}^2}{14-1}\right)}^2}$

$\approx 25$

The end point of interval are,

$\left({\overline{x}}_1-{\overline{x}}_2\right)\pm t_{\frac{a}{2}}·\sqrt{\left(\frac{s_1^2}{n_1}\right)+\left(\frac{s_2^2}{n_2}\right)}=(82.1-84.9)\pm 2.787$

$\sqrt{\left(\frac{1.{501}^2}{14}+\frac{1.{698}^2}{14}\right)}$

$=-4.488$ to$-1.111$