Question

Wing Length. Refer to Exercise 10.86 and find a 99 con fidence interval for the difference between the mean wing lengths c the two subspecies.

Short answer

-4.488 is the interval in text. from text to -1.111 text

Step-by-step solution

Given Information

Consider the table,

Explanation

For 99% confidence interval,

$\mathbf{df}=\frac{{\left[\left(\frac{s_1^2}{n_1}\right)+\left(\frac{s_2^2}{n_2}\right)\right]}^2}{\frac{{\left(\frac{s_1^2}{n_1}\right)}^2}{{\left(\frac{x_2^2}{n_2}\right)}^2}}$

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

= 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