Chapter 10: Q.10.92 (page 429)

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

Expert verified

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
${\bar{x}}_{1} = 82.1$	${\bar{x}}_{2} = 84.9$
$s_{1} = 1.501$	$s_{2} = 1.698$
$n_{1} = 14$	$n_{2} = 14$

For $99 %$ confidence interval,

$df = \frac{{[(\frac{s_{1}^{2}}{n_{1}}) + (\frac{s_{2}^{2}}{n_{2}})]}^{2}}{\frac{{(\frac{s_{1}^{2}}{n_{1}})}^{2}}{n_{1} - 1} + \frac{{(\frac{s_{2}^{2}}{n_{2}})}^{2}}{n_{2} - 1}}$

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

$\approx 25$

The end point of interval are,

$({\bar{x}}_{1} - {\bar{x}}_{2}) \pm t_{\frac{a}{2}} \cdot \sqrt{(\frac{s_{1}^{2}}{n_{1}}) + (\frac{s_{2}^{2}}{n_{2}})} = (82.1 - 84.9) \pm 2.787$

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

$= - 4.488$ to $- 1.111$

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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

Step by step solution

Given information

Explanation

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

Short Answer

Step by step solution

Given information

Explanation

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Geometry

Probability and Statistics

Discrete Mathematics

Applied Mathematics

Statistics

Mechanics Maths

Study anywhere. Anytime. Across all devices.

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