Question

Breathing rates, in breaths per minute, were measured for a group of 6 subjects at rest, then during moderate exercise. The results were as follows: Subject Rest Exercise 16 37 21 39 17 37 18 40 1539 1934 Find a 95% confidence interval for the increase in breathing rate due to exercise le. Exercise - Rest). (18.33, 2145) (7.83,31.95) (15.78, 19.551 (16.68, 23.32) (17.52.22261

Answer 1

We Can Use Confidence interval of Paired t test,:

Computational Table:

441 324 Rest Exercise D=Exercise-Rest D^2 16.0 37 21.0 17.0 18.0 15.0 19.0 Total 400 484 576 225 2450

95% Confidence interval:

$\bar{d}-t_{\alpha /2, df} * \frac{S_{d}}{\sqrt{n}} \leq \mu _{d} \leq \bar{d}+t_{\alpha /2, df} * \frac{S_{d}}{\sqrt{n}}$

Calculation:

$\bar{d} = \frac{\sum Di}{n} = \frac{120}{6} = 20$

$S_{d} = \sqrt{\frac{\sum d^2 - \frac{(\sum d)^2}{n}}{n-1}}= \sqrt{\frac{2450 - \frac{(120)^2}{6}}{6-1}} = 3.16$

Degrees of Freedom = n-1 = 6-1 = 5

Critical value:

$t_{\alpha ,df} = t_{0.05 ,6-1} = \pm 2.5706$           .................using t table

95% Confidence interval:

$\bar{d}-t_{\alpha /2, df} * \frac{S_{d}}{\sqrt{n}} \leq \mu _{d} \leq \bar{d}+t_{\alpha /2, df} * \frac{S_{d}}{\sqrt{n}}$

20 - 2.5706 * 3.16 3.16 <Hd <20 +2.5706*

$16.68\leq \mu _{d} \leq 23.32$

ANSWER: D

(16.68, 23.32)