Question

9.3.2 Listed below are body temperatures from five different subjects measured at 8 AM and again at 12 AM. Find the values of d overbar and s Subscript d. In​ general, what does mu Subscript d ​represent? Temperature (degrees Upper F )at 8 AM 98.1 98.8 97.3 97.5 97.9 Temperature (degrees Upper F )at 12 AM 98.7 99.4 97.7 97.1 98.0 Let the temperature at 8 AM be the first​ sample, and the temperature at 12 AM be the second sample. Find the values of d overbar and s Subscript d.

Answer 1

Answer:

$\frac{}{d}$ = −0.26

$s_{d}$ = 0.4219

Step-by-step explanation:

Given:

Sample1:  98.1  98.8  97.3  97.5  97.9

Sample2: 98.7  99.4  97.7  97.1  98.0

Sample 1           Sample 2              Difference d

98.1                        98.7                       -0.6  

98.8                       99.4                       -0.6

97.3                        97.7                       -0.4

97.5                        97.1                         0.4

97.9                        98.0                       -0.1

To find:

Find the values of $\frac{}{d}$ and $s_{d}$

d overbar ( $\frac{}{d}$)  is the sample mean of the differences which is calculated by dividing the sum of all the values of difference d with the number of values i.e. n = 5

$\frac{}{d}$ = ∑d/n

 = (−0.6 −0.6 −0.4 +0.4 −0.1) / 5

 = −1.3 / 5

$\frac{}{d}$ = −0.26

s Subscript d is the sample standard deviation of the difference which is calculated as following:

$s_{d}$ = √∑($d_{i}$ - $\frac{}{d}$)²/ n-1

$s_{d}$ =

√ $(-0.6 - (-0.26))^{2} + (-0.6 - (-0.26))^{2} + (-0.4 - (-0.26))^{2} + (0.4-(-0.26))^{2} + (-0.1 - (-0.26))^{2} / 5-1$

    =  √ (−0.6 − (−0.26 ))² + (−0.6 − (−0.26))² + (−0.4 − (−0.26))² + (0.4 −  

                                                                     (−0.26))² + (−0.1 − (−0.26))² / 5−1

=  $\sqrt{\frac{0.1156 + 0.1156 + 0.0196 + 0.4356 + 0.0256}{4} }$

= $\sqrt{\frac{0.712}{4} }$

= $\sqrt{0.178}$

= 0.4219

$s_{d}$ = 0.4219

Subscript d ​represent

μ$_{d}$ represents the mean of differences in body temperatures measured at 8 AM and at 12 AM of population.