College

Body Temperature Problem

Listed below are body temperatures from five different subjects measured at 8 AM and again at 12 PM.

[tex]
\[
\begin{array}{|c|c|c|c|c|c|}
\hline
\text{Temp at 8 AM} & 97.9 & 98.4 & 97.3 & 97.6 & 97.3 \\
\hline
\text{Temp at 12 PM} & 98.5 & 99.3 & 97.6 & 97.6 & 97.5 \\
\hline
\end{array}
\]
[/tex]

1. Find the value of [tex]$\bar{d}$[/tex] (Use 1-var-stats with L3). Do not round the answer.

[tex]$\bar{d} =$[/tex] [tex]$\square$[/tex]

2. Find the value of [tex]$s_d$[/tex] (Use 1-var-stats with L3). Round the answer to two decimal places as needed.

[tex]$s_d = $[/tex] [tex]$\square$[/tex]

Answer :

To solve this problem, we need to find the mean and standard deviation of the differences between body temperatures measured at 8 AM and 12 AM for five different subjects. Here's a step-by-step explanation:

1. List the Temperatures:
- Temperatures at 8 AM: 97.9, 98.4, 97.3, 97.6, 97.3
- Temperatures at 12 AM: 98.5, 99.3, 97.6, 97.6, 97.5

2. Calculate the Differences:
For each subject, subtract the temperature at 8 AM from the temperature at 12 AM to find the differences.
- 98.5 - 97.9 = 0.6
- 99.3 - 98.4 = 0.9
- 97.6 - 97.3 = 0.3
- 97.6 - 97.6 = 0.0
- 97.5 - 97.3 = 0.2

The differences are: 0.6, 0.9, 0.3, 0.0, 0.2

3. Calculate the Mean of the Differences ([tex]\(\bar{d}\)[/tex]):
To find the mean difference, sum all the differences and divide by the number of differences (5 in this case).

[tex]\[
\bar{d} = \frac{0.6 + 0.9 + 0.3 + 0.0 + 0.2}{5} = \frac{2.0}{5} = 0.4
\][/tex]

So, [tex]\(\bar{d} = 0.4\)[/tex].

4. Calculate the Standard Deviation of the Differences ([tex]\(s_d\)[/tex]):
The standard deviation measures how spread out the differences are from the mean.

Here, use the formula for the sample standard deviation:

[tex]\[
s_d = \sqrt{\frac{\sum (x_i - \bar{d})^2}{n - 1}}
\][/tex]

Substitute the calculated differences and the mean difference:

[tex]\[
s_d = \sqrt{\frac{(0.6-0.4)^2 + (0.9-0.4)^2 + (0.3-0.4)^2 + (0.0-0.4)^2 + (0.2-0.4)^2}{4}}
\][/tex]

[tex]\[
s_d = \sqrt{\frac{0.04 + 0.25 + 0.01 + 0.16 + 0.04}{4}} = \sqrt{\frac{0.5}{4}}
\][/tex]

[tex]\[
s_d = \sqrt{0.125} \approx 0.35
\][/tex]

So, [tex]\(s_d \approx 0.35\)[/tex] rounded to two decimal places.

Therefore, the mean of the differences ([tex]\(\bar{d}\)[/tex]) is 0.4, and the standard deviation of the differences ([tex]\(s_d\)[/tex]) is 0.35.