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Answer :
To solve the problem of finding [tex]\(\overline{d}\)[/tex] and [tex]\(s_d\)[/tex] for the listed body temperatures, follow these steps:
1. Identify the Paired Differences:
- We have temperatures for five subjects at 8 AM and 12 AM.
- Calculate the difference for each subject by subtracting the 8 AM temperature from the 12 AM temperature.
Differences:
- Subject 1: [tex]\(98.7 - 98.2 = 0.5\)[/tex]
- Subject 2: [tex]\(99.7 - 99.1 = 0.6\)[/tex]
- Subject 3: [tex]\(97.5 - 97.3 = 0.2\)[/tex]
- Subject 4: [tex]\(97.3 - 97.6 = -0.3\)[/tex]
- Subject 5: [tex]\(97.7 - 97.4 = 0.3\)[/tex]
2. Calculate the Mean of the Differences ([tex]\(\overline{d}\)[/tex]):
- Add up all the differences and divide by the number of subjects.
[tex]\[
\overline{d} = \frac{0.5 + 0.6 + 0.2 - 0.3 + 0.3}{5} = \frac{1.3}{5} = 0.26
\][/tex]
3. Calculate the Standard Deviation of the Differences ([tex]\(s_d\)[/tex]):
- First, find the squared differences from the mean for each subject.
- Then, find the average of these squared differences.
- Lastly, take the square root of the result to find [tex]\(s_d\)[/tex].
Squared Differences:
- Subject 1: [tex]\((0.5 - 0.26)^2 = 0.0576\)[/tex]
- Subject 2: [tex]\((0.6 - 0.26)^2 = 0.1156\)[/tex]
- Subject 3: [tex]\((0.2 - 0.26)^2 = 0.0036\)[/tex]
- Subject 4: [tex]\((-0.3 - 0.26)^2 = 0.3136\)[/tex]
- Subject 5: [tex]\((0.3 - 0.26)^2 = 0.0016\)[/tex]
Mean of Squared Differences:
[tex]\[
\text{Mean} = \frac{0.0576 + 0.1156 + 0.0036 + 0.3136 + 0.0016}{4} = \frac{0.492}{4} = 0.123
\][/tex]
[tex]\(s_d\)[/tex]:
[tex]\[
s_d = \sqrt{0.123} \approx 0.35
\][/tex]
Thus, the mean of the differences is [tex]\(\overline{d} = 0.26\)[/tex] and the standard deviation is [tex]\(s_d \approx 0.35\)[/tex].
4. Interpretation of [tex]\(\mu_{d}\)[/tex]:
- [tex]\(\mu_{d}\)[/tex] represents the mean population difference between the paired temperatures. It tells us about the average change or difference in temperatures at 8 AM and 12 AM across the entire population from which the samples were drawn.
1. Identify the Paired Differences:
- We have temperatures for five subjects at 8 AM and 12 AM.
- Calculate the difference for each subject by subtracting the 8 AM temperature from the 12 AM temperature.
Differences:
- Subject 1: [tex]\(98.7 - 98.2 = 0.5\)[/tex]
- Subject 2: [tex]\(99.7 - 99.1 = 0.6\)[/tex]
- Subject 3: [tex]\(97.5 - 97.3 = 0.2\)[/tex]
- Subject 4: [tex]\(97.3 - 97.6 = -0.3\)[/tex]
- Subject 5: [tex]\(97.7 - 97.4 = 0.3\)[/tex]
2. Calculate the Mean of the Differences ([tex]\(\overline{d}\)[/tex]):
- Add up all the differences and divide by the number of subjects.
[tex]\[
\overline{d} = \frac{0.5 + 0.6 + 0.2 - 0.3 + 0.3}{5} = \frac{1.3}{5} = 0.26
\][/tex]
3. Calculate the Standard Deviation of the Differences ([tex]\(s_d\)[/tex]):
- First, find the squared differences from the mean for each subject.
- Then, find the average of these squared differences.
- Lastly, take the square root of the result to find [tex]\(s_d\)[/tex].
Squared Differences:
- Subject 1: [tex]\((0.5 - 0.26)^2 = 0.0576\)[/tex]
- Subject 2: [tex]\((0.6 - 0.26)^2 = 0.1156\)[/tex]
- Subject 3: [tex]\((0.2 - 0.26)^2 = 0.0036\)[/tex]
- Subject 4: [tex]\((-0.3 - 0.26)^2 = 0.3136\)[/tex]
- Subject 5: [tex]\((0.3 - 0.26)^2 = 0.0016\)[/tex]
Mean of Squared Differences:
[tex]\[
\text{Mean} = \frac{0.0576 + 0.1156 + 0.0036 + 0.3136 + 0.0016}{4} = \frac{0.492}{4} = 0.123
\][/tex]
[tex]\(s_d\)[/tex]:
[tex]\[
s_d = \sqrt{0.123} \approx 0.35
\][/tex]
Thus, the mean of the differences is [tex]\(\overline{d} = 0.26\)[/tex] and the standard deviation is [tex]\(s_d \approx 0.35\)[/tex].
4. Interpretation of [tex]\(\mu_{d}\)[/tex]:
- [tex]\(\mu_{d}\)[/tex] represents the mean population difference between the paired temperatures. It tells us about the average change or difference in temperatures at 8 AM and 12 AM across the entire population from which the samples were drawn.
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