Thank you for visiting Listed below are body temperatures from five different subjects measured at 8 AM and again at 12 AM Find the values of tex bar d. This page is designed to guide you through key points and clear explanations related to the topic at hand. We aim to make your learning experience smooth, insightful, and informative. Dive in and discover the answers you're looking for!
Answer :
To solve the problem, we need to calculate two statistical measures based on the temperature differences between two times: 8 AM and 12 AM. These measures are the mean difference, denoted as [tex]\(\bar{d}\)[/tex], and the standard deviation of the differences, denoted as [tex]\(s_d\)[/tex].
Here are the steps to find [tex]\(\bar{d}\)[/tex] and [tex]\(s_d\)[/tex]:
1. Calculate the Differences:
We start by finding the difference between each corresponding pair of temperatures.
- Difference for Subject 1: [tex]\(99.1 - 98.3 = 0.8\)[/tex]
- Difference for Subject 2: [tex]\(99.1 - 98.6 = 0.5\)[/tex]
- Difference for Subject 3: [tex]\(98.0 - 97.8 = 0.2\)[/tex]
- Difference for Subject 4: [tex]\(96.8 - 97.3 = -0.5\)[/tex]
- Difference for Subject 5: [tex]\(97.9 - 97.6 = 0.3\)[/tex]
The differences are: [tex]\(0.8, 0.5, 0.2, -0.5, 0.3\)[/tex].
2. Calculate the Mean of the Differences ([tex]\(\bar{d}\)[/tex]):
To find the mean difference, sum up all the differences and divide by the number of differences.
[tex]\[
\bar{d} = \frac{0.8 + 0.5 + 0.2 - 0.5 + 0.3}{5} = \frac{1.3}{5} = 0.26
\][/tex]
3. Calculate the Standard Deviation of the Differences ([tex]\(s_d\)[/tex]):
First, compute the deviations by subtracting the mean difference from each original difference:
- [tex]\(0.8 - 0.26 = 0.54\)[/tex]
- [tex]\(0.5 - 0.26 = 0.24\)[/tex]
- [tex]\(0.2 - 0.26 = -0.06\)[/tex]
- [tex]\(-0.5 - 0.26 = -0.76\)[/tex]
- [tex]\(0.3 - 0.26 = 0.04\)[/tex]
Next, square each deviation:
- [tex]\(0.54^2 = 0.2916\)[/tex]
- [tex]\(0.24^2 = 0.0576\)[/tex]
- [tex]\((-0.06)^2 = 0.0036\)[/tex]
- [tex]\((-0.76)^2 = 0.5776\)[/tex]
- [tex]\(0.04^2 = 0.0016\)[/tex]
Add the squared deviations:
[tex]\[
0.2916 + 0.0576 + 0.0036 + 0.5776 + 0.0016 = 0.932
\][/tex]
Finally, divide by [tex]\(n-1\)[/tex] (where [tex]\(n\)[/tex] is the number of subjects, which is 5):
[tex]\[
s_d = \sqrt{\frac{0.932}{4}} = \sqrt{0.233} \approx 0.483
\][/tex]
Therefore, the mean difference [tex]\(\bar{d}\)[/tex] is approximately 0.26, and the standard deviation of the differences [tex]\(s_d\)[/tex] is approximately 0.483.
Finally, [tex]\(\mu_d\)[/tex] represents the hypothetical population mean of the differences in body temperature if we could measure the differences for every subject in the population at these times.
Here are the steps to find [tex]\(\bar{d}\)[/tex] and [tex]\(s_d\)[/tex]:
1. Calculate the Differences:
We start by finding the difference between each corresponding pair of temperatures.
- Difference for Subject 1: [tex]\(99.1 - 98.3 = 0.8\)[/tex]
- Difference for Subject 2: [tex]\(99.1 - 98.6 = 0.5\)[/tex]
- Difference for Subject 3: [tex]\(98.0 - 97.8 = 0.2\)[/tex]
- Difference for Subject 4: [tex]\(96.8 - 97.3 = -0.5\)[/tex]
- Difference for Subject 5: [tex]\(97.9 - 97.6 = 0.3\)[/tex]
The differences are: [tex]\(0.8, 0.5, 0.2, -0.5, 0.3\)[/tex].
2. Calculate the Mean of the Differences ([tex]\(\bar{d}\)[/tex]):
To find the mean difference, sum up all the differences and divide by the number of differences.
[tex]\[
\bar{d} = \frac{0.8 + 0.5 + 0.2 - 0.5 + 0.3}{5} = \frac{1.3}{5} = 0.26
\][/tex]
3. Calculate the Standard Deviation of the Differences ([tex]\(s_d\)[/tex]):
First, compute the deviations by subtracting the mean difference from each original difference:
- [tex]\(0.8 - 0.26 = 0.54\)[/tex]
- [tex]\(0.5 - 0.26 = 0.24\)[/tex]
- [tex]\(0.2 - 0.26 = -0.06\)[/tex]
- [tex]\(-0.5 - 0.26 = -0.76\)[/tex]
- [tex]\(0.3 - 0.26 = 0.04\)[/tex]
Next, square each deviation:
- [tex]\(0.54^2 = 0.2916\)[/tex]
- [tex]\(0.24^2 = 0.0576\)[/tex]
- [tex]\((-0.06)^2 = 0.0036\)[/tex]
- [tex]\((-0.76)^2 = 0.5776\)[/tex]
- [tex]\(0.04^2 = 0.0016\)[/tex]
Add the squared deviations:
[tex]\[
0.2916 + 0.0576 + 0.0036 + 0.5776 + 0.0016 = 0.932
\][/tex]
Finally, divide by [tex]\(n-1\)[/tex] (where [tex]\(n\)[/tex] is the number of subjects, which is 5):
[tex]\[
s_d = \sqrt{\frac{0.932}{4}} = \sqrt{0.233} \approx 0.483
\][/tex]
Therefore, the mean difference [tex]\(\bar{d}\)[/tex] is approximately 0.26, and the standard deviation of the differences [tex]\(s_d\)[/tex] is approximately 0.483.
Finally, [tex]\(\mu_d\)[/tex] represents the hypothetical population mean of the differences in body temperature if we could measure the differences for every subject in the population at these times.
Thank you for reading the article Listed below are body temperatures from five different subjects measured at 8 AM and again at 12 AM Find the values of tex bar d. We hope the information provided is useful and helps you understand this topic better. Feel free to explore more helpful content on our website!
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