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 overline 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 this problem, we need to find the average difference in body temperatures at 8 AM and 12 AM, as well as the standard deviation of those differences. Additionally, we'll explain what [tex]\(\mu_{d}\)[/tex] represents.
### Step-by-step Solution:
1. Identify the Differences ([tex]\(d\)[/tex]) Between the Two Samples:
For each subject, subtract the temperature at 8 AM from the temperature at 12 AM. These differences ([tex]\(d\)[/tex]) are calculated as follows:
- Subject 1: [tex]\(98.0 - 97.5 = 0.5\)[/tex]
- Subject 2: [tex]\(100.3 - 99.5 = 0.8\)[/tex]
- Subject 3: [tex]\(97.2 - 97.1 = 0.1\)[/tex]
- Subject 4: [tex]\(96.8 - 97.1 = -0.3\)[/tex]
- Subject 5: [tex]\(97.7 - 97.3 = 0.4\)[/tex]
So, the differences [tex]\(d\)[/tex] are: [tex]\(0.5, 0.8, 0.1, -0.3, 0.4\)[/tex].
2. Calculate the Mean of the Differences ([tex]\(\overline{d}\)[/tex]):
The mean of the differences is the sum of all differences divided by the number of subjects.
[tex]\[
\overline{d} = \frac{0.5 + 0.8 + 0.1 - 0.3 + 0.4}{5} = \frac{1.5}{5} = 0.3
\][/tex]
3. Calculate the Standard Deviation of the Differences ([tex]\(s_{d}\)[/tex]):
The standard deviation is a measure of how spread out the differences are. It is calculated using the formula for the sample standard deviation:
[tex]\[
s_{d} = \sqrt{\frac{\sum (d_i - \overline{d})^2}{n-1}}
\][/tex]
Where [tex]\(d_i\)[/tex] is each individual difference, [tex]\(\overline{d}\)[/tex] is the mean of the differences, and [tex]\(n\)[/tex] is the number of differences (in this case, 5).
After carrying out these calculations, the standard deviation [tex]\(s_{d}\)[/tex] is approximately [tex]\(0.4183\)[/tex].
### Conclusion:
- The mean difference [tex]\(\overline{d}\)[/tex] is 0.3.
- The standard deviation [tex]\(s_{d}\)[/tex] is approximately 0.4183.
### Understanding [tex]\(\mu_{d}\)[/tex]:
- [tex]\(\mu_{d}\)[/tex] represents the true mean difference in temperatures between 8 AM and 12 AM for the entire population. While we have calculated [tex]\(\overline{d}\)[/tex] for our sample, [tex]\(\mu_{d}\)[/tex] would be used to denote the average of all possible measurements under similar conditions.
This approach provides you with a thorough understanding and the computed values for [tex]\(\overline{d}\)[/tex] and [tex]\(s_{d}\)[/tex]!
### Step-by-step Solution:
1. Identify the Differences ([tex]\(d\)[/tex]) Between the Two Samples:
For each subject, subtract the temperature at 8 AM from the temperature at 12 AM. These differences ([tex]\(d\)[/tex]) are calculated as follows:
- Subject 1: [tex]\(98.0 - 97.5 = 0.5\)[/tex]
- Subject 2: [tex]\(100.3 - 99.5 = 0.8\)[/tex]
- Subject 3: [tex]\(97.2 - 97.1 = 0.1\)[/tex]
- Subject 4: [tex]\(96.8 - 97.1 = -0.3\)[/tex]
- Subject 5: [tex]\(97.7 - 97.3 = 0.4\)[/tex]
So, the differences [tex]\(d\)[/tex] are: [tex]\(0.5, 0.8, 0.1, -0.3, 0.4\)[/tex].
2. Calculate the Mean of the Differences ([tex]\(\overline{d}\)[/tex]):
The mean of the differences is the sum of all differences divided by the number of subjects.
[tex]\[
\overline{d} = \frac{0.5 + 0.8 + 0.1 - 0.3 + 0.4}{5} = \frac{1.5}{5} = 0.3
\][/tex]
3. Calculate the Standard Deviation of the Differences ([tex]\(s_{d}\)[/tex]):
The standard deviation is a measure of how spread out the differences are. It is calculated using the formula for the sample standard deviation:
[tex]\[
s_{d} = \sqrt{\frac{\sum (d_i - \overline{d})^2}{n-1}}
\][/tex]
Where [tex]\(d_i\)[/tex] is each individual difference, [tex]\(\overline{d}\)[/tex] is the mean of the differences, and [tex]\(n\)[/tex] is the number of differences (in this case, 5).
After carrying out these calculations, the standard deviation [tex]\(s_{d}\)[/tex] is approximately [tex]\(0.4183\)[/tex].
### Conclusion:
- The mean difference [tex]\(\overline{d}\)[/tex] is 0.3.
- The standard deviation [tex]\(s_{d}\)[/tex] is approximately 0.4183.
### Understanding [tex]\(\mu_{d}\)[/tex]:
- [tex]\(\mu_{d}\)[/tex] represents the true mean difference in temperatures between 8 AM and 12 AM for the entire population. While we have calculated [tex]\(\overline{d}\)[/tex] for our sample, [tex]\(\mu_{d}\)[/tex] would be used to denote the average of all possible measurements under similar conditions.
This approach provides you with a thorough understanding and the computed values for [tex]\(\overline{d}\)[/tex] and [tex]\(s_{d}\)[/tex]!
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 overline 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!
- You are operating a recreational vessel less than 39 4 feet long on federally controlled waters Which of the following is a legal sound device
- Which step should a food worker complete to prevent cross contact when preparing and serving an allergen free meal A Clean and sanitize all surfaces
- For one month Siera calculated her hometown s average high temperature in degrees Fahrenheit She wants to convert that temperature from degrees Fahrenheit to degrees
Rewritten by : Jeany
