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 find the mean difference ([tex]$\bar{d}$[/tex]) and the standard deviation of the differences ([tex]$s_d$[/tex]) between the body temperatures at 8 AM and 12 AM, we will follow these steps:
### Step 1: List the Temperatures
We have the temperatures recorded as follows:
- At 8 AM: [tex]\( [98.4, 99.5, 97.3, 97.8, 97.3] \)[/tex]
- At 12 AM: [tex]\( [99.0, 100.2, 97.4, 97.6, 97.5] \)[/tex]
### Step 2: Calculate the Differences
Calculate the difference [tex]\(d_i\)[/tex] between each pair of temperatures:
[tex]\[
d_i = \text{Temperature at 12 AM} - \text{Temperature at 8 AM}
\][/tex]
Let's work through each pair:
[tex]\[
d_1 = 99.0 - 98.4 = 0.6
\][/tex]
[tex]\[
d_2 = 100.2 - 99.5 = 0.7
\][/tex]
[tex]\[
d_3 = 97.4 - 97.3 = 0.1
\][/tex]
[tex]\[
d_4 = 97.6 - 97.8 = -0.2
\][/tex]
[tex]\[
d_5 = 97.5 - 97.3 = 0.2
\][/tex]
So, the differences are:
[tex]\[ d = [0.6, 0.7, 0.1, -0.2, 0.2] \][/tex]
### Step 3: Calculate the Mean of the Differences ([tex]$\bar{d}$[/tex])
To find the mean difference [tex]$\bar{d}$[/tex], sum all the differences and then divide by the number of pairs (5 in this case):
[tex]\[
\bar{d} = \frac{\sum_{i=1}^5 d_i}{5}
\][/tex]
Find the sum of the differences:
[tex]\[
0.6 + 0.7 + 0.1 - 0.2 + 0.2 = 1.4
\][/tex]
Then, divide by the number of samples:
[tex]\[
\bar{d} = \frac{1.4}{5} = 0.28
\][/tex]
### Step 4: Calculate the Standard Deviation of the Differences ([tex]$s_d$[/tex])
The formula for the sample standard deviation [tex]$s_d$[/tex] of the differences is:
[tex]\[
s_d = \sqrt{\frac{\sum_{i=1}^n (d_i - \bar{d})^2}{n-1}}
\][/tex]
Using the numerical values calculated:
- Mean difference [tex]$\bar{d}$[/tex] is [tex]\( 0.28 \)[/tex]
Calculate each squared deviation from the mean:
[tex]\[
(0.6 - 0.28)^2 = 0.1024
\][/tex]
[tex]\[
(0.7 - 0.28)^2 = 0.1764
\][/tex]
[tex]\[
(0.1 - 0.28)^2 = 0.0324
\][/tex]
[tex]\[
(-0.2 - 0.28)^2 = 0.2304
\][/tex]
[tex]\[
(0.2 - 0.28)^2 = 0.0064
\][/tex]
Sum these squared deviations:
[tex]\[
0.1024 + 0.1764 + 0.0324 + 0.2304 + 0.0064 = 0.548
\][/tex]
Then, divide by the degrees of freedom [tex]\(n - 1\)[/tex] (which is 5 - 1 = 4):
[tex]\[
\frac{0.548}{4} = 0.137
\][/tex]
Finally, take the square root to find [tex]$s_d$[/tex]:
[tex]\[
s_d = \sqrt{0.137} \approx 0.37
\][/tex]
### Summary of Results
- The mean difference ([tex]$\bar{d}$[/tex]) is:
[tex]\[
\bar{d} = 0.28
\][/tex]
- The standard deviation of the differences ([tex]$s_d$[/tex]) is:
[tex]\[
s_d = 0.37
\][/tex]
### Interpretation of [tex]$\mu_d$[/tex]
- [tex]\(\mu_d\)[/tex] represents the population mean difference. It could signify the average change in body temperature from 8 AM to 12 AM if we had the entire population's data instead of just a sample.
Thus, the values are:
[tex]\[
\bar{d} = 0.28 \quad \text{and} \quad s_d = 0.37
\][/tex]
### Step 1: List the Temperatures
We have the temperatures recorded as follows:
- At 8 AM: [tex]\( [98.4, 99.5, 97.3, 97.8, 97.3] \)[/tex]
- At 12 AM: [tex]\( [99.0, 100.2, 97.4, 97.6, 97.5] \)[/tex]
### Step 2: Calculate the Differences
Calculate the difference [tex]\(d_i\)[/tex] between each pair of temperatures:
[tex]\[
d_i = \text{Temperature at 12 AM} - \text{Temperature at 8 AM}
\][/tex]
Let's work through each pair:
[tex]\[
d_1 = 99.0 - 98.4 = 0.6
\][/tex]
[tex]\[
d_2 = 100.2 - 99.5 = 0.7
\][/tex]
[tex]\[
d_3 = 97.4 - 97.3 = 0.1
\][/tex]
[tex]\[
d_4 = 97.6 - 97.8 = -0.2
\][/tex]
[tex]\[
d_5 = 97.5 - 97.3 = 0.2
\][/tex]
So, the differences are:
[tex]\[ d = [0.6, 0.7, 0.1, -0.2, 0.2] \][/tex]
### Step 3: Calculate the Mean of the Differences ([tex]$\bar{d}$[/tex])
To find the mean difference [tex]$\bar{d}$[/tex], sum all the differences and then divide by the number of pairs (5 in this case):
[tex]\[
\bar{d} = \frac{\sum_{i=1}^5 d_i}{5}
\][/tex]
Find the sum of the differences:
[tex]\[
0.6 + 0.7 + 0.1 - 0.2 + 0.2 = 1.4
\][/tex]
Then, divide by the number of samples:
[tex]\[
\bar{d} = \frac{1.4}{5} = 0.28
\][/tex]
### Step 4: Calculate the Standard Deviation of the Differences ([tex]$s_d$[/tex])
The formula for the sample standard deviation [tex]$s_d$[/tex] of the differences is:
[tex]\[
s_d = \sqrt{\frac{\sum_{i=1}^n (d_i - \bar{d})^2}{n-1}}
\][/tex]
Using the numerical values calculated:
- Mean difference [tex]$\bar{d}$[/tex] is [tex]\( 0.28 \)[/tex]
Calculate each squared deviation from the mean:
[tex]\[
(0.6 - 0.28)^2 = 0.1024
\][/tex]
[tex]\[
(0.7 - 0.28)^2 = 0.1764
\][/tex]
[tex]\[
(0.1 - 0.28)^2 = 0.0324
\][/tex]
[tex]\[
(-0.2 - 0.28)^2 = 0.2304
\][/tex]
[tex]\[
(0.2 - 0.28)^2 = 0.0064
\][/tex]
Sum these squared deviations:
[tex]\[
0.1024 + 0.1764 + 0.0324 + 0.2304 + 0.0064 = 0.548
\][/tex]
Then, divide by the degrees of freedom [tex]\(n - 1\)[/tex] (which is 5 - 1 = 4):
[tex]\[
\frac{0.548}{4} = 0.137
\][/tex]
Finally, take the square root to find [tex]$s_d$[/tex]:
[tex]\[
s_d = \sqrt{0.137} \approx 0.37
\][/tex]
### Summary of Results
- The mean difference ([tex]$\bar{d}$[/tex]) is:
[tex]\[
\bar{d} = 0.28
\][/tex]
- The standard deviation of the differences ([tex]$s_d$[/tex]) is:
[tex]\[
s_d = 0.37
\][/tex]
### Interpretation of [tex]$\mu_d$[/tex]
- [tex]\(\mu_d\)[/tex] represents the population mean difference. It could signify the average change in body temperature from 8 AM to 12 AM if we had the entire population's data instead of just a sample.
Thus, the values are:
[tex]\[
\bar{d} = 0.28 \quad \text{and} \quad s_d = 0.37
\][/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 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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