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]
