91Ó°ÊÓ

The mean difference, denoted as \(\bar{d}\), represents the average change between paired observations. In this problem, we are examining the body temperature of subjects at two different times: 8 AM and 12 AM. By calculating the difference for each individual and averaging those differences, we can determine the overall trend.

**Why is it important?**
Knowing the mean difference helps us understand if there is a general increase, decrease, or no change between the measurements over time.

**How is it calculated?**
Take the sum of all individual differences and divide it by the number of subjects. For example, using our data:

\[ \bar{d} = \frac{0.8 + 0.5 + 0.1 - 0.1 + 0.1}{5} = \frac{1.4}{5} = 0.28 \]
This calculation tells us that, on average, body temperature increased by 0.28 degrees Fahrenheit from 8 AM to 12 AM.

Standard deviation, denoted as \(s_d\), measures the spread or dispersion of a set of data from its mean. In the context of this exercise, it tells us how much individual body temperature differences (between 8 AM and 12 AM) vary around our calculated mean difference.

**Why is it important?**
It helps us understand how consistent the measurements are. A small standard deviation means the differences are close to the mean difference, while a large one indicates more variability.

**How do we find it?**
1. **Calculate the Variance:** First, sum the squared deviations from the mean and divide by the number of subjects minus one:
\[ s_d^2 = \frac{0.2704 + 0.0484 + 0.0324 + 0.1444 + 0.0324}{5-1} = \frac{0.528}{4} = 0.132 \]
2. **Take the Square Root:** The square root of the variance gives the standard deviation:
\[ s_d = \sqrt{0.132} \approx 0.3633 \]
Therefore, the standard deviation of the differences in body temperature is approximately 0.3633 degrees Fahrenheit.

Variance, often labeled as \(s_{d}^{2}\), measures the average degree to which each number is different from the mean. It's an intermediate step in calculating standard deviation and provides insight into the variability of the data set.

**Why is it important?**
While standard deviation gives a sense of spread in units we can easily understand, variance helps us in computations, especially in statistical testing. A high variance indicates that numbers are far from the mean and from each other, while a low variance indicates the opposite.

**Steps to calculate variance:**
1. **Find the Mean Difference:** Using our example, it was found to be \(\bar{d} = 0.28\).
2. **Calculate each squared deviation from the mean difference:**
\[ (0.8 - 0.28)^2 = 0.2704 \]
\[ (0.5 - 0.28)^2 = 0.0484 \]
\[ (0.1 - 0.28)^2 = 0.0324 \]
\[ (-0.1 - 0.28)^2 = 0.1444 \]
\[ (0.1 - 0.28)^2 = 0.0324 \]
3. **Sum these squared deviations:**
\[ 0.2704 + 0.0484 + 0.0324 + 0.1444 + 0.0324 = 0.528 \]
4. **Divide by the number of subjects minus one:**
\[ s_{d}^{2} = \frac{0.528}{4} = 0.132 \]
This result indicates that the variance of the differences in temperatures is 0.132 degrees Fahrenheit squared.