To calculate the sample mean ($\bar{x}$) and sample standard deviation ($s$), we use the following formulas:

Sample Mean ($\bar{x}$) Formula:

$\bar{x} = \frac{1}{n} \sum_{i=1}^{n} x_i$ where $n$ is the sample size (in this case, $n = 6$) and $x_i$ are the individual sample values.

Sample Standard Deviation ($s$) Formula:

$s = \sqrt{\frac{1}{n-1} \sum_{i=1}^{n} (x_i - \bar{x})^2}$ where $x_i$ is each data point, $\bar{x}$ is the sample mean, and $n$ is the sample size.

Given Data:

The RBC counts from the six tests are: $4.9, 4.2, 4.5, 4.1, 4.4, 4.3$

Let's calculate the sample mean and sample standard deviation.

1. Calculate the Sample Mean ($\bar{x}$):

$\bar{x} = \frac{4.9 + 4.2 + 4.5 + 4.1 + 4.4 + 4.3}{6} = \frac{26.4}{6} = 4.4000$

2. Calculate the Sample Standard Deviation ($s$):

First, calculate each squared deviation from the mean:

• $(4.9 - 4.4)^2 = (0.5)^2 = 0.25$
• $(4.2 - 4.4)^2 = (-0.2)^2 = 0.04$
• $(4.5 - 4.4)^2 = (0.1)^2 = 0.01$
• $(4.1 - 4.4)^2 = (-0.3)^2 = 0.09$
• $(4.4 - 4.4)^2 = (0)^2 = 0.00$
• $(4.3 - 4.4)^2 = (-0.1)^2 = 0.01$

Sum of squared deviations: $0.25 + 0.04 + 0.01 + 0.09 + 0.00 + 0.01 = 0.40$

Now, calculate the sample standard deviation: $s = \sqrt{\frac{0.40}{6 - 1}} = \sqrt{\frac{0.40}{5}} = \sqrt{0.08} \approx 0.2828$

Final Results:

• Sample mean ($\bar{x}$) = 4.4000
• Sample standard deviation ($s$) = 0.2828

Let me know if you need further clarification or additional details!

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Here are some related questions for further exploration:

1. What are the differences between a sample mean and a population mean?
2. How do you calculate the population standard deviation?
3. What is the significance of rounding in statistics, especially in medical testing?
4. How would the sample mean and standard deviation change if one of the RBC values was much higher than the others?
5. What are the assumptions for a data set to be considered "approximately normal"?

Tip: When calculating the standard deviation, be careful not to divide by $n$ (the sample size), but by $n-1$ to get an unbiased estimate for the population standard deviation!