Questions: An adult patient has been treated for tetany (severe muscle spasms). This condition is associated with an average total calcium level below 6 mg / dl. A patient had a series of total calcium tests that resulted in the following readings: (in mg / dl.) (Use t-distribution) 9.3, 8.8, 10.1, 8.9, 9.4, 9.8, 10.0, 9.9, 11.2, 12.1 a) Calculate the mean and standard deviation for these data.

Solution Steps

To find the mean \( \mu \) of the calcium levels, we use the formula:

\[ \mu = \frac{\sum_{i=1}^N x_i}{N} \]

where \( N \) is the number of observations and \( x_i \) are the individual calcium level readings.

Given the readings:

\[ \sum_{i=1}^{10} x_i = 9.3 + 8.8 + 10.1 + 8.9 + 9.4 + 9.8 + 10.0 + 9.9 + 11.2 + 12.1 = 99.5 \]

Thus, the mean is calculated as:

\[ \mu = \frac{99.5}{10} = 9.95 \]

The variance \( \sigma^2 \) is calculated using the formula:

\[ \sigma^2 = \frac{\sum (x_i - \mu)^2}{n-1} \]

First, we find \( (x_i - \mu)^2 \) for each reading:

• \( (9.3 - 9.95)^2 = 0.4225 \)
• \( (8.8 - 9.95)^2 = 1.3225 \)
• \( (10.1 - 9.95)^2 = 0.0225 \)
• \( (8.9 - 9.95)^2 = 1.1025 \)
• \( (9.4 - 9.95)^2 = 0.3025 \)
• \( (9.8 - 9.95)^2 = 0.0225 \)
• \( (10.0 - 9.95)^2 = 0.0025 \)
• \( (9.9 - 9.95)^2 = 0.0025 \)
• \( (11.2 - 9.95)^2 = 1.5625 \)
• \( (12.1 - 9.95)^2 = 4.6225 \)

Now, summing these squared differences:

\[ \sum (x_i - \mu)^2 = 0.4225 + 1.3225 + 0.0225 + 1.1025 + 0.3025 + 0.0225 + 0.0025 + 0.0025 + 1.5625 + 4.6225 = 9.5 \]

Now, substituting into the variance formula:

\[ \sigma^2 = \frac{9.5}{10 - 1} = \frac{9.5}{9} \approx 1.04 \]

The standard deviation \( \sigma \) is the square root of the variance:

\[ \sigma = \sqrt{\sigma^2} = \sqrt{1.04} \approx 1.02 \]

Final Answer