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Chapter 7: Problem 11

A volumetric calcium analysis on triplicate samples of the blood serum of a patient believed to be suffering from a hyperparathyroid condition produced the following data: \(\mathrm{mmol} \mathrm{Ca} / \mathrm{L}=3.15,3.25,3.26\). What is the \(95 \%\) confidence interval for the mean of the data, assuming (a) no prior information about the precision of the analysis? (b) \(s \rightarrow \sigma=0.056 \mathrm{mmol} \mathrm{Ca} / \mathrm{L}\) ?

Short Answer

Expert verified
(a) CI: [3.07, 3.37] (b) CI: [3.160, 3.280]

Step by step solution

01

Calculate the Sample Mean

To find the mean, sum all the values and divide by the number of values. The formula is: \( \bar{x} = \frac{\sum x_i}{n} \). For our data: \( \bar{x} = \frac{3.15 + 3.25 + 3.26}{3} = 3.22 \ \mathrm{mmol} \ \mathrm{Ca} / \mathrm{L} \)
02

Calculate the Sample Standard Deviation

Use the formula for the sample standard deviation: \( s = \sqrt{\frac{\sum (x_i - \bar{x})^2}{n-1}} \). Substituting in the values: \( s = \sqrt{\frac{(3.15 - 3.22)^2 + (3.25 - 3.22)^2 + (3.26 - 3.22)^2}{2}} = 0.058 \ \mathrm{mmol} \ \mathrm{Ca} / \mathrm{L} \)
03

Determine t-value for 95% Confidence Interval

With a sample size (n) of 3, degrees of freedom (df) is \(n-1 = 2\). Look up t-value for 95% confidence with 2 degrees of freedom, which is approximately 4.303.
04

Calculate the 95% Confidence Interval (a)

The formula is: \( \bar{x} \pm t \frac{s}{\sqrt{n}} \). Substituting the known values: \( 3.22 \pm 4.303 \times \frac{0.058}{\sqrt{3}} \), which calculates to a range of approximately \( [3.07, 3.37] \).
05

Calculate the 95% Confidence Interval (b)

Assuming \(s \rightarrow \sigma = 0.056 \ \mathrm{mmol} \ \mathrm{Ca} / \mathrm{L}\), use the z-value for 95% confidence, which is 1.96. The interval is: \( \bar{x} \pm z \frac{\sigma}{\sqrt{n}} \). Substituting the values: \( 3.22 \pm 1.96 \times \frac{0.056}{\sqrt{3}} \), which calculates to a range of approximately \( [3.160, 3.280] \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Volumetric Calcium Analysis
Volumetric calcium analysis is a method used to determine the concentration of calcium in a given sample, often used in medical tests such as blood serum analysis. This process involves taking multiple samples and measuring the amount of calcium in each one.
For those suspected of conditions like hyperparathyroidism, such analyses are crucial as calcium imbalance is a key indicator.
In our exercise, triplicate samples were measured. This ensures accuracy and reliability in the results by providing a better picture of the concentration through averaging and statistical measures, such as confidence intervals.
Sample Standard Deviation
The sample standard deviation is a measure of the amount of variation or dispersion in a set of values. In this context, it's used to assess how much individual calcium levels in the serum deviate from the mean.
The formula for calculating the sample standard deviation (\[s\]) is: \[s = \sqrt{\frac{\sum (x_i - \bar{x})^2}{n-1}}\]
Here,
  • \(x_i\) represents each individual data point,
  • \(\bar{x}\) is the sample mean, and
  • \(n\) is the number of samples.
A small standard deviation indicates that the values are close to the mean, indicating precision, while a larger standard deviation shows more variability.
t-value
When dealing with small sample sizes, as in our case with three samples, the t-distribution is preferable over the standard normal distribution (z-distribution).
The t-value helps determine the width of the confidence interval around the mean by taking into account the sample size and variability.
The degrees of freedom for calculating the t-value is \(n-1\), which equals 2 for our sample size of 3. For a 95% confidence interval, this gives us a t-value of approximately 4.303, providing a wider interval to capture the variability inherent in smaller samples.
Mean Calculation
Calculating the mean provides an average value of the calcium concentration among the samples, giving an estimate of the central tendency of data.
The formula for the mean is: \[\bar{x} = \frac{\sum x_i}{n}\] where:
  • \(\sum x_i\) is the sum of all sample values, and
  • \(n\) is the number of samples.
In the exercise, the mean calcium concentration was calculated from the values 3.15, 3.25, and 3.26 mmol/L, resulting in a mean of 3.22 mmol/L.
This mean then serves as a central point for calculating confidence intervals, helping to understand the reliability and expected range of calcium concentrations.

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Most popular questions from this chapter

Determination of phosphorous in blood serum gave results of \(4.40,4.42,4.60,4.48\), and \(4.50 \mathrm{ppm}\) P. Determine whether the \(4.60 \mathrm{ppm}\) result is an outlier or should be retained at the \(95 \%\) confidence level.

A standard method for the determination of glucose in serum is reported to have a standard deviation of \(0.38 \mathrm{mg} / \mathrm{dL}\). If \(s=0.38\) is a good estimate of \(\sigma\), how many replicate determinations should be made in order for the mean for the analysis of a sample to be within *(a) \(0.3 \mathrm{mg} / \mathrm{dL}\) of the true mean \(99 \%\) of the time? (b) \(0.3 \mathrm{mg} / \mathrm{dL}\) of the true mean \(95 \%\) of the time? (c) \(0.2 \mathrm{mg} / \mathrm{dL}\) of the true mean \(90 \%\) of the time?

Consider the following sets of replicate measurements: $$ \begin{array}{lccccc} { }^{*} \mathbf{A} & \mathbf{B} & { }^{*} \mathbf{C} & \mathbf{D} & { }^{*} \mathbf{E} & \mathbf{F} \\ \hline 2.7 & 0.514 & 70.24 & 3.5 & 0.812 & 70.65 \\ 3.0 & 0.503 & 70.22 & 3.1 & 0.792 & 70.63 \\ 2.6 & 0.486 & 70.10 & 3.1 & 0.794 & 70.64 \\ 2.8 & 0.497 & & 3.3 & 0.900 & 70.21 \\ 3.2 & 0.472 & & 2.5 & & \\ \hline \end{array} $$ Calculate the mean and the standard deviation for each of these six data sets. Calculate the \(95 \%\) confidence interval for each set of data. What does this interval mean?

Apply the \(Q\) test to the following data sets to determine whether the outlying result should be retained or rejected at the \(95 \%\) confidence level. (a) \(85.10,84.62,84.70\) (b) \(85.10,84.62,84.65,84.70\)

Discuss how the size of the confidence interval for the mean is influenced by the following (all the other factors are constant): (a) the standard deviation \(\sigma\). (b) the sample size \(N\). (c) the confidence level.
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