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Over the past several months, an adult patient has been treated for tetany (severe muscle spasms). This condition is associated with an average total calcium level below \(6 \mathrm{mg} / \mathrm{dl}(\) Reference: Manual of Laboratory and Diagnostic Tests, F. Fischbach). Recently, the patient's total calcium tests gave the following readings (in \(\mathrm{mg} / \mathrm{dl}\) ). \(9.3\) \(\begin{array}{llllll}8.8 & 10.1 & 8.9 & 9.4 & 9.8 & 10.0\end{array}\) \(\begin{array}{lll}9.9 & 11.2 & 12.1\end{array}\) (a) Use a calculator to verify that \(\bar{x}=9.95\) and \(s \approx 1.02\). (b) Find a \(99.9 \%\) confidence interval for the population mean of total calcium in this patient's blood. (c) Based on your results in part (b), do you think this patient still has a calcium deficiency? Explain.

Short Answer

Expert verified
The patient probably does not have a calcium deficiency, as the confidence interval shows the mean level is above the deficiency threshold of 6 mg/dL.

Step by step solution

01

Verify the mean

To confirm the mean \( \bar{x} \), calculate the total sum of the calcium readings and divide by the number of readings. Given readings are: \(9.3, 8.8, 10.1, 8.9, 9.4, 9.8, 10.0, 9.9, 11.2, 12.1\). Sum them to get \(99.5\). The mean is then \( \bar{x} = \frac{99.5}{10} = 9.95\). This confirms \( \bar{x} = 9.95 \).
02

Verify the standard deviation

Calculate the sample standard deviation \( s \) using the formula: \[ s = \sqrt{\frac{\sum (x_i - \bar{x})^2}{n-1}} \] Calculate each \((x_i - \bar{x})^2\), sum them, and divide by \( n-1 = 9 \). Compute the square root to verify \( s \approx 1.02 \).
03

Find the 99.9% confidence interval

For a 99.9% confidence interval, use the formula: \[ \bar{x} \pm t \frac{s}{\sqrt{n}} \] where \( t \) is the t-value from the t-distribution table with \(9\) degrees of freedom for 99.9% confidence. The approximate \( t \) value is \( 4.297 \). Calculate: \[ CI = 9.95 \pm 4.297 \times \frac{1.02}{\sqrt{10}} \] This results in the confidence interval: \[ CI = 9.95 \pm 1.385 \] Thus, the interval is \([8.565, 11.335]\).
04

Interpret the confidence interval

Since the calculated confidence interval \([8.565, 11.335]\) lies entirely above 6, which is the deficiency threshold, the patient's mean calcium level is statistically above the deficiency level. This suggests that there is no current calcium deficiency.

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Key Concepts

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

Confidence Interval
A confidence interval provides a range of values within which we can be reasonably certain the true population parameter lies. In simpler terms, it's like saying, "We are sure the actual value is between here and here, 99.9% of the time." To construct a confidence interval, we utilize the sample mean, the standard deviation, and a critical value from the t-distribution, which reflects our desired confidence level.

Given our scenario, we're constructing a 99.9% confidence interval for the mean calcium level in a patient. This high confidence level requires a larger range of values. The wider the interval, the more confident we are about capturing the true mean. Here, we calculated the interval as \[ CI = \bar{x} \pm t \frac{s}{\sqrt{n}} \] with a range of [8.565, 11.335]. Because this interval sits comfortably above the 6 mg/dl deficiency level, we can strongly suspect the patient does not currently suffer from a calcium deficiency.
Mean Calculation
The mean is a measure of central tendency — it's where data points gather around. To calculate the mean, sum up all your data, then divide by the number of data points. It's like finding an average score.

In our exercise, the patient's calcium readings were:
  • 9.3
  • 8.8
  • 10.1
  • 8.9
  • 9.4
  • 9.8
  • 10.0
  • 9.9
  • 11.2
  • 12.1
To discover the mean, the sum of these readings is divided by the total number of readings, 10. This results in a mean (\(\bar{x}\)) of 9.95. This average gives us a peek into the patient’s typical calcium level during tests.
Standard Deviation
Standard deviation represents how much individual data points differ from the mean. In other words, it tells us about the spread of the data points. If many data points are clustered closely around the mean, the standard deviation is small; if they are scattered far and wide, it’s bigger.

Calculating it involves:
  • Subtracting the mean from each data point to get the deviation.
  • Squaring each deviation to make them positive.
  • Averaging these squared deviations (divide the sum by one less than the number of data points).
  • Finally, taking the square root of that average to get the standard deviation.

For this exercise, the standard deviation was approximately 1.02, suggesting that most readings fall within about 1 mg/dl of the mean. This indicates a moderate spread around the mean reading of 9.95.
T-Distribution
The t-distribution is a probability distribution that resembles the normal distribution but has heavier tails. It's particularly useful when dealing with smaller sample sizes or when the population standard deviation is unknown, just like in our scenario.

This distribution allows for the calculation of confidence intervals and hypothesis testing for small to medium-sized data sets. In our case, the t-distribution is used to determine the critical value or t-value required to construct the confidence interval at a 99.9% confidence level.

With 9 degrees of freedom (total readings minus one), the critical t-value of approximately 4.297 was used. This value is essential because it modifies our confidence interval, making sure it’s sufficiently wide to give us the certainty we need. The t-distribution effectively aids in making strong and accurate statistical inferences, especially when data is limited.

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Most popular questions from this chapter

If the original \(x\) distribution has a relatively small standard deviation, the confidence interval for \(\mu\) will be relatively short.

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