Math Problem Statement

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 mg/dl. Recently, the patient's total calcium tests gave the following readings (in mg/dl). Assume that the population of x values has an approximately normal distribution. 10.10 9.00 10.90 9.10 9.40 9.80 10.00 9.90 11.20 12.10 Find a 99.9% confidence interval for the population mean of total calcium in this patient's blood. (Round your answer to two decimal places.) Find upper and lower limit

Solution

To find a 99.9% confidence interval for the population mean of total calcium levels, we will follow these steps:

  1. Calculate the sample mean (xˉ\bar{x}).
  2. Calculate the sample standard deviation (s).
  3. Determine the sample size (n).
  4. Find the t-value for 99.9% confidence.
  5. Calculate the margin of error.
  6. Construct the confidence interval.

Step 1: Calculate the Sample Mean (xˉ\bar{x})

Given the values: 10.10,9.00,10.90,9.10,9.40,9.80,10.00,9.90,11.20,12.1010.10, 9.00, 10.90, 9.10, 9.40, 9.80, 10.00, 9.90, 11.20, 12.10 xˉ=10.10+9.00+10.90+9.10+9.40+9.80+10.00+9.90+11.20+12.1010=10.10+9.00+10.90+9.10+9.40+9.80+10.00+9.90+11.20+12.1010\bar{x} = \frac{10.10 + 9.00 + 10.90 + 9.10 + 9.40 + 9.80 + 10.00 + 9.90 + 11.20 + 12.10}{10} = \frac{ 10.10 + 9.00 + 10.90 + 9.10 + 9.40 + 9.80 + 10.00 + 9.90 + 11.20 + 12.10 }{ 10 } xˉ=10.10+9.00+10.90+9.10+9.40+9.80+10.00+9.90+11.20+12.1010=10.05 mg/dl\bar{x} = \frac{ 10.10 + 9.00 + 10.90 + 9.10 + 9.40 + 9.80 + 10.00 + 9.90 + 11.20 + 12.10 }{ 10 } = 10.05 \text{ mg/dl}

Step 2: Calculate the Sample Standard Deviation (s)

First, calculate the variance: s2=(xixˉ)2n1s^2 = \frac{\sum (x_i - \bar{x})^2}{n-1} Where xix_i are the sample values and nn is the number of samples.

Calculating deviations: (10.1010.05)2,(9.0010.05)2,(10.9010.05)2,(9.1010.05)2,(9.4010.05)2,(10.10 - 10.05)^2, (9.00 - 10.05)^2, (10.90 - 10.05)^2, (9.10 - 10.05)^2, (9.40 - 10.05)^2, (9.8010.05)2,(10.0010.05)2,(9.9010.05)2,(11.2010.05)2,(12.1010.05)2(9.80 - 10.05)^2, (10.00 - 10.05)^2, (9.90 - 10.05)^2, (11.20 - 10.05)^2, (12.10 - 10.05)^2

Calculating the squared deviations: 0.0025,1.1025,0.7225,0.9025,0.4225,0.0625,0.0025,0.0225,1.3225,4.20250.0025, 1.1025, 0.7225, 0.9025, 0.4225, 0.0625, 0.0025, 0.0225, 1.3225, 4.2025 Summing them up: (xixˉ)2=0.0025+1.1025+0.7225+0.9025+0.4225+0.0625+0.0025+0.0225+1.3225+4.2025=8.5\sum (x_i - \bar{x})^2 = 0.0025 + 1.1025 + 0.7225 + 0.9025 + 0.4225 + 0.0625 + 0.0025 + 0.0225 + 1.3225 + 4.2025 = 8.5 Calculating variance: s2=8.5101=8.590.9444s^2 = \frac{8.5}{10-1} = \frac{8.5}{9} \approx 0.9444 Calculating standard deviation: s0.94440.9724 mg/dls \approx \sqrt{0.9444} \approx 0.9724 \text{ mg/dl}

Step 3: Determine the Sample Size (n)

n=10n = 10

Step 4: Find the t-value for 99.9% Confidence

Using a t-table for 9 degrees of freedom at a 99.9% confidence level: t0.999,93.250t_{0.999, 9} \approx 3.250

Step 5: Calculate the Margin of Error

ME=t(sn)ME = t \cdot \left(\frac{s}{\sqrt{n}}\right) ME=3.250(0.972410)3.2500.30771.000ME = 3.250 \cdot \left(\frac{0.9724}{\sqrt{10}}\right) \approx 3.250 \cdot 0.3077 \approx 1.000

Step 6: Construct the Confidence Interval

Lower limit=xˉME10.051.0009.05\text{Lower limit} = \bar{x} - ME \approx 10.05 - 1.000 \approx 9.05 Upper limit=xˉ+ME10.05+1.00011.05\text{Upper limit} = \bar{x} + ME \approx 10.05 + 1.000 \approx 11.05

Final Result

The 99.9% confidence interval for the population mean of total calcium is approximately:

  • Lower Limit: 9.05 mg/dl
  • Upper Limit: 11.05 mg/dl

Would you like to know more details or have any questions? Here are some related questions:

  1. How does the t-distribution differ from the normal distribution?
  2. What are the implications of a confidence interval in medical studies?
  3. How would you interpret a confidence interval of a different confidence level, such as 95%?
  4. What factors can affect the width of a confidence interval?
  5. How can sample size influence the reliability of a confidence interval?

Tip: Always check for normality in your data when using the t-distribution for confidence intervals.

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Math Problem Analysis

Mathematical Concepts

Statistics
Confidence Intervals
Normal Distribution

Formulas

Sample Mean (\bar{x}) = \frac{\sum x_i}{n}
Sample Standard Deviation (s) = \sqrt{\frac{\sum (x_i - \bar{x})^2}{n-1}}
Margin of Error (ME) = t \cdot \left(\frac{s}{\sqrt{n}}\right)
Confidence Interval = \bar{x} \pm ME

Theorems

Central Limit Theorem
t-distribution properties

Suitable Grade Level

Grades 11-12