Math Problem Statement

A data set includes 108 body temperatures of healthy adult humans having a mean of 98.2degreesF and a standard deviation of 0.64degreesF. Construct a 99​% confidence interval estimate of the mean body temperature of all healthy humans. What does the sample suggest about the use of 98.6degreesF as the mean body​ temperature? Click here to view a t distribution table.LOADING... Click here to view page 1 of the standard normal distribution table.LOADING... Click here to view page 2 of the standard normal distribution table.LOADING... Question content area bottom Part 1 What is the confidence interval estimate of the population mean mu​?    enter your response heredegreesFless thanmuless than    enter your response heredegreesF ​(Round to three decimal places as​ needed.)

Solution

To construct a 99% confidence interval for the mean body temperature of healthy adults, we'll use the following formula:

Confidence Interval=xˉ±z(σn)\text{Confidence Interval} = \bar{x} \pm z \left(\frac{\sigma}{\sqrt{n}}\right)

where:

  • xˉ\bar{x} = sample mean = 98.2°F
  • σ\sigma = sample standard deviation = 0.64°F
  • nn = sample size = 108
  • zz is the z-score corresponding to a 99% confidence level.

Step 1: Determine the z-score for a 99% Confidence Level

For a 99% confidence interval, the z-score is approximately 2.576.

Step 2: Calculate the Margin of Error (ME)

Margin of Error (ME)=z×(σn)\text{Margin of Error (ME)} = z \times \left(\frac{\sigma}{\sqrt{n}}\right)

Substitute the values:

ME=2.576×(0.64108)\text{ME} = 2.576 \times \left(\frac{0.64}{\sqrt{108}}\right)

Let's calculate that:

  1. Calculate 10810.3923\sqrt{108} \approx 10.3923.
  2. Then, 0.6410.39230.0616\frac{0.64}{10.3923} \approx 0.0616.
  3. Finally, 2.576×0.06160.15872.576 \times 0.0616 \approx 0.1587.

So, the margin of error is approximately 0.159°F.

Step 3: Construct the Confidence Interval

Confidence Interval=xˉ±ME\text{Confidence Interval} = \bar{x} \pm \text{ME} =98.2±0.159= 98.2 \pm 0.159

This gives us:

98.20.159=98.041and98.2+0.159=98.35998.2 - 0.159 = 98.041 \quad \text{and} \quad 98.2 + 0.159 = 98.359

Confidence Interval

Thus, the 99% confidence interval for the mean body temperature is:

98.041F<μ<98.359F98.041^\circ \text{F} < \mu < 98.359^\circ \text{F}

Interpretation Regarding 98.6°F as a Mean Body Temperature

Since 98.6°F is outside this confidence interval, the sample suggests that the true mean body temperature of healthy humans might be lower than the traditionally accepted value of 98.6°F.


Would you like further details on any part of this calculation?

Here are some related questions to explore:

  1. How does sample size impact the width of a confidence interval?
  2. Why is a z-score used instead of a t-score in this example?
  3. How would the confidence interval change if the confidence level were 95% instead of 99%?
  4. What are the implications of finding a different mean for health-related norms?
  5. How does standard deviation affect the margin of error in confidence intervals?

Tip: For larger samples (n > 30), the z-score is typically used for confidence intervals, even if the population standard deviation is unknown.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Statistics
Confidence Intervals
Standard Deviation
Z-Scores

Formulas

Confidence Interval formula: CI = x̄ ± z * (σ / √n)
Margin of Error formula: ME = z * (σ / √n)

Theorems

Central Limit Theorem (implied in large sample size)

Suitable Grade Level

Undergraduate/College Level