Math Problem Statement

A random sample of nn measurements was selected from a population with standard deviation σ=10.5σ=10.5 and unknown mean μμ. Calculate a 9999 % confidence interval for μμ for each of the following situations:

(a) n=35, x⎯⎯⎯=86.7n=35, x¯=86.7

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≤μ≤≤μ≤

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(b) n=55, x⎯⎯⎯=86.7n=55, x¯=86.7

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≤μ≤≤μ≤

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(c) n=70, x⎯⎯⎯=86.7n=70, x¯=86.7

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≤μ≤≤μ≤

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(d) In general, we can say that for the same confidence level, increasing the sample size

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the margin of error (width) of the confidence interval. (Enter: ''DECREASES'', ''DOES NOT CHANGE'' or ''INCREASES'', without the quotes.)

Solution

To calculate the confidence intervals for the population mean μ\mu, we use the formula for the confidence interval when the population standard deviation σ\sigma is known:

Confidence Interval: xˉ±zα/2(σn)\text{Confidence Interval: } \bar{x} \pm z_{\alpha/2} \left(\frac{\sigma}{\sqrt{n}}\right)

where:

  • xˉ\bar{x} is the sample mean,
  • zα/2z_{\alpha/2} is the z-score corresponding to the desired confidence level (which is based on the standard normal distribution),
  • σ\sigma is the population standard deviation,
  • nn is the sample size.

For a 99.99% confidence level, the critical z-value zα/2z_{\alpha/2} is approximately 3.890 (from standard normal distribution tables).

The population standard deviation σ=10.5\sigma = 10.5, and the sample means are provided as xˉ=86.7\bar{x} = 86.7. We will calculate the confidence interval for each scenario:


(a) When n=35n = 35:

Margin of Error=zα/2(σn)=3.890(10.535)6.895\text{Margin of Error} = z_{\alpha/2} \left(\frac{\sigma}{\sqrt{n}}\right) = 3.890 \left(\frac{10.5}{\sqrt{35}}\right) \approx 6.895

So the confidence interval is: 86.7±6.89579.805μ93.59586.7 \pm 6.895 \quad \Rightarrow \quad 79.805 \leq \mu \leq 93.595


(b) When n=55n = 55:

Margin of Error=3.890(10.555)5.509\text{Margin of Error} = 3.890 \left(\frac{10.5}{\sqrt{55}}\right) \approx 5.509

So the confidence interval is: 86.7±5.50981.191μ92.20986.7 \pm 5.509 \quad \Rightarrow \quad 81.191 \leq \mu \leq 92.209


(c) When n=70n = 70:

Margin of Error=3.890(10.570)4.882\text{Margin of Error} = 3.890 \left(\frac{10.5}{\sqrt{70}}\right) \approx 4.882

So the confidence interval is: 86.7±4.88281.818μ91.58286.7 \pm 4.882 \quad \Rightarrow \quad 81.818 \leq \mu \leq 91.582


(d) In general, when the sample size nn increases, the margin of error DECREASES.

This happens because the margin of error is inversely proportional to the square root of the sample size (n\sqrt{n}), meaning that as the sample size increases, the width of the confidence interval shrinks, resulting in a more precise estimate.


Would you like more details or explanations on any part? Here are some questions that expand on this topic:

  1. How does the z-score change if we switch to a different confidence level, like 95%?
  2. Why is the sample size nn important in determining the width of the confidence interval?
  3. How does this method change if the population standard deviation σ\sigma is unknown?
  4. What are the implications of a very small or very large confidence interval in practical terms?
  5. Can you derive the formula for confidence intervals if the sample follows a t-distribution instead of a normal distribution?

Tip: When increasing the sample size, the reduction in margin of error follows a square root relationship, so doubling the sample size won't halve the margin of error.

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Math Problem Analysis

Mathematical Concepts

Statistics
Confidence Intervals
Normal Distribution

Formulas

Confidence Interval Formula: x̄ ± zα/2 (σ/√n)
Margin of Error: zα/2 (σ/√n)

Theorems

Central Limit Theorem
Standard Normal Distribution

Suitable Grade Level

College/Advanced High School