Math Problem Statement

2.3 You are required to provide a 91.98% confidence interval for a mean, it is important that the interval be no longer than 2 cm . Assume that the population is normally distributed with the variance of 2 25 cm . What is the smallest sample size needed? [8]

Solution

Let's carefully analyze the problem and find the smallest sample size required.

Given Data:

  • Confidence level: 91.98%91.98\%
  • Desired margin of error EE: 2cm2 \, \text{cm}
  • Population variance σ2\sigma^2: 2×25cm=50cm22 \times 25 \, \text{cm} = 50 \, \text{cm}^2
  • Population standard deviation σ\sigma: 50cm\sqrt{50} \, \text{cm}

Step 1: Determine the z-value corresponding to the confidence level.

The confidence level is 91.98%, so the significance level α=10.9198=0.0802\alpha = 1 - 0.9198 = 0.0802.

Since the confidence interval is symmetric, α/2=0.0401\alpha/2 = 0.0401.

Now, we find the z-value corresponding to 1α/2=0.95991 - \alpha/2 = 0.9599.

This z-value is approximately: zα/21.75z_{\alpha/2} \approx 1.75

Step 2: Use the margin of error formula.

The margin of error EE for the confidence interval is given by: E=zα/2σnE = z_{\alpha/2} \cdot \frac{\sigma}{\sqrt{n}} Given that E=2cmE = 2 \, \text{cm}, σ=50cm\sigma = \sqrt{50} \, \text{cm}, and zα/2=1.75z_{\alpha/2} = 1.75, we set up the equation: 2=1.7550n2 = 1.75 \cdot \frac{\sqrt{50}}{\sqrt{n}}

Step 3: Solve for nn (sample size).

Rearrange the equation to solve for nn: n=1.75502\sqrt{n} = 1.75 \cdot \frac{\sqrt{50}}{2} n=1.757.0711212.374426.1872\sqrt{n} = \frac{1.75 \cdot 7.0711}{2} \approx \frac{12.3744}{2} \approx 6.1872

n(6.1872)238.28n \approx (6.1872)^2 \approx 38.28

Since the sample size nn must be a whole number, we round up to the next whole number.

Final Answer:

The smallest sample size needed is 39\boxed{39}.

Would you like any additional details or have any questions?


Here are five related questions you might find interesting:

  1. How does the z-value change for different confidence levels?
  2. How does increasing the desired margin of error affect the sample size?
  3. What if the variance was 4×25cm24 \times 25 \, \text{cm}^2? How would that affect the sample size?
  4. What happens to the sample size if the confidence level is increased to 95%?
  5. How do we determine the sample size if the population variance is unknown?

Tip: When calculating sample size, always ensure you round up to the next whole number to maintain the desired confidence level and margin of error.

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Math Problem Analysis

Mathematical Concepts

Confidence Intervals
Normal Distribution
Sample Size Calculation

Formulas

Margin of Error formula
Z-value calculation

Theorems

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Suitable Grade Level

Advanced Undergraduate