Math Problem Statement

An educational researcher is interested in estimating the average GPA of all UCF students. The GPA is collected from 150 random students. A 98% confidence interval for µ is: (2.71, 3.01). The researcher wants to reduce the margin of error to 0.1. How many students should be sampled to be 98% confident that the true mean is within 0.1 of the sample mean? Assume the standard deviation is 0.78.

Solution

To find the number of students that should be sampled to reduce the margin of error to 0.1, we can use the formula for the margin of error in a confidence interval for the mean:

E=Zα/2×σnE = Z_{\alpha/2} \times \frac{\sigma}{\sqrt{n}}

Where:

  • EE is the margin of error.
  • Zα/2Z_{\alpha/2} is the critical value for the desired confidence level.
  • σ\sigma is the population standard deviation.
  • nn is the sample size.

Given:

  • Desired margin of error E=0.1E = 0.1
  • Confidence level = 98%, so Zα/2Z_{\alpha/2} corresponds to 98% confidence.
  • σ=0.78\sigma = 0.78

Step 1: Find the critical value Zα/2Z_{\alpha/2} for a 98% confidence level.

A 98% confidence level corresponds to α=0.02\alpha = 0.02. The critical value Zα/2Z_{\alpha/2} can be found using a Z-table or standard normal distribution.

For a 98% confidence level: Zα/22.33Z_{\alpha/2} \approx 2.33

Step 2: Use the formula for the margin of error to solve for nn.

0.1=2.33×0.78n0.1 = 2.33 \times \frac{0.78}{\sqrt{n}}

Step 3: Solve for nn.

n=2.33×0.780.1\sqrt{n} = \frac{2.33 \times 0.78}{0.1}

n=1.81740.1=18.174\sqrt{n} = \frac{1.8174}{0.1} = 18.174

n=(18.174)2n = (18.174)^2

n330.31n \approx 330.31

Since nn must be an integer, we round up to the nearest whole number:

n=331n = 331

Conclusion:

The researcher should sample 331 students to be 98% confident that the true mean is within 0.1 of the sample mean.


Would you like more details on the steps? Here are some related questions to consider:

  1. What does the margin of error represent in a confidence interval?
  2. How is the critical value Zα/2Z_{\alpha/2} determined?
  3. Why do we round up the sample size nn to the nearest whole number?
  4. How does changing the confidence level affect the required sample size?
  5. What is the impact of the population standard deviation σ\sigma on the margin of error?
  6. How would the required sample size change if the desired margin of error was smaller or larger?
  7. What is the difference between a confidence interval and a prediction interval?
  8. Can the margin of error be reduced without increasing the sample size?

Tip: Increasing the sample size is a common method to reduce the margin of error, but improving data collection techniques can also enhance precision without necessarily requiring a larger sample.