A scientist studying babies born prematurely would like to obtain an estimate for the mean birth weight, u, of babies born during the 24th week of the gestation period. She plans to select a random sample of birth weights of such babies and use the mean of the sample to estimate u. Assuming that the population of birth weights of babies born during the 24th week has a standard deviation of 2.5 pounds, what is the minimum sample size needed for the scientist to be 90% confident that her estimate is within  0.6 pounds of u?
Carry your intermediate computations to at least three decimal places. Write your answer as a whole number (and make sure that it is the minimum whole number that satisfies the requirements).

To determine the minimum sample size needed for the scientist to be 90% confident that her estimate is within 0.6 pounds of the mean birth weight, we need to use the formula for the sample size \( n \) in estimating a population mean with a specified margin of error \( E \):

\[ n = \left( \frac{z \cdot \sigma}{E} \right)^2 \]

where:
- \( z \) is the z-value corresponding to the desired confidence level,
- \( \sigma \) is the population standard deviation,
- \( E \) is the margin of error.

Given:
- Confidence level = 90%
- Population standard deviation \( \sigma = 2.5 \) pounds
- Margin of error \( E = 0.6 \) pounds

First, we find the z-value for a 90% confidence level. The z-value corresponds to the critical value that leaves 5% in each tail of the standard normal distribution (since 90% confidence level leaves 10% total in the two tails, 5% in each). We can look up this value in a standard normal distribution table or use a calculator.

The z-value for 90% confidence is approximately 1.645.

Now, we can plug in the values into the sample size formula:

\[ n = \left( \frac{1.645 \cdot 2.5}{0.6} \right)^2 \]

Calculating the intermediate steps:

\[ \frac{1.645 \cdot 2.5}{0.6} = \frac{4.1125}{0.6} \approx 6.854 \]

Next, we square this value:

\[ (6.854)^2 \approx 47.001 \]

Since the sample size \( n \) must be a whole number, we round up to the next whole number to ensure the margin of error requirement is met:

\[ n \approx 48 \]

Therefore, the minimum sample size needed is:

\[ \boxed{48} \]

Would you like details or have any questions?

Here are 8 related questions that expand on this information:
1. What is the margin of error if the sample size is increased to 100?
2. How does changing the confidence level to 95% affect the required sample size?
3. What is the impact of a larger standard deviation on the required sample size?
4. Can you explain the concept of a confidence interval in the context of this problem?
5. How is the z-value determined for different confidence levels?
6. What assumptions are made in this sample size calculation?
7. How would you calculate the sample size if the standard deviation was not known?
8. What is the significance of using a random sample in this study?

**Tip:** Always round up the sample size to ensure the desired margin of error is not exceeded.

Learn how to calculate the minimum sample size needed to estimate the mean birth weight of babies born during the 24th week of gestation with a specified margin of error. This problem involves statistical concepts including confidence intervals, sample size calculations, and standard deviation. Suitable for advanced undergraduate students.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation