A doctor at a local hospital is interested in estimating the birth weight of infants. How large a sample must she select if she desires to be 

9999​% 

confident that her estimate is within 

44 

ounces of the true​ mean? Assume that 

sequals=99 

ounces based on earlier studies. Use the standard normal​ z-distribution in your calculations.

Question content area bottom

Part 1

A.32

3232

B.34

3434

C.6

66

D.4

4

We are tasked with determining the sample size (\( n \)) necessary for the doctor to estimate the birth weight of infants with a certain confidence level and margin of error. Here's the information provided:

- Confidence level: 99.99% (0.9999)
- Margin of error (\( E \)): 44 ounces
- Population standard deviation (\( \sigma \)): 99 ounces
- We will use the standard normal distribution (z-distribution).

### Step 1: Find the critical value (\( z_{\alpha/2} \)) for a 99.99% confidence level
For a confidence level of 99.99%, the area in both tails of the normal distribution is \( 1 - 0.9999 = 0.0001 \). Therefore, \( \alpha/2 = 0.00005 \), and we need the \( z \)-score that corresponds to the upper 0.00005 area.

Looking this up in standard normal tables, the \( z_{\alpha/2} \) value for a 99.99% confidence level is approximately 3.890.

### Step 2: Use the sample size formula
The formula for determining the sample size \( n \) is:

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

Where:
- \( z_{\alpha/2} \) is the critical value (3.890),
- \( \sigma \) is the standard deviation (99 ounces),
- \( E \) is the margin of error (44 ounces).

Substitute the values:

\[
n = \left( \frac{3.890 \times 99}{44} \right)^2
\]

\[
n = \left( \frac{385.11}{44} \right)^2
\]

\[
n = (8.753)^2
\]

\[
n = 76.61
\]

### Step 3: Round up to the nearest whole number
Since the sample size must be a whole number, round 76.61 up to 77.

Therefore, the doctor must select a sample size of **77** infants to estimate the birth weight with 99.99% confidence and a margin of error of 44 ounces.

However, none of the answer choices (32, 34, 66, 4) seem to match. It seems there might be an issue with the given data or options. Based on the calculations, the correct answer is **77**, but it's not in the provided options.

Would you like more details on the steps or any specific clarification? Here are some additional questions you can consider:

1. How is the margin of error related to the sample size in hypothesis testing?
2. Why do we use the standard normal \( z \)-distribution for confidence intervals?
3. What happens to the confidence interval if the sample size increases?
4. How does increasing the confidence level affect the sample size needed?
5. Why do we round up the sample size in these calculations?

**Tip:** When working with confidence intervals, always remember to adjust for smaller sample sizes using the t-distribution if the population standard deviation is unknown or the sample size is small.

A doctor at a local hospital is interested in estimating the birth weight of infants. How large a sample must she select if she desires to be 99.99% confident that her estimate is within 44 ounces of the true mean? Assume that the population standard deviation is 99 ounces based on earlier studies. Use the standard normal z-distribution in your calculations.

Learn how to calculate the sample size needed for a 99.99% confidence interval in estimating the birth weight of infants. This problem uses key statistical concepts such as the z-distribution and margin of error. It provides a detailed solution using the formula n = (z * σ / E)^2, and is suitable for undergraduate statistics students.

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