Math Problem Statement

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

Solution

We are tasked with determining the sample size (nn) 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 (EE): 44 ounces
  • Population standard deviation (σ\sigma): 99 ounces
  • We will use the standard normal distribution (z-distribution).

Step 1: Find the critical value (zα/2z_{\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 10.9999=0.00011 - 0.9999 = 0.0001. Therefore, α/2=0.00005\alpha/2 = 0.00005, and we need the zz-score that corresponds to the upper 0.00005 area.

Looking this up in standard normal tables, the zα/2z_{\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 nn is:

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

Where:

  • zα/2z_{\alpha/2} is the critical value (3.890),
  • σ\sigma is the standard deviation (99 ounces),
  • EE is the margin of error (44 ounces).

Substitute the values:

n=(3.890×9944)2n = \left( \frac{3.890 \times 99}{44} \right)^2

n=(385.1144)2n = \left( \frac{385.11}{44} \right)^2

n=(8.753)2n = (8.753)^2

n=76.61n = 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 zz-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.

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Math Problem Analysis

Mathematical Concepts

Statistics
Confidence Intervals
Sample Size Determination

Formulas

n = (z * σ / E)^2

Theorems

Z-distribution
Margin of Error Theorem

Suitable Grade Level

Undergraduate (Statistics)