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Chapter 7: Problem 38

Who Give Birth An epidemiologist plans to conduct a survey to estimate the percentage of women who give birth. How many women must be surveyed in order to be \(99 \%\) confident that the estimated percentage is in error by no more than two percentage points? a. Assume that nothing is known about the percentage to be estimated. b. Assume that a prior study conducted by the U.S. Census Bureau showed that \(82 \%\) of women give birth. c. What is wrong with surveying randomly selected adult women?

Short Answer

Expert verified
a. The sample size needed is 4161 women. b. The sample size needed is 869 women. c. Randomly sampling all adult women may include those who cannot give birth, skewing results.

Step by step solution

01

Determine the Confidence Level and Margin of Error

The first step in estimating the sample size is to determine the confidence level and the margin of error. In this case, the confidence level is 99%, and the margin of error is 2 percentage points or 0.02.
02

- When Nothing is Known

If nothing is known about the percentage to be estimated, assume the worst-case scenario where the sample proportion, \( p \), is 0.5. Use the formula for the sample size \( n \) in estimating a proportion: \[ n = \frac{{Z^2 \cdot p(1 - p)}}{{E^2}} \] where \( Z \) is the Z-value for the confidence level (2.576 for 99%), \( p \) is the estimated proportion (0.5), and \( E \) is the margin of error (0.02). Thus, \[ n = \frac{{(2.576)^2 \cdot 0.5 \cdot 0.5}}{{0.02^2}} \] Calculate the result to find the required sample size.
03

- Use Prior Study Results

When a prior study shows that 82% (or 0.82) of women give birth, use this value instead of 0.5. The formula remains the same: \[ n = \frac{{Z^2 \cdot p(1 - p)}}{{E^2}} \] Substitute \( Z \) with 2.576, \( p \) with 0.82, and \( E \) with 0.02: \[ n = \frac{{(2.576)^2 \cdot 0.82 \cdot (1 - 0.82)}}{{0.02^2}} \] Calculate the result to find the required sample size.
04

- Identify Issues with Sampling Method

When randomly surveying adult women, it is possible that the sample may include women who are not of childbearing age or due to other demographic factors. This can lead to biased results that do not accurately reflect the target population of women who are of childbearing capability.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Confidence Level
In statistics, the confidence level indicates the degree of certainty that your survey results fall within the specified margin of error. For example, a 99% confidence level means that if you were to conduct 100 identical surveys, the results of 99 of them would fall within the margin of error of the true population parameter. Confidence levels are typically represented as a percentage and common values are 90%, 95%, and 99%. For this exercise, a 99% confidence level is used which corresponds to a Z-value of 2.576. The higher the confidence level, the larger the sample size needed to achieve it while maintaining the same margin of error.
Margin of Error
The margin of error represents the range that your survey results are expected to fall within, considering the true population parameter. It reflects the amount of random sampling error in a survey's results. For example, a 2% margin of error means the survey results could fall within plus or minus 2 percentage points of the true population proportion. In the provided problem, a margin of error of 2% (or 0.02) is specified, meaning the estimated proportion should be accurate within 2 percentage points. The formula to incorporate margin of error in determining sample size ensures that your survey yields reliable and precise results.
Proportion Estimation

Proportion estimation in statistics aims to determine the ratio of a particular outcome within a population.
In this exercise, we are estimating the proportion of women who give birth.
Proportion is often denoted by \( p \)
.

When nothing is known about the proportion, assuming a 50% split (p = 0.5) gives the maximum possible variance (and thus the largest required sample size).
A prior study indicating 82% sets p = 0.82.
This reduces the required sample size since variability is less than at 50%
.
The formula is given by:

n = \(\frac{{(Z^2 \cdot p(1 - p))}}{{E^2}}.\)Here Z is the Z-value (2.576 for 99%) and E is the margin of error (0.02 in this example).
Sampling Bias
Sampling bias occurs when some members of a population are more likely to be included in the sample than others. This leads to results that are not representative of the entire population and hence, not valid. In the problem, randomly surveying adult women without considering age or childbearing capability can introduce bias. It may include women who are not of childbearing age, skewing the survey results. To avoid bias, you should focus on the target population that matches the criteria you are studying. For this survey on women who give birth, you should selectively sample women who are of childbearing age.

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