Question 1169827: For the specified margin of error and confidence level, obtain a sample size that will ensure a margin of error of at most the one specified.
margin of error=0.01; confidence level=90%
n=
Answer by CPhill(1959)   (Show Source):
You can put this solution on YOUR website! Alright, let's calculate the required sample size for the given margin of error and confidence level.
**Understanding the Problem**
We need to find the minimum sample size (n) that will ensure a margin of error of at most 0.01 with a 90% confidence level.
**Formula for Sample Size (Proportion Unknown)**
When the population proportion (p) is unknown, we use the following formula:
$$n = \left(\frac{z_{\alpha/2}}{E}\right)^2 \cdot 0.25$$
Where:
* n is the sample size.
* zα/2 is the z-critical value corresponding to the desired confidence level.
* E is the margin of error.
* 0.25 is used because it represents the maximum possible value of p(1-p), which occurs when p = 0.5.
**Steps**
1. **Find the z-critical value (zα/2):**
* For a 90% confidence level, α = 1 - 0.90 = 0.10.
* α/2 = 0.10 / 2 = 0.05.
* We need to find the z-value that corresponds to an area of 0.95 (0.90 + 0.05) in the cumulative standard normal distribution.
* Using a z-table or calculator, the z-critical value for a 90% confidence level is approximately 1.645.
2. **Plug the values into the formula:**
* n = (1.645 / 0.01)^2 * 0.25
* n = (164.5)^2 * 0.25
* n = 27060.25 * 0.25
* n = 6765.0625
3. **Round up to the nearest whole number:**
* Since we need a whole number for the sample size, and we want to ensure the margin of error is at most 0.01, we always round up.
* n = 6766
**Final Answer**
The required sample size is 6766.
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