To calculate the margin of error (ME) at an 80% confidence level, we use the following formula:

$ME = z \times \frac{\sigma}{\sqrt{n}}$

where:

• $z$ is the z-score corresponding to the 80% confidence level,
• $\sigma$ is the standard deviation (which is 2 in this case),
• $n$ is the sample size (which is 23).

Step 1: Find the z-score for 80% confidence level.

For an 80% confidence level, the area in each tail of the standard normal distribution is $\frac{1 - 0.80}{2} = 0.10$. We look for the z-score corresponding to an area of 0.90 (i.e., 80% confidence in the middle).

From the standard normal table or a z-score calculator, the z-score for 80% confidence is approximately:

$z = 1.28$

Step 2: Calculate the margin of error.

Now, we substitute the known values into the formula:

$ME = 1.28 \times \frac{2}{\sqrt{23}} = 1.28 \times \frac{2}{4.7958} = 1.28 \times 0.417 = 0.534$

Step 3: Round the result.

Rounding to two decimal places:

$ME \approx 0.53$

So, the margin of error is approximately $0.53.

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Do you want further details or have any questions on the calculation?

Here are 5 related questions to expand your understanding:

1. How would the margin of error change if the confidence level were 95% instead of 80%?
2. What would happen to the margin of error if the sample size were increased to 50?
3. How does a larger standard deviation affect the margin of error?
4. How can the margin of error be used to interpret the results of this survey?
5. What is the difference between the margin of error and the standard error?

Tip: The margin of error helps to understand the precision of a sample statistic, and it's important to choose an appropriate confidence level for your analysis.