M E

To calculate the margin of error (ME) for the given data, we can use the formula for the margin of error at a confidence level:

$ME = Z_{\alpha/2} \times \frac{\sigma}{\sqrt{n}}$

Where:

• $Z_{\alpha/2}$ is the z-value corresponding to the 80% confidence level,
• $\sigma$ is the standard deviation,
• $n$ is the sample size.

Step 1: Find the Z-value for an 80% confidence level

For an 80% confidence level, the remaining 20% is split between both tails of the normal distribution, leaving 10% (0.10) in each tail. From the Z-table, the critical z-value for 80% confidence is approximately 1.28.

Step 2: Apply the formula

Given:

• $\sigma = 5$,
• $n = 19$,
• $Z_{\alpha/2} = 1.28$,

The margin of error is calculated as:

$ME = 1.28 \times \frac{5}{\sqrt{19}} \approx 1.28 \times 1.148 = 1.470$

Final Answer:

$ME \approx \pm 1.47$

Would you like a more detailed explanation of any step or have any other questions?

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Here are 5 related questions you might find useful:

1. How is the z-value determined for different confidence levels?
2. What factors influence the margin of error in a survey?
3. How does sample size affect the margin of error?
4. What is the relationship between confidence level and the width of a confidence interval?
5. How can the margin of error be reduced in surveys?

Tip: The margin of error decreases as the sample size increases, making your estimates more precise.