In a survey, 19 people were asked how much they spent on their child's last birthday gift. The results were roughly bell-shaped with a mean of $46 and standard deviation of $5. Find the margin of error at a 80% confidence level.

Give your answer to two decimal places.

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.

In a survey, 19 people were asked how much they spent on their child's last birthday gift. The results were roughly bell-shaped with a mean of $46 and standard deviation of $5. Find the margin of error at a 80% confidence level. Give your answer to two decimal places.

Learn how to calculate the margin of error for a survey where 19 people were asked about birthday gift expenses. The problem involves using a normal distribution with a mean of $46 and a standard deviation of $5. Find the margin of error at an 80% confidence level.

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