To construct a 99% confidence interval for the average amount spent on a child’s birthday gift, we'll use the formula:

$\text{Confidence Interval} = \text{mean} \pm (z \times \frac{\sigma}{\sqrt{n}})$

where:

• mean = $34
• standard deviation ($\sigma$) = $9
• sample size ($n$) = 14
• z-score for a 99% confidence level = 2.576 (for a two-tailed test)

Step 1: Calculate the Standard Error

$\text{Standard Error (SE)} = \frac{\sigma}{\sqrt{n}} = \frac{9}{\sqrt{14}} \approx 2.41$

Step 2: Multiply by the z-score

$\text{Margin of Error (ME)} = 2.576 \times 2.41 \approx 6.2$

Step 3: Construct the Confidence Interval

$\text{Confidence Interval} = 34 \pm 6.2$

This results in: $\text{Confidence Interval} = (34 - 6.2, 34 + 6.2) = (27.8, 40.2)$

Final Answer

The 99% confidence interval is: $34 \pm 6.2 \quad \text{or} \quad (27.8, 40.2)$

Would you like further clarification on any part of this process?

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Here are some related questions for further exploration:

1. What would the confidence interval be if the confidence level were 95% instead?
2. How does increasing the sample size affect the confidence interval?
3. What would happen to the confidence interval if the standard deviation were higher?
4. Why do we use the z-score rather than the t-score in this scenario?
5. What does a 99% confidence interval imply about the survey results?

Tip: When constructing confidence intervals, always check if the sample size is large enough or if the population is normally distributed to decide between using z-scores or t-scores.