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To construct a 99% confidence interval for the mean amount spent, given that we have a sample mean ($\bar{x}$) of $47, sample standard deviation (s) of $16, and a sample size (n) of 13, we'll use the following steps:

### Step 1: Determine the $ t $-value
Since the sample size is small (n = 13), and we don’t know the population standard deviation, we use the t-distribution. For a 99% confidence level and $ n - 1 = 12 $ degrees of freedom, we find the critical $ t $-value ($ t_{0.005, 12} $).

Looking up this value, we get:
\[
t_{0.005, 12} \approx 3.055
\]

### Step 2: Calculate the Standard Error (SE)
The standard error (SE) of the mean is given by:
\[
\text{SE} = \frac{s}{\sqrt{n}} = \frac{16}{\sqrt{13}} \approx 4.44
\]

### Step 3: Compute the Margin of Error (ME)
The margin of error (ME) is calculated as:
\[
\text{ME} = t \times \text{SE} = 3.055 \times 4.44 \approx 13.57
\]

### Step 4: Construct the Confidence Interval
The confidence interval is given by:
\[
\bar{x} \pm \text{ME} = 47 \pm 13.6
\]

So, the 99% confidence interval is approximately:
\[
(33.4, 60.6)
\]

### Final Answer:
- Lower limit: 33.4
- Upper limit: 60.6

Would you like further details or have any questions?

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Here are 5 related questions to deepen understanding:

1. How does the confidence level affect the width of the confidence interval?
2. What is the impact of sample size on the standard error and confidence interval?
3. Why do we use the t-distribution instead of the z-distribution in this case?
4. How would the confidence interval change if we were only using a 95% confidence level?
5. What would happen to the confidence interval if the standard deviation was larger?

**Tip:** For small sample sizes, especially when the population standard deviation is unknown, use the t-distribution to calculate confidence intervals.

In a survey, 13 people were asked how much they spent on their child’s last birthday gift. The results were roughly bell-shaped with a mean of $47 and standard deviation of $16. Construct a confidence interval at a 99% confidence level.

Calculate a 99% confidence interval for the mean spending on a child's birthday gift based on survey data. The solution involves using the sample mean of $47, standard deviation of $16, and t-distribution for a sample size of 13.

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