Question

In a survey, 22 people were asked how much they spent on their child's last birthday gift. The results were roughly bell-shaped with a mean of $48 and standard deviation of $10. Construct a confidence interval at a 80% confidence level.

Give your answers to one decimal place.

Answer 1

To construct an 80% confidence interval for the mean spending on a child's birthday gift, we can use the formula for the confidence interval:

\[
\text{Confidence Interval} = \bar{x} \pm z \left(\frac{s}{\sqrt{n}}\right)
\]

Where:
- \(\bar{x}\) is the sample mean.
- \(z\) is the z-score corresponding to the desired confidence level.
- \(s\) is the sample standard deviation.
- \(n\) is the sample size.

Given:
- \(\bar{x} = 48\)
- \(s = 10\)
- \(n = 22\)

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

For an 80% confidence level, the area in each tail is \( (1 - 0.80) / 2 = 0.10 \). Looking up the z-value that corresponds to 0.10 in the right tail, we find:

\[
z \approx 1.28
\]

### Step 2: Calculate the standard error (SE).

\[
\text{SE} = \frac{s}{\sqrt{n}} = \frac{10}{\sqrt{22}} \approx \frac{10}{4.690} \approx 2.13
\]

### Step 3: Calculate the margin of error (ME).

\[
\text{ME} = z \cdot \text{SE} \approx 1.28 \cdot 2.13 \approx 2.73
\]

### Step 4: Construct the confidence interval.

\[
\text{Confidence Interval} = 48 \pm 2.73
\]

Calculating the lower and upper bounds:

\[
\text{Lower Bound} = 48 - 2.73 \approx 45.3
\]
\[
\text{Upper Bound} = 48 + 2.73 \approx 50.7
\]

### Final Result:

The 80% confidence interval for the amount spent on a child's last birthday gift is:

\[
\boxed{(45.3, 50.7)}
\]