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