To construct a confidence interval for the mean amount spent on a child's last birthday gift, we can use the following formula for the confidence interval (CI) for the mean when the population standard deviation is unknown and the sample size is small (n < 30). Since we have a small sample size of 11, we will use the t-distribution.
The formula for the confidence interval is:
\[ \text{CI} = \bar{x} \pm t^* \left(\frac{s}{\sqrt{n}}\right) \]
Where:
- \(\bar{x}\) is the sample mean
- \(t^*\) is the t-value for the desired confidence level and degrees of freedom
- \(s\) is the sample standard deviation
- \(n\) is the sample size
Step 1: Identify the components
- Sample mean (\(\bar{x}\)) = $33
- Sample standard deviation (\(s\)) = $5
- Sample size (\(n\)) = 11
- Confidence level = 98%
Step 2: Find the t-value
The degrees of freedom (df) can be calculated as: \[ df = n - 1 = 11 - 1 = 10 \]
For a 98% confidence level and 10 degrees of freedom, we can look up the t-value in a t-distribution table or use a calculator. The t-value for 98% confidence with 10 df is approximately 2.228.
Step 3: Calculate the standard error (SE)
\[ SE = \frac{s}{\sqrt{n}} = \frac{5}{\sqrt{11}} \approx \frac{5}{3.317} \approx 1.506 \]
Step 4: Compute the margin of error (ME)
\[ ME = t^* \times SE = 2.228 \times 1.506 \approx 3.357 \]
Step 5: Construct the confidence interval
\[ \text{CI} = \bar{x} \pm ME = 33 \pm 3.357 \]
Calculating the lower and upper bounds:
- Lower bound: \(33 - 3.357 \approx 29.643\)
- Upper bound: \(33 + 3.357 \approx 36.357\)
Final Result
The 98% confidence interval for the mean amount spent on a child's last birthday gift is approximately:
\[ (29.64, 36.36) \]
Therefore, we are 98% confident that the true mean amount spent on birthday gifts lies between $29.64 and $36.36.
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