To calculate the 95 percent confidence interval for the average savings account balance, we can use the formula for the confidence interval for the mean when the population standard deviation is unknown:
\[ \text{Confidence interval} = \bar{x} \pm t \times \frac{s}{\sqrt{n}} \]
Where:
- \(\bar{x}\) is the sample mean,
- \(t\) is the t-value for the desired confidence level (based on degrees of freedom),
- \(s\) is the sample standard deviation,
- \(n\) is the sample size.
- Sample Mean (\(\bar{x}\)) = $18,412
- Sample Standard Deviation (s) = $993
- Sample Size (n) = 187
First, we need to find the t-value. The degrees of freedom (df) is calculated as:
\[ \text{df} = n - 1 = 187 - 1 = 186 \]
For a 95% confidence level and 186 degrees of freedom, we can find the t-value using a t-table or calculator. The t-value for 95% confidence level with df = 186 is approximately 1.972.
Now we can calculate the standard error (SE):
\[ SE = \frac{s}{\sqrt{n}} = \frac{993}{\sqrt{187}} \approx \frac{993}{13.646} \approx 72.77 \]
Next, we can calculate the margin of error (ME):
\[ ME = t \times SE = 1.972 \times 72.77 \approx 143.34 \]
Now we can find the confidence interval:
\[ \text{Lower bound} = \bar{x} - ME = 18,412 - 143.34 \approx 18,268.66 \] \[ \text{Upper bound} = \bar{x} + ME = 18,412 + 143.34 \approx 18,555.34 \]
Thus, the 95% confidence interval for the average savings account balance is approximately:
\[ \text{Confidence interval} \approx 18,268.66 - 18,555.34 \]
Among the provided options, the closest interval is:
$18,269.67–$18,554.33
So the final answer is:
$18,269.67–$18,554.33