Question

A random sample of 187 adults was selected from a population of a town to determine their average savings account balance. The sample mean was equal to $18,412, with a sample standard deviation of $993. What is the 95 percent confidence interval?(1 point) A. 18225-18599 B. 806-1180 C. 18269.67-18554.33 D. 17419-19405

Answer 1

To calculate the 95 percent confidence interval for the average savings account balance, we can use the formula for the confidence interval of the mean:

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

Where:
- \(\bar{x}\) = sample mean
- \(z\) = z-value corresponding to the desired confidence level (for 95%, \(z \approx 1.96\))
- \(s\) = sample standard deviation
- \(n\) = sample size

Given data:
- \(\bar{x} = 18412\)
- \(s = 993\)
- \(n = 187\)

First, calculate the standard error (SE):

\[
SE = \frac{s}{\sqrt{n}} = \frac{993}{\sqrt{187}} \approx \frac{993}{13.64} \approx 72.8
\]

Now, calculate the margin of error (ME):

\[
ME = z \cdot SE = 1.96 \cdot 72.8 \approx 142.9
\]

Now, we can find the confidence interval:

\[
\text{Lower limit} = \bar{x} - ME = 18412 - 142.9 \approx 18269.1
\]
\[
\text{Upper limit} = \bar{x} + ME = 18412 + 142.9 \approx 18554.9
\]

So the 95% confidence interval is approximately \(18269.1\) to \(18554.9\).

The closest answer to our calculated confidence interval is **C. 18269.67 - 18554.33**.