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) Responses $17,419–$19,405 $17,419–$19,405 $18,225–$18,599 $18,225–$18,599 $18,269.67–$18,554.33 $18,269.67–$18,554.33 $806–$1,180

Answer 1

To calculate the 95 percent confidence interval for the average savings account balance, we can use the following formula for confidence intervals when dealing with a sample mean:

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

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

First, we calculate the standard error (SE):

\[
SE = \frac{s}{\sqrt{n}} = \frac{993}{\sqrt{187}} \approx \frac{993}{13.645} \approx 72.8
\]

Next, we calculate the margin of error (ME):

\[
ME = z \cdot SE \approx 1.96 \cdot 72.8 \approx 142.81
\]

Now we can find the confidence interval:

\[
\text{Lower limit} = \bar{x} - ME \approx 18412 - 142.81 \approx 18269.19
\]
\[
\text{Upper limit} = \bar{x} + ME \approx 18412 + 142.81 \approx 18554.81
\]

Thus, the 95 percent confidence interval is:

\[
(18269.19, 18554.81)
\]

Rounding to two decimal places, the final confidence interval is approximately:

\[
(18269.67, 18554.33)
\]

So the correct response is:

**$18,269.67–$18,554.33**