Question

Use the given degree of confidence and sample data to construct a confidence interval for the population mean μ. Assume that the population has a normal distribution.

A savings and loan association needs information concerning the checking account balances of its local customers. A random sample of 14 accounts was checked and yielded a mean balance of $664.14 and a standard deviation of $297.29. Find a 98% confidence interval for the true mean checking account balance for local customers.

A) $455.65 < μ < $872.63
B) $492.52 < μ < $835.76
C) $493.71 < μ < $834.57
D) $453.59 < μ < $874.69

Answer 1

To construct a confidence interval for the population mean \(\mu\) based on the sample data provided, we can use the formula for the confidence interval when the population standard deviation is unknown and the sample size is small (which is our case, as \(n = 14\)). We will use the t-distribution.

### Step 1: Identify the given data
- Sample size (\(n\)) = 14
- Sample mean (\(\bar{x}\)) = $664.14
- Sample standard deviation (\(s\)) = $297.29
- Confidence level = 98%

### Step 2: Find the t-critical value
Since the confidence level is 98%, the alpha level (\(\alpha\)) is:
\[
\alpha = 1 - 0.98 = 0.02
\]
For a two-tailed test, the critical value will be \(\alpha/2 = 0.01\).

Degrees of freedom (\(df\)):
\[
df = n - 1 = 14 - 1 = 13
\]

Using a t-table, or an online calculator, we find the t-critical value for:
- \(df = 13\)
- \(0.01\) (one-tailed for 98% confidence)

The t-critical value \(t_{0.01, 13} \approx 2.650\).

### Step 3: Calculate the margin of error (ME)
Using the formula for the margin of error:
\[
ME = t \times \frac{s}{\sqrt{n}}
\]

Substituting the known values:
\[
ME = 2.650 \times \frac{297.29}{\sqrt{14}}
\]
Calculating \(\sqrt{14} \approx 3.7417\):
\[
ME \approx 2.650 \times \frac{297.29}{3.7417} \approx 2.650 \times 79.43 \approx 210.85
\]

### Step 4: Construct the confidence interval
The confidence interval is calculated as:
\[
\text{CI} = \bar{x} \pm ME
\]

Calculating the lower and upper bounds:
\[
\text{Lower bound} = 664.14 - 210.85 \approx 453.29
\]
\[
\text{Upper bound} = 664.14 + 210.85 \approx 874.99
\]

Thus, the confidence interval is approximately:
\[
453.29 < \mu < 874.99
\]

### Conclusion
Among the given options, the closest match is:
**D) \(453.59 < μ < 874.69\)**