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)
Answers
GPT-4o mini
To calculate the 95% confidence interval for the average savings account balance, we can use the formula for the confidence interval of the mean when the population standard deviation is unknown.
The formula for the confidence interval is given by:
\[
\text{CI} = \bar{x} \pm t^* \left(\frac{s}{\sqrt{n}}\right)
\]
Where:
- \(\bar{x}\) = sample mean
- \(t^*\) = t-score corresponding to the desired confidence level (from the t-distribution table)
- \(s\) = sample standard deviation
- \(n\) = sample size
### Step 1: Identify the values
- Sample mean (\(\bar{x}\)) = $18,412
- Sample standard deviation (\(s\)) = $993
- Sample size (\(n\)) = 187
### Step 2: Find the t-score for a 95% confidence interval
For a two-tailed test with 95% confidence and \(n-1\) degrees of freedom (which is 186 in this case), we can look up the t-score in a t-distribution table or the appropriate calculator.
For 186 degrees of freedom:
- The t-score (\(t^*\)) for a 95% confidence interval is approximately **1.972**.
### Step 3: Calculate the standard error (SE)
\[
SE = \frac{s}{\sqrt{n}} = \frac{993}{\sqrt{187}} \approx \frac{993}{13.646} \approx 72.8
\]
### Step 4: Calculate the margin of error (ME)
\[
ME = t^* \times SE = 1.972 \times 72.8 \approx 143.5
\]
### Step 5: Calculate the confidence interval
\[
\text{CI} = \bar{x} \pm ME = 18412 \pm 143.5
\]
So, the confidence interval is:
- Lower bound: \(18412 - 143.5 \approx 18268.5\)
- Upper bound: \(18412 + 143.5 \approx 18555.5\)
### Final Result
The 95% confidence interval for the average savings account balance is approximately \(\$18,268.50\) to \(\$18,555.50\).
The formula for the confidence interval is given by:
\[
\text{CI} = \bar{x} \pm t^* \left(\frac{s}{\sqrt{n}}\right)
\]
Where:
- \(\bar{x}\) = sample mean
- \(t^*\) = t-score corresponding to the desired confidence level (from the t-distribution table)
- \(s\) = sample standard deviation
- \(n\) = sample size
### Step 1: Identify the values
- Sample mean (\(\bar{x}\)) = $18,412
- Sample standard deviation (\(s\)) = $993
- Sample size (\(n\)) = 187
### Step 2: Find the t-score for a 95% confidence interval
For a two-tailed test with 95% confidence and \(n-1\) degrees of freedom (which is 186 in this case), we can look up the t-score in a t-distribution table or the appropriate calculator.
For 186 degrees of freedom:
- The t-score (\(t^*\)) for a 95% confidence interval is approximately **1.972**.
### Step 3: Calculate the standard error (SE)
\[
SE = \frac{s}{\sqrt{n}} = \frac{993}{\sqrt{187}} \approx \frac{993}{13.646} \approx 72.8
\]
### Step 4: Calculate the margin of error (ME)
\[
ME = t^* \times SE = 1.972 \times 72.8 \approx 143.5
\]
### Step 5: Calculate the confidence interval
\[
\text{CI} = \bar{x} \pm ME = 18412 \pm 143.5
\]
So, the confidence interval is:
- Lower bound: \(18412 - 143.5 \approx 18268.5\)
- Upper bound: \(18412 + 143.5 \approx 18555.5\)
### Final Result
The 95% confidence interval for the average savings account balance is approximately \(\$18,268.50\) to \(\$18,555.50\).