Question

A random sample of 25 office workers was selected to determine how much the average person spends weekly on groceries. The sample mean was equal to $260, with a sample standard deviation of $8. What is the 95 percent confidence interval of this sample? Round the answer to the nearest hundredth

Answer 1

To calculate the 95% confidence interval for the mean spending on groceries, we can use the formula for the confidence interval based on the sample mean, sample standard deviation, and the t-distribution (since the sample size is small, n < 30).

The formula for the confidence interval is:

\[
CI = \bar{x} \pm t^* \left( \frac{s}{\sqrt{n}} \right)
\]

Where:
- \(\bar{x}\) = sample mean
- \(t^*\) = t-value from the t-distribution table corresponding to the desired confidence level and degrees of freedom
- \(s\) = sample standard deviation
- \(n\) = sample size

### Given values:
- Sample mean (\(\bar{x}\)) = $260
- Sample standard deviation (\(s\)) = $8
- Sample size (\(n\)) = 25

### Step 1: Calculate degrees of freedom
Degrees of freedom (df) = \(n - 1 = 25 - 1 = 24\)

### Step 2: Find the t-value
For a 95% confidence interval and 24 degrees of freedom, the critical t-value (using a t-table or calculator) is approximately \(t^* \approx 2.064\).

### Step 3: Calculate the standard error (SE)
\[
SE = \frac{s}{\sqrt{n}} = \frac{8}{\sqrt{25}} = \frac{8}{5} = 1.6
\]

### Step 4: Calculate the margin of error (ME)
\[
ME = t^* \times SE = 2.064 \times 1.6 \approx 3.3024
\]

### Step 5: Calculate the confidence interval
\[
CI = \bar{x} \pm ME = 260 \pm 3.3024
\]

Calculating the lower and upper bounds:
- Lower bound: \(260 - 3.3024 \approx 256.70\)
- Upper bound: \(260 + 3.3024 \approx 263.30\)

### Final Result
The 95% confidence interval for the average amount spent weekly on groceries is approximately:

\[
\boxed{(256.70, 263.30)}
\]