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 percent confidence interval for the mean spending on groceries, we will use the formula for the confidence interval based on the sample mean, standard deviation, and sample size. The formula is:

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

Where:
- \(\bar{x}\) is the sample mean
- \(z\) is the z-score corresponding to the desired confidence level
- \(s\) is the sample standard deviation
- \(n\) is the sample size

For a 95 percent confidence interval, the z-score is approximately 1.96.

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

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

2. Calculate the margin of error (ME):
\[
ME = z \cdot SE = 1.96 \cdot 1.6 = 3.136
\]

3. Determine the confidence interval:
\[
\text{Lower bound} = \bar{x} - ME = 260 - 3.136 = 256.864
\]
\[
\text{Upper bound} = \bar{x} + ME = 260 + 3.136 = 263.136
\]

Thus, the 95 percent confidence interval is:
\[
(256.864, 263.136)
\]

Rounding to the nearest hundredth, the confidence interval is:
\[
\boxed{(256.86, 263.14)}
\]