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
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Calculate the standard error (SE): \[ SE = \frac{s}{\sqrt{n}} = \frac{8}{\sqrt{25}} = \frac{8}{5} = 1.6 \]
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Calculate the margin of error (ME): \[ ME = z \cdot SE = 1.96 \cdot 1.6 = 3.136 \]
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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: \[ \boxed{(256.86, 263.14)} \]