To calculate the width of the 90% confidence interval for the population mean, we'll use the formula for the confidence interval when the population standard deviation is known. The formula is:
\[ CI = \bar{x} \pm z \left(\frac{\sigma}{\sqrt{n}}\right) \]
Where:
- \(\bar{x}\) = sample mean
- \(z\) = z-score corresponding to the desired confidence level
- \(\sigma\) = population standard deviation
- \(n\) = sample size
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Determine the parameters:
- Sample mean, \(\bar{x} = R10000\)
- Population standard deviation, \(\sigma = R500\)
- Sample size, \(n = 50\)
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Find the z-score for a 90% confidence level:
- For a 90% confidence interval, the z-score (which corresponds to the critical value) can be found using standard normal distribution tables. The z-score for 90% confidence level is approximately \(1.645\).
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Calculate the standard error (SE): \[ SE = \frac{\sigma}{\sqrt{n}} = \frac{500}{\sqrt{50}} \approx \frac{500}{7.0711} \approx 70.71 \]
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Calculate the margin of error (ME): \[ ME = z \cdot SE = 1.645 \cdot 70.71 \approx 116.67 \]
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Calculate the width of the confidence interval: The width of the confidence interval is \(2 \times ME\): \[ \text{Width} = 2 \times 116.67 \approx 233.34 \]
Thus, the width of the 90% confidence interval is approximately R233.34.