To calculate a 95% confidence interval for the population mean when the population standard deviation is known, we can use the formula:
\[ \text{CI} = \bar{x} \pm z \left( \frac{\sigma}{\sqrt{n}} \right) \]
where:
- \(\bar{x}\) is the sample mean,
- \(z\) is the z-score corresponding to the desired level of confidence,
- \(\sigma\) is the population standard deviation,
- \(n\) is the sample size.
Given:
- Sample mean \(\bar{x} = 68\)
- Population standard deviation \(\sigma = 3\)
- Sample size \(n = 36\)
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First, find the z-score for a 95% confidence level. The z-score for 95% confidence is approximately \(z = 1.96\).
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Calculate the standard error (SE): \[ SE = \frac{\sigma}{\sqrt{n}} = \frac{3}{\sqrt{36}} = \frac{3}{6} = 0.5 \]
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Now calculate the margin of error (ME): \[ ME = z \times SE = 1.96 \times 0.5 = 0.98 \]
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Finally, construct the confidence interval: \[ \text{Lower limit} = \bar{x} - ME = 68 - 0.98 = 67.02 \] \[ \text{Upper limit} = \bar{x} + ME = 68 + 0.98 = 68.98 \]
Thus, the 95% confidence interval estimate for the population mean of exam scores is:
\[ (67.02, 68.98) \]
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