To calculate the 99% confidence interval for the population mean \(\mu\), we can use the formula for the confidence interval based on the sample mean, the critical value from the standard normal distribution (Z-distribution), and the standard error of the mean.
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Calculate the standard error (SE): \[ SE = \frac{s}{\sqrt{n}} \] where \(s\) is the sample standard deviation, and \(n\) is the sample size.
Given:
- \(s = 10.7\)
- \(n = 45\)
\[ SE = \frac{10.7}{\sqrt{45}} \approx \frac{10.7}{6.708} \approx 1.592 \]
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Find the critical value for a 99% confidence level: For a two-tailed test at a 99% confidence level, we look up the Z-score that corresponds to an area of \(0.005\) in each tail (since \(0.01/2 = 0.005\)). This critical value is approximately \(Z \approx 2.576\).
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Calculate the margin of error (ME): \[ ME = Z \times SE \] \[ ME = 2.576 \times 1.592 \approx 4.102 \]
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Construct the confidence interval: The confidence interval can be calculated using the sample mean and the margin of error. \[ \text{Confidence Interval} = (\bar{x} - ME, \bar{x} + ME) \] Where \(\bar{x} = 78.2\).
\[ \text{Lower limit} = 78.2 - 4.102 \approx 74.1 \] \[ \text{Upper limit} = 78.2 + 4.102 \approx 82.3 \]
Thus, the 99% confidence interval for the population mean \(\mu\) is:
\[ \text{99% C.I.} = (74.1, 82.3) \]