To construct a 95% confidence interval for the population mean \( \mu \), we can use the formula:
\[ \text{CI} = \bar{x} \pm t^* \left( \frac{s}{\sqrt{n}} \right) \]
where:
- \( \bar{x} \) is the sample mean,
- \( t^* \) is the critical value from the t-distribution for \( n-1 \) degrees of freedom,
- \( s \) is the sample standard deviation,
- \( n \) is the sample size.
Given:
- Sample size \( n = 39 \)
- Sample mean \( \bar{x} = 55 \)
- Sample standard deviation \( s = 5.8 \)
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Calculate degrees of freedom: \[ df = n - 1 = 39 - 1 = 38 \]
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Determine the critical value \( t^ \)* for a 95% confidence level and 38 degrees of freedom. You can find this from a t-table or using software. The critical value \( t^* \approx 2.024 \).
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Calculate the standard error (SE): \[ SE = \frac{s}{\sqrt{n}} = \frac{5.8}{\sqrt{39}} \approx \frac{5.8}{6.244} \approx 0.928 \]
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Calculate the margin of error (ME): \[ ME = t^* \cdot SE \approx 2.024 \cdot 0.928 \approx 1.876 \]
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Construct the confidence interval: \[ \text{Lower limit} = \bar{x} - ME = 55 - 1.876 \approx 53.1 \] \[ \text{Upper limit} = \bar{x} + ME = 55 + 1.876 \approx 56.9 \]
Thus, the 95% confidence interval for the population mean \( \mu \) is approximately:
- Lower limit: 53.1
- Upper limit: 56.9
Finally, to summarize the answer:
- Lower limit: 53.1
- Upper limit: 56.9
- Mean: 55