To construct a 90% confidence interval for the true average nicotine content of the cigarettes, we can use the formula for the confidence interval (CI) when the sample size is small (n < 30) and the population standard deviation is unknown. In this case, we will use the t-distribution.
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Identify the sample statistics:
- Sample mean (\(\bar{x}\)) = 2.6 mg
- Sample standard deviation (s) = 0.9 mg
- Sample size (n) = 8
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Determine the critical value: Since the confidence level is 90%, we can find the critical value \( t \) for a 90% confidence level and 7 degrees of freedom (n - 1 = 8 - 1 = 7).
Using a t-table or calculator, we find:
- The critical value \( t \) for 90% confidence level with 7 degrees of freedom is approximately 1.895.
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Calculate the standard error (SE): \[ SE = \frac{s}{\sqrt{n}} = \frac{0.9}{\sqrt{8}} \approx \frac{0.9}{2.828} \approx 0.3182 \]
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Calculate the margin of error (ME): \[ ME = t \times SE = 1.895 \times 0.3182 \approx 0.603 \]
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Construct the confidence interval: \[ CI = \bar{x} \pm ME = 2.6 \pm 0.603 \] This gives us:
- Lower limit: \( 2.6 - 0.603 \approx 1.997 \)
- Upper limit: \( 2.6 + 0.603 \approx 3.203 \)
Thus, the 90% confidence interval for the true average nicotine content of this brand of cigarettes is approximately (1.997 mg, 3.203 mg).
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