To construct a 98% confidence interval for the population mean (μ), we can use the formula for the confidence interval when the population standard deviation is unknown and the sample size is large (n > 30), which allows us to use the t-distribution.
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Identify the sample statistics:
- Sample mean (\(\bar{x}\)) = 21 ounces
- Sample standard deviation (s) = 3.0 ounces
- Sample size (n) = 130
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Determine the critical value: For a 98% confidence level and a sample size of 130, we need to find the t-score that corresponds to \(\alpha/2\) (where \(\alpha = 0.02\)) for \(n - 1 = 129\) degrees of freedom. However, for large sample sizes, the z-score can be used as an approximation. The z-score for a 98% confidence level can be found using a z-table or normal distribution calculator.
The z-score for 98% confidence is approximately 2.33.
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Calculate the standard error (SE): \[ SE = \frac{s}{\sqrt{n}} = \frac{3.0}{\sqrt{130}} \approx \frac{3.0}{11.401} \approx 0.263 \]
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Calculate the margin of error (ME): \[ ME = z \times SE \approx 2.33 \times 0.263 \approx 0.614 \]
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Construct the confidence interval: \[ \text{Confidence interval} = \bar{x} \pm ME \] \[ \text{Confidence interval} = 21 \pm 0.614 \] This results in:
- Lower limit: \(21 - 0.614 \approx 20.386\)
- Upper limit: \(21 + 0.614 \approx 21.614\)
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Round and present the final confidence interval: Rounding to the same number of decimal places as the sample mean (which is 21 ounces), we get: \[ 20.4 \text{ oz} < μ < 21.6 \text{ oz} \]
Now, let's compare our result with the available options:
A) 20 oz < μ < 22 oz (This interval does include our result.) B) 21 oz < μ < 23 oz C) 19 oz < μ < 21 oz D) 20 oz < μ < 23 oz
The best option that contains our calculated interval (20.4 oz < μ < 21.6 oz) is:
A) 20 oz < μ < 22 oz
Thus, the answer is A.
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