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A sample of 49 eggs yields a mean weight of 1.69 ounces. Assuming that , find the margin of error in estimating μ at the 95% level of confidence.

A. 0.11oz
B. 0.13oz
C. 0.02oz
D. 0.47oz

User Lardois
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1 Answer

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Final answer:

The question asks for the margin of error at a 95% confidence level, which requires the standard deviation and the sample size to calculate using the formula Margin of Error = z * (σ/√n), but the standard deviation is missing.

Step-by-step explanation:

The question deals with the concept of margin of error in the context of constructing a 95% confidence interval for the mean weight of a population (eggs) using a sample. The margin of error can be calculated using the standard deviation and the sample size along with the z-score associated with the desired level of confidence.

Since the population standard deviation (σ) is known and the sample size (n) is 49, we can utilize the z-score for a 95% confidence level (which is approximately 1.96) to calculate the margin of error using the formula Margin of Error = z * (σ/√n). The standard deviation is not provided in the question. However, in a similar example given where the sample mean is 2 ounces, and standard deviation for a different scenario is known, which is 0.1 ounce, this is indicative that additional information might be needed to solve the problem accurately.

User Audie
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