ช. 1 Homework
Score: $1 / 7 \quad 1 / 7$ answered
Question 2
You measure 33 dogs' weights, and find they have a mean weight of 63 ounces. Assume the population standard deviation is 7.4 ounces. Based on this, what is the maximal margin of error associated with a confidence interval for the true population mean dog weight.
Give your answer as a decimal, to two places
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ounces
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Step 1 ：The question is asking for the maximal margin of error associated with a confidence interval for the true population mean dog weight. The margin of error can be calculated using the formula for the confidence interval, which is: \(E = Z \times \frac{\sigma}{\sqrt{n}}\)

Step 2 ：In this formula: E is the margin of error, Z is the Z-score, which depends on the desired level of confidence (for a 95% confidence interval, Z is approximately 1.96), \(\sigma\) is the population standard deviation, and n is the sample size.

Step 3 ：In this case, we know that \(\sigma = 7.4\) ounces, and n = 33. We are not given a desired level of confidence, but the question asks for the maximal margin of error, so we should use the Z-score for a 99% confidence interval, which is approximately 2.58.

Step 4 ：Let's plug these values into the formula and calculate the margin of error: \(E = 2.58 \times \frac{7.4}{\sqrt{33}}\)

Step 5 ：The calculation gives us the maximal margin of error to be approximately 3.32 ounces. This means that the true population mean dog weight is likely to be within 3.32 ounces of the sample mean weight of 63 ounces, with a 99% level of confidence.

Step 6 ：Final Answer: The maximal margin of error associated with a confidence interval for the true population mean dog weight is approximately \(\boxed{3.32}\) ounces.