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You measure a random sample of n = 29 watermelons' weights, and find they have a mean weight of X = 56 ounces. Assume the population standard deviation is known to be a = 4.1 ounces. Based on this, construct a 99% confidence interval for the true (unknown) population mean watermelon weight . Give your answer in the form of estimate margin of error to two decimal places.

User Lion
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Final answer:

To construct the 99% confidence interval for watermelon weights with a sample mean of 56 ounces and population standard deviation of 4.1 ounces from a sample size of 29, we use the z-distribution to find the margin of error 1.96, resulting in the interval 56 ± 1.96 ounces.

Step-by-step explanation:

Constructing a 99% Confidence Interval

To construct a 99% confidence interval for the true population mean weight of watermelons when the sample mean is 56 ounces and the population standard deviation is 4.1 ounces, with a sample size of n = 29, we utilize the z-distribution as the population standard deviation (σ) is known.

The formula for the confidence interval is given by:
X ± (z * (σ / √n))

Firstly, we must find the z-value that corresponds to a 99% confidence level. Using a z-table or calculator, we find that the z-value is approximately 2.576. The margin of error (MOE) is then calculated as:
MOE = z * (σ / √n)

Plugging in the values we get:
MOE = 2.576 * (4.1 / √29)
MOE ≈ 2.576 * (0.76196)
MOE ≈ 1.96

The 99% confidence interval is:
56 ± 1.96 ounces

The final step is to report this interval as an estimate plus or minus the margin of error. Therefore, the confidence interval for the true mean weight of the watermelons is 56 ± 1.96 ounces.

User Sugunan
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