Question

You measure the weights of 47 watermelons and find they have a mean weight of 77 ounces. Assume the population standard deviation is 14.4 ounces Based on this , construct a 95% confidence interval for the true population mean watermelon weight.

____<μ<___ , use 3 decimal places.

State in words what this means in context of the problem:

Answer 1

At 95% confidence level, the critical value is z_0.025 = 1.96

The 95% confidence interval is

Answer 2

To construct a 95% confidence interval for the true population mean watermelon weight, we can use the sample mean and the population standard deviation.

Given: Sample size (n) = 47 Sample mean (x̄) = 77 ounces Population standard deviation (σ) = 14.4 ounces

To calculate the confidence interval, we can use the formula: Confidence Interval = x̄ ± Z * (σ / √n)

Where: x̄ is the sample mean Z is the Z-score corresponding to the desired confidence level (95% confidence level corresponds to Z = 1.96) σ is the population standard deviation n is the sample size

Plugging in the values, we have: Confidence Interval = 77 ± 1.96 * (14.4 / √47)

Calculating the values: Confidence Interval = 77 ± 1.96 * (14.4 / √47) ≈ 77 ± 4.387

Rounding to 3 decimal places, the confidence interval is approximately: 73.613 < μ < 80.387

In the context of the problem, this means that we are 95% confident that the true population mean watermelon weight falls within the range of 73.613 to 80.387 ounces based on the given sample.