You measure 34 watermelons' weights, and find they have a mean weight of 74 ounces. Assume the population standard deviation is 11.1 ounces.
Based on this, what is the margin of error associated with a 95% confidence interval for the true population mean watermelon weight.
Round your answer to two decimal places.
To calculate the margin of error associated with a 95% confidence interval for the true population mean watermelon weight, we can use the formula:
Margin of Error = Z * (Standard Deviation / sqrt(Sample Size))
Where:
Z is the z-score associated with the desired confidence interval
Standard Deviation is the population standard deviation
Sample Size is the number of watermelons measured
For a 95% confidence interval, the z-score is 1.96 (which corresponds to the two-tailed test).
So, plugging in the given values:
Z = 1.96
Standard Deviation = 11.1 ounces
Sample Size = 34 watermelons
Margin of Error = 1.96 * (11.1 / sqrt(34))
To calculate this, we need to take the square root of 34:
sqrt(34) ≈ 5.83
Now, plugging in the values:
Margin of Error ≈ 1.96 * (11.1 / 5.83)
Simplifying the expression:
Margin of Error ≈ 1.96 * 1.903
Finally, calculating the result:
Margin of Error ≈ 3.727
Therefore, the margin of error associated with a 95% confidence interval for the true population mean watermelon weight is approximately 3.73 ounces (rounded to two decimal places).