Computing a Margin of Error: Confidence Interval for a Population Mean

Question

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.

Answer 1

Whenever calculating confidence intervals or margins of error one must always ask oneself, what is the required distribution and what is the required statistic.

Since we have a 'sample' mean we must use the student's t distribution.

Since we are finding a CI for a 'population mean', we use the standard deviation of the mean. (note: this is occasionally called standard error in some older texts)

Using a t distribution the critical t factor for α/2 (where α is 1-deg of confidence or 95%)

with 33 degrees of freedom we find t_crit(.975,33) = 2.035

Given σ = 11.1 and n = 34 the σ_xbar = σ /√34 = 0.3265

Thus the MOE will be 2.035*.3265

or