Question

You measure 40 watermelons' weights, and find they have a mean weight of 66 ounces. Assume the population standard deviation is 13.3 ounces. Based on this, what is the maximal margin of error associated with a 90% confidence interval for the true population mean watermelon weight.

Answer 1

The maximal margin of error associated with a 90% confidence interval for the true population mean watermelon weight is of 3.46 ounces.

Explanation:

We have that to find our
`\alpha` level, that is the subtraction of 1 by the confidence interval divided by 2. So:

`\alpha = (1-0.9)/(2) = 0.05`

Now, we have to find z in the Ztable as such z has a pvalue of
`1-\alpha`.

So it is z with a pvalue of
`1-0.05 = 0.95`, so
`z = 1.645`

Now, find the margin of error M as such

`M = z*(\sigma)/(√(n))`

In which
`\sigma` is the standard deviation of the population and n is the size of the sample.

In this problem:

`\sigma = 13.3, n = 40`

So

`M = z*(\sigma)/(√(n))`

`M = 1.645*(13.3)/(√(40))`

`M = 3.46`

The maximal margin of error associated with a 90% confidence interval for the true population mean watermelon weight is of 3.46 ounces.