Answer:
The maximal margin of error associated with a 90% confidence interval for the true population mean watermelon weight is of 3.46 ounces.
Step-by-step explanation:
We have that to find our [tex]\alpha[/tex] level, that is the subtraction of 1 by the confidence interval divided by 2. So:
[tex]\alpha = \frac{1-0.9}{2} = 0.05[/tex]
Now, we have to find z in the Ztable as such z has a pvalue of [tex]1-\alpha[/tex].
So it is z with a pvalue of [tex]1-0.05 = 0.95[/tex], so [tex]z = 1.645[/tex]
Now, find the margin of error M as such
[tex]M = z*\frac{\sigma}{\sqrt{n}}[/tex]
In which [tex]\sigma[/tex] is the standard deviation of the population and n is the size of the sample.
In this problem:
[tex]\sigma = 13.3, n = 40[/tex]
So
[tex]M = z*\frac{\sigma}{\sqrt{n}}[/tex]
[tex]M = 1.645*\frac{13.3}{\sqrt{40}}[/tex]
[tex]M = 3.46[/tex]
The maximal margin of error associated with a 90% confidence interval for the true population mean watermelon weight is of 3.46 ounces.
