Question

You measure the weight of 38 bags of nuts, and find they have a mean weight of 78 ounces. Assume the population standard deviation is 14.5 ounces. Based on this, what is the maximal margin of error associated with a 96% confidence interval for the true population mean bags of nuts weight.

Answer 1

Answer: The maximal margin of error is 4.822.

Step-by-step explanation:

Since we have given that

Number of bags = 38

Sample mean = 78 ounces

Standard deviation = 14.5 ounces

At 96% confidence interval for the true population mean bags of nut weight.

So, z = 2.05

As we know that formula for "Margin of error":

$z\times \dfrac{\sigma}{\sqrt{n}}\\\\=2.05\times \dfrac{14.5}{\sqrt{38}}\\\\=4.822$

Hence, the maximal margin of error is 4.822.