You measure 47 backpacks' weights, and find they have a mean weight of 79 ounces. Assume the population standard deviation is 13.1 ounces. Based on this, what is the maximal margin of error associated with a 90% confidence interval for the true population mean backpack weight. Give your answer as a decimal, to two places

3.14

Step-by-step explanation:

Given that:

Mean weight (m) = 79 ounces

Population standard deviation (s) = 13.1

Sample size = 47

Maximal margin of error associated with 90% confidence interval.

The margin of error is given by:

Zcritical * (standard deviation / sqrt(sample size)

Z critical at 90% confidence interval = 1.645

Hence,

Zcritical * (standard deviation / sqrt(sample size)

1.645 * 13.1 / sqrt(47)

1.645 * (13.1 / 6.8556546)

1.645 * 1.9108313

Hence, the margin of error is :

3.1433174885

= 3.14

what i would say the best answer is a. 40.5