Question

You measure 33 backpacks' weights, and find they have a mean weight of 36 ounces. Assume the population standard deviation is 12.1 ounces. Based on this, what is the maximal margin of error associated with a 90% confidence interval for the true population mean backpack weight.

Answer 1

`E=3.4363`

Explanation:

#Given a significance level of
`\alpha=1-CI=1-0.90=0.1` and

`\bar x=36, \ \sigma=12.1 \ n=41,` the critical value is,
`z_\alpha_/_2=z_\alpha_/_0_._0_5=1.645`

#the margin of error can then be calculated as:

`E=z_\alpha_/_2*(\sigma)/(\sqrt n)=1.645* (12.1)/(\sqrt 33)\\\approx 3.4363`

Hence the margin of error is 3.4363