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

Answer:

$E=3.4363$

Step-by-step 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:

$E=z_\alpha_/_2\times\frac{\sigma}{\sqrt n}=1.645\times \frac{12.1}{\sqrt 33}\\\approx 3.4363$

Hence the margin of error is 3.4363