Question

You measure the weight of 49 backpacks, and find they have a mean weight of 67 ounces. Assume the population standard deviation is 8.8 ounces?. Based on this, what is the maximal margin of error associated with a 94% confidence interval for the true population mean backpack weight.

Answer 1

Answer: 2.36

Step-by-step explanation:

The formula to find the margin of error is given by :-

$E=z_{\alpha/2}\dfrac{\sigma}{\sqrt{n}}$

Given : Significance level : $\alpha: 1-0.94=0.06$

Critical value : $z_{\alpha/2}=1.88$   [Using standard normal distribution table]

Sample size : n=49

Standard deviation : 8.8 ounces

Then , the margin of error will be :-

$E=(1.88)\dfrac{8.8}{\sqrt{49}}\\\\\Rightarrow\ E=2.36342857143\approx2.36$

Hence, the margin of error associated with a 94% confidence interval for the true population mean backpack weight = 2.36