Question

You measure 44 backpacks' weights, and find they have a mean weight of 33 ounces. Assume the population standard deviation is 2.5 ounces. Based on this, what is the maximal margin of error associated with a 99% confidence interval for the true population mean backpack weight.

As in the reading, in your calculations:
--Use z = 1.645 for a 90% confidence interval
--Use z = 2 for a 95% confidence interval
--Use z = 2.576 for a 99% confidence interval.

Give your answer as a decimal, to two places

m =  ounces

Answer 1

Solution :

Given that,

Point estimate = sample mean = = 33

Population standard deviation =    = 2.5

Sample size = n = 44

At 99% confidence level

= 1 - 99%  

= 1 - 0.99 =0.01

/2 = 0.005

Z/2 = Z0.005 = 2.576

Margin of error = E = Z/2 * ( /n)

E = 2.576 * ( 2.5 /  44 )

E = 0.97 ounces.