Question

Grandma​ Gertrude's Chocolates, a family owned​ business, has an opportunity to supply its product for distribution through a large coffee house chain.​ However, the coffee house chain has certain specifications regarding cacao content as it wishes to advertise the health benefits​ (antioxidants) of the chocolate products it sells. In order to determine the mean​ % cacao in its dark chocolate​ products, quality inspectors sample 36 pieces. They find a sample mean of​ 55% with a standard deviation of​ 4%. What is the margin of error at​ 95% confidence?

Answer 1

= Solution :

Given that,

Point estimate = sample mean = = 55%=0.55

Population standard deviation =    = 4%=0.04

Sample size n =36

At 95% confidence level the z is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

Z/2 = Z0.025 = 1.96   ( Using z table )

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

= 1.96 * ( 0.05 /  36 )

=0.01633