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 30 pieces. They find a sample mean of 4% with a standard deviation of 2%. What is the margin of error at 90% confidence?
a. 1.49%
b. 1.24%
c. 0.62%
d. 0.75%
e. 1.36%

Answer 1

Answer:

c. 0.62%

Step-by-step explanation:

We have the standard deviation for the sample, which means that the t-distribution is used to solve this question.

The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So

df = 30 - 1 = 29

90% confidence interval

Now, we have to find a value of T, which is found looking at the t table, with 29degrees of freedom(y-axis) and a confidence level of $1 - \frac{1 - 0.9}{2} = 0.95$. So we have T = 1.6991

The margin of error is:

$M = T\frac{s}{\sqrt{n}}$

Sample of 30, standard deviation of 2%.

Then

$M = T\frac{s}{\sqrt{n}} = 1.6991\frac{2}{\sqrt{30}} = 0.62$

0.62%, so option c.