Question

Suppose that a jewelry store tracked the amount of emeralds they sold each week to more accurately estimate how many emeralds to keep in stock. At the end of one year, the store used the weekly sales information to construct a 95% confidence interval for the mean number of emeralds sold per week. The confidence interval they calculated was ( 2.13 , 3.37 ) . Use the confidence interval to find the point estimate for the mean and the margin of error. Give your answers precise to two decimal places.

Answer 1

Answer:

The mean is 2.75.

The margin of error is 0.62.

Step-by-step explanation:

In this case, we know the 95% confidence interval, and we want to know the mean and the margin of error.

The mean can be calculated as the average between the upper limit and lower limit:

$\mu=\frac{UL-LL}{2}=\frac{3.37+2.13}{2}=\frac{5.5}{2}=2.75$

The margin of error is the difference between the upper (or lower limit) and the mean. It equals the product between the z-value and the standard deviation divided by the square of the sample size.

It can also be calculated as half the difference of the limits of the confidence interval.

We only have the limits of the confidence interval to calculate the margin of error:

$ME=z\sigma/\sqrt{n}=\frac{UL-LL}{2}=\frac{3.37-2.13}{2}=\frac{1.24}{2}= 0.62$