Question

A National Retail Foundation survey found households intend to spend an average of $469 during the December holiday season. Assume that the survey included 600 households and that the sample variance was $375. Calculate the margin of error for a 99% confidence interval.

Answer 1

Answer:

The margin of error for a 99% confidence interval. is of $2.25.

Step-by-step explanation:

We have the sample's variance, so we use the t-distribution 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 = 600 - 1 = 599

99% confidence interval

Now, we have to find a value of T, which is found looking at the t table, with 600 degrees of freedom(y-axis) and a confidence level of $1 - \frac{1 - 0.99}{2} = 0.995$. So we have T = 2.84.

The margin of error is:

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

In which s is the standard deviation of the sample(square root of the variance) and n is the size of the sample.

In this question:

$s = \sqrt{375}, n = 600$. So, the margin of error is of:

$M = 2.84\frac{\sqrt{375}}{\sqrt{600}} = 2.25$

The margin of error for a 99% confidence interval. is of $2.25.