A National Retail Federation 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.

The margin of error for a 99% confidence interval was of $1.84.

Step-by-step explanation:

We have that to find our level, that is the subtraction of 1 by the confidence interval divided by 2. So:

Now, we have to find z in the Ztable as such z has a pvalue of .

So it is z with a pvalue of , so

Now, find the margin of error M as such

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

So

The margin of error for a 99% confidence interval was of $1.84.

2500

step-by-step explanation:

step-by-step explanation:

divide, then simplify your answer.

1. make -15.66 ÷   5.8 a fraction. then divide.

2. simplify.  

3. final answer.