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.

$35.59 in pounds

step-by-step explanation:

hello from mrbilldoesmath!

answer:

the domain is the set { -2, 0, 2, 3}

discussion:

the domain is the set of x values in the ordered pairs (x,y). taking the x, or first component, of each ordered pair gives   -2, 0, 2, 3.

you,

mrb