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.

step-by-stp

the correct answer is the section c

3-4

step-by-step explanation:

the answer is 3-4 because both 45 and 60 has a multiple of 15.

45/3=15

60/4=15

therefore the odds against selecting a red is 4-3