Examine the required sample size needed to be able to
estimatethe mean dollars that each consumer will spend each month.
desireis to be within plus or minus $10 of the true mean with a
98%confidence level. the standard deviation is thought to
be500.
what are the trade-offs that will occur when you lower
thesample level to 2000?

And if we use a sample level of n =2000 the margin of error would be higher as we can see here:

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

represent the sample mean for the sample  

population mean (variable of interest)

represent the population standard deviation assumed

n represent the sample size  (variable of interest)

Confidence =98% or 0.98

ME = 10 represent the margin of error desired

The margin of error is given by this formula:

   (a)

And on this case we have that ME =10 and we are interested in order to find the value of n, if we solve n from equation (a) we got:

  (b)

The critical value for 98% of confidence interval now can be founded using the normal distribution. And in excel we can use this formla to find it:"=-NORM.INV(0.01;0;1)", and we got , replacing into formula (b) we got:

So the answer for this case would be n=13573 rounded up to the nearest integer

And if we use a sample level of n =2000 the margin of error would be higher as we can see here:

step-by-step explanation:

you can add the two numbers and divide the result by 2. or you can find the gap between the two numbers and add half of this to the smaller number.

step-by-step explanation: