Annual starting salaries for college graduates is unknown. Assume that a 95% confidence interval estimate of the population mean annual starting salary is desired. If the population standard deviation is $3,750 how large should the sample be if margin of error is $500

A sample of 217 is needed.

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 of the population and n is the size of the sample.

If the population standard deviation is $3,750 how large should the sample be if margin of error is $500

We have that .

We need a sample of n, and n is found when . So

Rounding up

A sample of 217 is needed.

step-by-step explanation:

because 1/3 is a fraction. we can tell because there is a divisor symbol between the 1 and the 3. if we were to add 2 fractions together, we would add the numerators, which is the number on top, but not the denominators, which is the number on the bottom

1/3+1/3=2/3

hope this !

its   c, f, a

step-by-step explanation: