You are planning a survey of starting salaries for recent computer science majors. In a recent survey by the National Association of Colleges and Employers, the average starting salary was reported to be $61,467. Grade: 5 4 3 2 1 Communication strong ←− − − − − − −→ weak ×1 If the standard deviation is known to be $12,000, what sample size do you need to have a margin of error equal to $1000, with 95% confidence

A sample size of 554 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 .

That is z with a pvalue of , so Z = 1.96.

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.

Standard deviation is known to be $12,000

This means that

What sample size do you need to have a margin of error equal to $1000, with 95% confidence?

This is n for which M = 1000. So

Dividing both sides by 1000:

Rounding up:

A sample size of 554 is needed.

$450.00 for 6 weeks

or

$300

step-by-step explanation:

$200.00 for first week $50.00 x5 equals $250 plus $200 equals $450.00 for six weeks.

$200 for first week

$200 x 0.10 = $20

$20 x 5 = $100

$200 + $100 = $300