You are planning a survey of starting salaries for recent computer science majors. In the latest survey by the National Association of Colleges and Employers, the average starting salary was reported to be $61,238. If you assume that the standard deviation is $3850, what sample size do you need to have a margin of error equal to $500 with 95% confidence

A sample size of at least 228 must be needed.

Step-by-step explanation:

We are given that in the latest survey by the National Association of Colleges and Employers, the average starting salary was reported to be $61,238. Assume that the standard deviation is $3850.

And we have to find that what sample size do we need to have a margin of error equal to $500 with 95% confidence.

As we know that the Margin of error formula is given by;

 Margin of error =

where, = significance level = 1 - 0.95 = 0.05 and   = 0.025.

             = standard deviation = $3,850

            n = sample size

Also, at 0.025 significance level the z table gives critical value of 1.96.

So, margin of error is ;

                          = 15.092

 Squaring both sides we get,

                            n = = 227.8 ≈ 228

So, we must need at least a sample size of 228 to have a margin of error equal to $500 with 95% confidence.

$53.50

step-by-step explanation:

you have to subtract the amount of money chuck earned ($72.50) by the cost of the gas for the mower ($11.25) and the new string for the trimmer he bought ($7.75).

72.50-11.25=61.25

and then subtract $61.25 by the second thing he bought (the new string for the trimmer)

61.25-7.75=53.5

so in total it's $53.50 (53 dollars and 50 cents)

step-by-step explanation: