# determining margin of error?

having a little trouble on this problem, got pretty far but could use some help.

a local university administers a comprehensive examination to the recipients of a b.s degree in business administraion. a sample of 5 exams are selected at random and scored. the scores are

54

68

75

80

98

a) calculate the sample mean (75)

b) compute the standard deviation (16.16)

c) determine the margin of error at 95% confidence

i'm a little confused on this but from what i read online i should be doing 95/100 = .95

.95*2 = .475

1 - .475 = .525

.525*16.16 = 8.484 ???????

d) at 95% confidence determine an interval for the mean of the population

not sure because i'm not sure if my margin of error is correct

any help guys?

Relevance

c) The way my stats teacher taught us to calculate the margin of error was:

ME = z* x SE(x)

I found the z* using invNorm(.025) on my calculator (a TI-83), however you might do it some other way. Then SE, or standard error, is your standard deviation divided by the square root of your sample size. So, 16.16 / √5 = 7.2250. Substitute these values and you have your answer.

ME = z* x SE(x)

ME = 1.96 x 7.2250

ME = 14.1609

However, if you have learned about the t model, it would be smarter to use this since your sample size of 5 is so small. If not, ignore this part. In this case, you would have to find t* which is invT on a calculator, or you can use a table to get the value of 2.776. It would be solved almost exactly the same after this.

ME = z* x SE(x)

ME = 2.776 x 7.2250

ME = 20.0565

d) A confidence interval is found using the following equation:

x ± (ME) where x is 75.

If you used the z*, you would get a value of 60.8391 to 89.1609.

If you used the t*, you would get a value of 54.9435 to 95.0565.

Hope this helps!