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






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?

1 Answer

  • 9 years ago
    Best Answer

    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!

Still have questions? Get your answers by asking now.