Question

Question: How can I determine the sample size given the standard error (SE <= 70), and in the confidence intervals of 95% and 85% using t-tables.

tve tAle

Side Note: I am studying for my finals and have been asked to answer the following question but I am not completely sure how to do so. I believe I would have to have the standard deviation provided to answer this, but am not completely sure if that is the case. If this does require a standard deviation value please use SD = 10 and please answer for both the 95% confidence interval and 85% confidence interval.

Answer 1

n = (Z * sd/e)^2

here z = 1.96 for 95 % confidence interval

sd = 10

e is the desired margin of error

Now standard error = sd/sqrt(n)

sd/sqrt(n) < 70

hence sqrt(n) > sd/70

n > (Sd/70)^2

since sd = 10

n > (10/70)^2

n > 0.02040

n = 1 suffices

I think numbers in your question is wrong

generally standard error is much than sd

so value of sd = 10 is not correctly assumed given that you required standard error =70

as standard error = sd/sqrt(n)