Anonymous
Anonymous asked in Science & MathematicsMathematics · 6 months ago

# Confidence intervals - what would the required sample size be?

Update:

Assume that you have a 95% confidence intervaI for a popuIation mean based on a sample size of 60. lf you wish to have a confidence intervaI of the same confidence IeveI but with a Iength which is one fifth of the one you aIready have, then what would the required sampIe size be?

Relevance
• Alan
Lv 7
6 months ago

CI_low = mean - Z_critical* standard deviation/sqrt(N)

CI_High = mean + Z_critical*standard deviation/sqrt(N)

length of Interval = 2*Z_critical*standard deviation/sqrt(N)

original length = 2*Z_critical*standard deviation/sqrt(60)

so 2 , Z_critical, and standard deviation don't change

2*Z_critical*standard deviation is a constant

(1/5)*k/sqrt(60) = k/sqrt(new size)

1/5sqrt(60) = 1/sqrt(new_size)

multiply both side by sqrt(new_size)

sqrt(new_size)/5*sqrt(60) = 1

multiply both sides by 5*sqrt(60)

sqrt(New_size) = 5*sqrt(60)

square both sides

New_size = 25*60 = 1500

sample size = 1500

----- checking

old size =k/sqrt(60) = k /0.129099445

new_size = k/sqrt(1500) = k /0.025819889

new_size/old size= sqrt(60)/sqrt(1500) = sqrt(1)/sqrt(25) = 1/5

standard deviation/sqrt(60)