Example 1:
Find the minimum sample size required to estimate the mean height of students within 2 cm of it’s value, with confidence level of 95%, considering that the heights of students are normally distributed with a standard deviation = 20 cm.
Solution:
Here
SD = 20, ε (absolute error) = 2, Confidence level = 95%
After putting these values, we get required sample size = 384
What will happen after selecting this sample size, considering that the mean height is 165 cm and SD of 20 cm.
SE of mean = SQRT(SD^2/N) = SQRT (20^2/384) = 1.02
95 % CI of mean = mean ± 1.96 * SE
95% CI of mean = 165 ± 1.96 * 1.02
95% CI of mean = 165 ± 2 = 163 – 167 (allowable error of 2 towards both sides of mean, at 95% confidence level)
Example 2:
Find the minimum sample size required to estimate the mean height of students within 2% of it’s value, with confidence level of 95%, considering that the heights of students are normally distributed with approximate mean of 160 cm and standard deviation = 10 cm.
Solution:
Here
SD = 10, mean = 160 cm, ε (relative error) = 2 %, Confidence level = 95%
After putting these values, we get required sample size = 38
What will happen after selecting this sample size.
SE of mean = SQRT(SD^2/N) = SQRT (10^2/38) = 1.62
95 % CI of mean = mean ± 1.96 * SE
95% CI of mean = 160 ± 1.96 * 1.62
95% CI of mean = 160 ± 3.18 = 157 – 163 (allowable error of 3, which is approximately equal to 2% of 160, after removing fractions)
@ Sachin Mumbare