20 independent observations on a random variable X with unknown distribution function F(x) are:
142 134 98 119 131 103 154 122 93 137 86 119 161 144 158 165 81 117 128 103. find the 95% confidence interval for F(100).
i found that the sum of all 20 numbers are 2595. hence sample mean is 124.75.
after that what should i do?
i thought to use Z value, the sample size has to be large since the population variance is unknown? the sample size of 20 is small..
i think you should that (X − μ)2 is calculated, but it is longer time.
but can i just approx to normal since sample size is small?
(X − μ)2 would be the calculation of sample variance right? not population
No matter because these values are populations. You should read carefully this question.
using the method of normal distribution,
i got that sample mean is 124.75 and variance is 635.1447.
at Z(0.025)= 1.96, my upper confidence interval is 124.75+ 1.96( 635.1447/ 20)^(1/2)
but the answer should be (0.057, 0.437) which is way smaller.
i think that there might be another way to do it because in usual cases, the confidence intervals are values around the the sample mean... however in this question, the confidence interval is very small..which makes me wonder why this is such the case..
you should table of t (n-1, α/2)
for your problem, value of t is 2,093