Confidence Intervals (CI)

Confidence intervals (CI)

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Interval (a,b) is CI with a parameter of a population, q, with confidence level  if

(3,4) and ,

CI for the mean (μ)

Sample size n

sample mean

S      sample standard deviation

CI for the mean

Suppose standard deviation of the population is given (σ), and the underlying population is normally distributed or sample size is large, then CJ with  is the following

is the value of standard normal random variable such that the area to its right is equal to α/2

Area = α/2

Suppose underlying population is normal, standard deviation not known, then  CI for μ is as follows

is the value of t random variable with n – 1 degrees of freedom (DOF), such that are to its right is α/2

Area = α/2

CI for a proportion, p,

proportion in the sample

CJ for p

CLA1 Guide: Estimate the number of devices left in taxicabs over 6 monhs in Chicago, and to estimate a bound on the error, for this purpose we construct CI with confidence level 0.95 on average of devices left and then we multiply by fleet size to find bound on error in estimating total number of devices left, note that the researchers multiplied point estimate on average of devices left by fleet size in order to get point estimate on total numbers of devices left, this approach may bear an error, we want to find the bound on the error

Point estimate of mean of the number of devvices left per taxicab

Mean = 3.42

Fleet size in Chicago = 85619/3.42 = 25035

Use t statistic to CI 95% Confidence level

Assume numbers left is normally distributed

Assume S = 2.5

N = 100

=T.INV(0.975,99)

=E4-E7*(E5/SQRT(100))

Left limit = 2.923946

Right limit = 3.916054

CI = (2.923946, 3.916054)

Bound on error in CI = 0.496054

Bound on error in estimating total number of devices left =12419

You can reduce size of the error by increasing n,

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