Question

Business Statistics

A more practical method presents itself::

Let's redo our broker exercise with this modification: A. how many standard deviations are the lower and upper bounds from the mean (of the sampled means)? B. Can we use this thought to solve any confidence interval with a flick of the algorithmic finger? C. What happens if we do not know the population mean or standard deviation (oh my!)?

Answer 1

A.) Add/subtract 1/2∗std(x) in population mean (mu) and you'll get the result for lower and upper bounds.

Also, if you want to find it for sample mean, then you should proceed as follows:-

X(bar)~N(μ,σ/root(n))

The distribution of sample means is normally distributed with mean equal to the population mean and standard deviation given by the population standard deviation divided by the square root of the sample size.

The general form for a confidence interval for a single population mean, known standard deviation, normal distribution is given by
(lower bound, upper bound) = (point estimate – EBM, point estimate + EBM)
= (x¯−EBM, x¯+EBM)
= (x¯−zσ/(root(n)),x¯+zσ/root(n))

EBM = zσn√zσn = the error bound for the mean, or the margin of error for a single population mean; this formula is used when the population standard deviation is known.

B.) yes, we can use  this thought to solve any confidence interval with a flick of the algorithmic finger because

General form of a confidence interval

(lower value, upper value) = (point estimate−error bound, point estimate + error bound)

To find the error bound when you know the confidence interval

error bound = upper value−point estimate OR error bound = (upper value−lower value)/2

Single Population Mean, Known Standard Deviation, Normal Distribution

Use the Normal Distribution for Means, Population Standard Deviation is Known EBM = Zα2⋅σ/root(n)

Therefore, The confidence interval has the format (x¯ − EBM, x¯ + EBM).

C.) If we do not know the population mean or standard deviation, it's not a very big problem because in real life, it almost always happen and hence, in such cases we use Sample mean and Sampling Variance and then take student's t distribution into account instead of normal.

Here is, how we proceed:-

s = the standard deviation of sample values.

t = ( x¯−μ)/(s/root(n)) is the formula for the t-score which measures how far away a measure is from the population mean in the Student’s t-distribution

df = n - 1; the degrees of freedom for a Student’s t-distribution where n represents the size of the sample

T~tdf the random variable, T, has a Student’s t-distribution with df degrees of freedom

EBM= tα2.s/root(n)= the error bound for the population mean when the population standard deviation is unknown

tα/2 is the t-score in the Student’s t-distribution with area to the right equal to α/2.

The general form for a confidence interval for a single mean, population standard deviation unknown, Student's t is given by (lower bound, upper bound)
= (point estimate – EBM, point estimate + EBM)
= (x¯–ts./(√n),x¯+ ts/(√n))

Hope i answered your question well.

Thanks.