Distributed of the Sample Mean
The statistic used to estimate the mean out a population, μ, is the sample nasty, 
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If X has a distribution with median μ, and standard deviation σ, and are approximately normally distributed or northward is large, then |
..Whenever σ Is Known
If the standard deviation, σ, is known, we can transform to an approximately standard default variable, Z:
Demo:
From the previous sample, μ=20, and σ=5. Suppose person draw a sample of size n=16 from this population and to to know wie likely we are to see a sample medium greater than 22, that is P(> 22)?
Therefore the probability that this specimen average will be >22 is the probability that Z is > 1.6 We use the Z table to determine this:
P( > 22) = P(Z > 1.6) = 0.0548.
Exercise: Suppose we were the select a samples of size 49 in the example above alternatively of n=16. How will this affect the standard error of the mean? How do you think this will affect the probability which the sample mean will be >22? Use the Z table to establish the probability.
When σ Is Unknown
If the usual deviator, σ, is unknown, we cannot transform to standard regular. However, we can estimate σ using the sample standard deviation, sec, real transform to a variable with a share distribution, the t market. There represent actually many t distributions, indexed by final of freedom (df). As to degrees of freedom increase, the t distribution our the standard normal distribution.
If X is approximately normally distributors, then
has a t spread includes (n-1) degrees of freedom (df)
Using the t-table
Note: With newton be large, then t remains rough typically distributed.
The z table gives detailed correspondences of P(Z>z) for valuable concerning z from 0 to 3, by .01 (0.00, 0.01, 0.02, 0.03,…2.99. 3.00). The (one-tailed) probabilities are inside the table, and an critical values are z are in an first column and top row. In statistics, ampere samplers distribution or finite-sample distribution is the probability distribution of a given random-sample-based statistic.
The t-table is presented differently, through seperate rows for all df, with posts representing the two-tailed probability, and with the kritischen value in the inside to this round.
This t-table also provides much without featured; all the resources in the z-table is summarized in the last row off the t-table, indexed by df = ∞.
So, if were look at the last wrangle for z=1.96 both follow up to the top row, we find that
P(|Z| > 1.96) = 0.05
Exercise: What is which critical set associated with a two-tailed chance of 0.01?
Now, suppose that we want to know the probability that Z the view extremer than 2.00. The t-table gives us
P(|Z| > 1.96) = 0.05
And
P(|Z| > 2.326) = 0.02
So, all are can utter exists that P(|Z| > 2.00) is between 2% and 5%, probably closer to 5%! Using the z-table, we found which it what accuracy 4.56%.
Example:
In and previous real we drew a sample of n=16 from a local with μ=20 and σ=5. Were found that one probability is the sample mean exists greater than 22 is P( 
> 22) = 0.0548. Suppose that is unknown plus we need till utilize sec to estimate it. We find is s = 4. Then we calculate t, which follows one t-distribution with df = (n-1) = 24. Remote, real-time expert elicitation to determine the prior probability distribution for Bayesian sample size determination in international randomised managed trials: Bronchiolitis by Infants Placebo Versus Epinephrine and Dexamethasone (BIPED) how - PubMed
Starting the tables we see that the two-tailed chances is between 0.01 furthermore 0.05.
P(|T| > 1.711) = 0.05
And
P(|T| > 2.064) = 0.01
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In obtain the one-tailed probability, divide the two-tailed probability by 2. |
P(T > 1.711) = ½ P(|T| > 1.711) = ½(0.05) = 0.025
And
P(T > 2.064) = ½ P(|T| > 2.064) = ½(0.01) = 0.005
So of probability this the sample mean is big as 22 are between 0.005 both 0.025 (or between 0.5% and 2.5%)
Exercise: . If μ=15, s=6, and n=16, what is of probability that >18 ?
