If the variance of the population is 10, what is the variance of the sampling distribution brainly

The statistic used to estimate the mean of a population, μ, is the sample mean,

If X has a distribution with mean μ, and standard deviation σ, and is approximately normally distributed or n is large, then is approximately normally distributed with mean μ and standard error ..

When σ Is Known

If the standard deviation, σ, is known, we can transform

Example:

From the previous example, μ=20, and σ=5. Suppose we draw a sample of size n=16 from this population and want to know how likely we are to see a sample average greater than 22, that is P(

So the probability that the sample mean 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 to select a sample of size 49 in the example above instead of n=16. How will this affect the standard error of the mean? How do you think this will affect the probability that the sample mean will be >22? Use the Z table to determine the probability.

Answer

When σ Is Unknown

If the standard deviation, σ, is unknown, we cannot transform

If X is approximately normally distributed, then

has a t distribution with (n-1) degrees of freedom (df)

Using the t-table

Note: If n is large, then t is approximately normally distributed.

The z table gives detailed correspondences of P(Z>z) for values of 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 the critical values of z are in the first column and top row.

The t-table is presented differently, with separate rows for each df, with columns representing the two-tailed probability, and with the critical value in the inside of the table.

The t-table also provides much less detail; all the information in the z-table is summarized in the last row of the t-table, indexed by df = ∞.

So, if we look at the last row for z=1.96 and follow up to the top row, we find that

 P(|Z| > 1.96) = 0.05

Exercise: What is the critical value associated with a two-tailed probability of 0.01?

Answer

Now, suppose that we want to know the probability that Z is more extreme than 2.00. The t-table gives us

P(|Z| > 1.96) = 0.05

And

P(|Z| > 2.326) = 0.02

So, all we can say is that P(|Z| > 2.00) is between 2% and 5%, probably closer to 5%! Using the z-table, we found that it was exactly 4.56%.

Example:

In the previous example we drew a sample of n=16 from a population with μ=20 and σ=5. We found that the probability that the sample mean is greater than 22 is P(

From the tables we see that the two-tailed probability is between 0.01 and 0.05.

P(|T| > 1.711) = 0.05

And

P(|T| > 2.064) = 0.01

To 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 the probability that the sample mean is greater than 22 is between 0.005 and 0.025 (or between 0.5% and 2.5%)

Exercise: . If μ=15, s=6, and n=16, what is the probability that

Answer

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