When a sample is selected at random from a population the sample mean x will likely be

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

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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When a sample is selected at random from a population?

When given the population mean what would the sample mean be?

Can the sample mean be larger than the population mean?

What is the probability that the sample mean will be more than?