8.2

Estimating μ When σ is Unknown

What if it is impossible or impractical to use a large sample?

Apply the Student’s t distribution:

It is important to note that the t distribution uses a degrees of freedom (d.f.) = n – 1

The shape of the t distribution depends only on the sample size, n, if the basic variable x has a normal distribution.

When using the t distribution, we will assume that the x distribution is normal.

Student’s t Distributions

• In order to use the normal distribution to find confidence intervals for a population mean μ we need to know the value of σ. However, much of the time, we don’t know either. In such cases, we use the sample standard deviation s to approximate σ.

• When we use s to approximate σ, the sampling distribution for x bar follows a distribution known as Student’s t distribution.

More about t distribution

• The graph is always symmetrical about its mean, which (as with z dist) is 0.

• t distribution is similar to normal z distribution except it has somewhat thicker tails. (as seen on next slide)

• As the degrees of freedom increase, the t distribution approaches the standard normal distribution.– As n increases… what happens to the distribution of

any sample?

The t Distribution has a shape similar to that of the the Normal Distribution

A Normal distribution

A “t” distribution

Maximal Margin of Error, E

Similar to what we saw in the previous section, we can determine the confidence interval by

using the margin of error:

Example

A company has a new process for manufacturing large artificial sapphires. In a trial run, 37 sapphires are

produced. The mean weight for these 37 gems is 6.75 carats and σ = 0.33 carats. Let μ be the mean weight for

distribution of all.

Why is the t distribution the preferred method here?• Find E for a 95% confidence interval. • Then find the confidence interval.• Interpret the confidence interval in the context of the

problem.

The mean weight of eight fish caught in a local lake is 15.7 ounces with a standard deviation of

2.3 ounces.

Construct a 90% confidence interval for the mean weight of the population of fish in the lake.

Another Example

Mean = 15.7 ounces Standard deviation = 2.3 ounces.

• n = 8, so d.f. = n – 1 = 7• For c = 0.90, Table 6 in Appendix II gives

t0.90 = 1.895.

Mean = 15.7 ounces Standard deviation = 2.3 ounces.

E = 1.54

The 90% confidence interval is:

15.7 - 1.54 < < 15.7 + 1.54

14.16 < < 17.24

ExEx

Summary

Examine problem Statement

σ is known

Use normal distribution with margin of error

σ is unknown

Use Student t’s distribution with

margin of error (don’t forget about d.f.)