Chapter 8: STATISTICAL INTERVALSFOR A SINGLE SAMPLE

Part 2: Confidence Interval on µfrom a normal distribution with anUNKNOWN population variance σ

and small sampleSection 8-2

NOTE: The title of this section if very specific.We’re trying to estimate µ (so, µ is unknown)from a normally distributed pop’n with a un-known σ2 with not necessarily a large samplesize.

This is a realistic situation.

1

• To form a confidence interval (CI) for µ whenσ2 is unknown and our sample isn’t verylarge, we NEED the original distribution tobe normal (or at least nearly normal).

This is an assumption that gets us the neededbehavior of our statistic of interest.

•Many populations are normal, so this isn’ttoo much of a limitation.

• After we collect our data, we can check thisassumption of normality by using a normalprobability plot (recall section 6-6).

• If the data isn’t normal, we have to use adifferent approach. Something that doesn’tdepend on this normality assumption, suchmethods are called nonparametric methods(which we won’t cover in here).

2

• In section 8-1, we wanted a CI on the meanµ and we knew the value of σ2. In that case,when n ≥ 30, we had

X̄ − µσ/√n∼ N(0, 1)

(this is also true for n < 30 when originalpop’n is normal)

•When σ2 was unknown and estimated withS2 and we had a LARGE SAMPLE (n ≥60), we also had X̄−µ

S/√n∼ N(0, 1).

The 95% CI for µ in this case was:

X̄ ± z0.05/2 · S√n

or X̄ ± 1.96 · S√n

• But what happens when we don’t know σ2

and we don’t have a very large n (like ≥ 60)?How do we form a CI for µ?

3

• Suppose we have a small sample from anormal population and σ2 is unknown.

We do have an unbiased estimator for σ2,which is the sample variance S2.

Can I plug-in S2 to get my Z distribution?

Which means...

~N(0,1)

when n isn’t very large.

4

But, it turns out that this quantity does follow adifferent and very useful distribution called thet distribution. So, for smaller n, drawn from anormal population

X̄ − µS/√n∼ tn−1

where tn−1 is a t-distribution with n−1 degreesof freedom.

• t DistributionLetX1, X2, . . . , Xn be a random sample froma normal distribution (so, original populationis normal) with unknown mean µ and un-known variance σ2. The random variable

T =X̄ − µS/√n

has a t distribution with n − 1 degrees offreedom.

5

•What does the t-distribution look like?

It looks a lot like the N(0, 1), but with heav-ier tails.

The heaviness of the tails depends on the de-grees of freedom (the subscript on the tdf ),so it depends on the sample size.

•When df = n−1 is very large, the t looks justlike the N(0, 1). So, N(0, 1) is the limitingdistribution for t as n→∞.

6

• At n ≥ 60, the tn−1 and N(0, 1) distribu-tions are very similar.

This is why we say X̄−µS/√n∼ N(0, 1)

for n ≥ 60.

• For smaller n, the t has heavier (fatter) tailsthan the N(0, 1).

7

A confidence interval for µ using thet distribution

• Using the same reasoning as in the previouslydiscussed creation of confidence intervals, wehave

P (−tα/2,n−1 ≤X̄ − µS/√n≤ tα/2,n−1) = 1−α

• Rearranging gives the 100(1-α)% CI for µ:

P (X̄− tα/2,n−1S√n≤ µ ≤ X̄ + tα/2,n−1︸ ︷︷ ︸ S√

n) = 1−α

↑Instead of using a z percentile asthe multiplier, we have t percentilemultiplier.

• How do I get the tα/2,n−1 value?

8

• How do I get the tα/2,n−1 value?

• Similar to getting a z-value. Look at t-tablein your book p.745.

When α = 0.05 and sample size is n = 10,tα/2,n−1 = t0.025,9

This is the t-value for a t9 distribution with2.5% above and 97.5% below.

t0.025,9 = 2.262

9

• Example: CI for µ using t-distribution

Suppose a sample of size n = 10 is takenfrom a normal population and x̄ = 8.94 ands = 4.3. Construct a 95% CI for the popu-lation mean.

Upper bound: x̄ + tα/2,n−1 · s√n

= x̄ + t0.025,9 · s√n

= 8.94 + 2.262(

4.3√10

)= 10.02

Lower bound: x̄− tα/2,n−1 · s√n

= 8.94− 2.262(

4.3√10

)= 5.86

The 95% confidence interval for µ is[5.86, 10.02].

10