1. Student’s t-Distribution
A Lecture for the Intro Stat Course 
Each slide has its own narration in an audio file. For the explanation of any slide click on the audio icon to start it.
Professor Friedman's Statistics Course by H & L Friedman is licensed under a  Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. 

2. Student’s t-Distribution
In the previous lectures on statistical inference (estimation, hypothesis testing) about the mean, we used the Z statistic even when we did not know σ. In that case, we used s as a point estimate of σ for “large enough” n.
What do we do if (a) we don’t know σ and (b) n is small? If the population of interest is normally distributed, we can use the Student’s t-distribution in place of the standard normal distribution, Z.
William S. Gossett, who developed the t-distribution, wrote under the name “Student” since, as an employee of the Guiness Brewery in Dublin, he was required by the firm to use a pseudonym in publishing his results.
In a nutshell, we use the t-Distribution when we have small samples. As you recall, the central limit theorem tells us that the sample means follow a normal distribution when n is large (some say at least 30; others say 50).
The t Distribution

3. When to use the t-Distribution
We will use Z for EITHER (1) known σ OR (2) large n
Use t for (1) Small sample AND (2) Taken from N.D. population** AND (3) Unknown σ
** What can we do if the population is not normally distributed and we have a small sample? Always use non-parametric statistical methods in this case. (We will not study non-parametric statistics in this course.   
Large sample (n)
Small sample (n)
σ known
Z test
σ unknown
t test – but only if population is ND**
The t Distribution

4. The t-Distribution
The t-distribution looks like the normal distribution, except that it has more spread.
It is still symmetrical about the mean; mean=median=mode.
It also goes from -∞ to +∞.
The Student’s t distribution is not a single distribution (like the standardized normal (Z) distribution.
It is a series of distributions, one for each number of degrees of freedom.
The t distribution with 10 degrees of freedom has a slightly different shape than a t distribution with 20 degrees of freedom.
As n gets larger, student’s t distribution approaches the normal distribution.
Degrees of freedom (df) = n -1. We will shortly see why we lose the degree of freedom (hint: remember we divided by (n-1) when computing the sample standard deviation).
The t Distribution

5. Using the t statistic for testing hypotheses (n-1 degrees of freedom):
µH0 or μ0 denotes the claimed population mean (for example, a company might be making a claim about this parameter).
(1-)% Confidence Interval Estimator:
The t Distribution

6. Losing a degree of freedom
The number of degrees of freedom is equal to the sample size minus 1.
We “lose” a degree of freedom each time a statistic computed from the sample is used as a point estimator in place of the parameter.
In this case, we do not know the population σ, so instead of using σ (which comes from a census, not a sample) we use s, the sample standard deviation.
As you already know, in the formula to compute s, we divide by (n-1), a loss of a degree of freedom. If we did not divide by (n-1) in computing the standard deviation, it would be biased when used as a point estimator of the population standard deviation, σ
In a nutshell, we have n-1 degrees of freedom for reasons that have to do with mathematically ensuring that our formulas all work properly without any bias.
The t Distribution

7. Example 1
A consulting firm claims that its consultants earn on average exactly $260 an hour. You decide to test the claim using a sample of 16 consultants. You find that X̅ = $200 and s = $96. Test the claim at α = .05 level. Assume that the population follows a normal distribution.
We follow the same steps in hypothesis testing as before (see the virtual handout in question).
Step 1: Formulate the null and alternate hypotheses
H0: μ = $260
H1: μ ≠ $260
The t Distribution

8. Example 1 (cont’d) Step 2: =.05 Step 3: Choose the test statistic
We are going to use the t-distribution since n is small (only 16) and we do not know σ
Step 4: Establish the critical value or values of the test statistic needed to reject H0.
This is a t15 so we cannot use the critical values of ±1.96 which we used for a Z test. If you go to the t-table (next slide) and examine the column that has .025 on the top and the row for 15 degrees of freedom, you will find that the critical values are ±2.1315
Please do not forget that you have (n-1) degrees of freedom (d.f.); n = 16, but d.f. = 15.
The t Distribution

9. Example 1 (cont’d)
The t Distribution

10. Example 1 (cont’d)
Step 5: Determine the actual value (computed value) of the test statistic.
t15 = =
Step 6: Make the decision (Reject H0 or Do Not Reject H0).
Reject H0
The t Distribution

11. Example 1 (cont’d)
Now, how do we construct a confidence interval estimator (CIE), using t?
Suppose there had not been a claim about the population mean and you simply wanted to construct a 95% CIE for μ. Using the sample evidence, what should you do?
200±
$200 ± $51.16
$ ↔ $251.16
Interpretation: We have 95% confidence that this interval ($ ↔ $251.16) really does contain the true population mean, µ.
The t Distribution

12. Example 2
A company claims that its soup vending machines deliver (on average) exactly ounces of soup. The company statistician finds:
n=25
X̅=3.97 ounces
s=.04 ounces
(a) Test the claim at α = .02.
(b) Supposing no claim was made, construct a two-sided 98% confidence interval estimator using the sample evidence.
The t Distribution

13. Example 2 (cont’d) (a) Test the claim (=.02) H0: µ=4.0 oz.
Reject H0
The t Distribution

14. Example 2 (cont’d) (b) 98% CIE of µ
3.97 ± (.008)
3.97 ± .02
3.95 oz ←———→ 3.99 oz
Note that 4.00 ounces is not in the 98% CIE. How is this consistent with (a)?
The t Distribution

15. Example 3
A school claims that the average reading score of its students is at least 70.
A sample of 16 students is randomly selected (n=16) to test this claim.
X̅ = 68 and s = 9
(a) Test claim at α = .05.
(b) Supposing no claim was made. Use sample evidence to construct a two-sided 95% confidence interval estimate (CIE) of μ.
The t Distribution

16. Example 3 (cont’d) Test the claim at =.05 H0: µ≥70 H1: µ<70
Do Not Reject H0
The t Distribution

17. Example 3 (cont’d) (b) Two-sided 95% CIE. 68  2.1315(2.25) 68  4.8
We will always construct two-sided CIEs in this course.
68  (2.25)
68  4.8
63.2 ——— 72.8
The t Distribution

18. Example 4
The Vandelay Water Company claims that at most there is1 ppm (part per million) of benzene in their terribly expensive horrid- tasting bottled water.
(a) Test the claim at α=.05
(b) Supposing no claim was made. Construct a two- tailed 95% C.I.E of μ
SampleData:
n=25 randomly selected bottles of water
X̅= 1.16 ppm and s = .20 ppm
The t Distribution

19. Example 4 (cont’d) Test the claim at α=.05 H0: µ≤1.0 ppm
Reject H0
The t Distribution

20. Example 4 (cont’d) (b) 2-sided 95% CIE of μ 1.16 2.0639(0.04)
1.16.083
1.077 ppm —— ppm
The t Distribution

21. Do Your Homework Practice, practice, practice.
Do lots and lots of problems. You can find these in the online lecture notes.
The t Distribution