State the difference between the shape of the t distribution and the normal distribution

State how the difference between the shape of the t distribution and normal distribution
is affected by the degrees of freedom

Use a t table to find the value of t to use in a confidence interval

Use the t calculator to find the value of t to use in a confidence
interval

In the introduction to normal distributions it
was shown that 95% of the area of a normal distribution is within
1.96 standard deviations of the mean. Therefore, if you randomly
sampled a value from a normal distribution with a mean of 100,
the probability it would be within 1.96σ of 100 is 0.95.
Similarly, if you sample N values from the population, the probability
that the sample mean (M) will be within 1.96 σ_{M}
of 100 is 0.95.

Now consider the case in which you have a normal
distribution but you do not know the standard deviation. You
sample N values and compute the sample mean (M) and estimate
the standard error of the mean (σ_{M})
with s_{M}.
What is the probability that M will be within 1.96 s_{M} of the population mean (μ)? This is a difficult problem because
there are two ways in which M could be more than 1.96 s_{M} from μ: (1) M could, by chance, be either very high or very
low and (2) s_{M} could, by chance, be
very low. Intuitively, it makes sense that the probability
of being within 1.96 standard errors of the mean should be
smaller than in the case when the standard deviation is known
(and cannot be underestimated). But exactly how much smaller?
Fortunately, the way to work out this type of problem was solved
in the early 20th century by W. S. Gosset who determined the
distribution of a mean divided by an estimate
of its standard error. This distribution is called the Student's
t distribution or sometimes just the t
distribution. Gosset worked out the t distribution
and associated statistical tests while working for a brewery
in Ireland. Because of a contractual agreement with the brewery,
he published the article under the pseudonym "Student." That
is why the t test is called the "Student's t test."

The t distribution is very similar to the normal
distribution when the estimate of variance
is based on many degrees
of freedom, but has relatively more scores in its tails
when there are fewer degrees of freedom. Figure 1 shows t distributions with 2, 4, and 10 degrees of freedom and the standard normal distribution. Notice that the normal distribution has relatively more scores in the center of the distribution and the t distribution has relatively more in the tails. The t distribution is therefore leptokurtic. The t distribution approaches the normal distribution as the degrees of freedom increase.

Figure 1. A comparison of t distributions with 2, 4, and 10 df and the standard normal distribution. The distribution with the lowest peak is the 2 df distribution, the next lowest is 4 df, the lowest after that is 10 df, and the highest is the standard normal distribution.

Since the t distribution is leptokurtic, the
percentage of the distribution within 1.96 standard deviations
of the mean is less than the 95% for the normal distribution.
Table 1 shows the number of standard deviations from the mean
required to contain 95% and 99% of the area of the t distribution
for various degrees of freedom. These are the values of t that
you use in a confidence interval. The corresponding values for
the normal distribution are 1.96 and 2.58 respectively. Notice
that with few degrees of freedom, the values of t are much higher
than the corresponding values for a normal distribution and that
the difference decreases as the degrees of freedom increase.
The values in Table 1 can be obtained from the "Find
t for a confidence interval" calculator.

Table 1. Abbreviated t table.

df

0.95

0.99

2

4.303

9.925

3

3.182

5.841

4

2.776

4.604

5

2.571

4.032

8

2.306

3.355

10

2.228

3.169

20

2.086

2.845

50

2.009

2.678

100

1.984

2.626

Returning to the problem posed at the beginning of this section,
suppose you sampled 9 values from a normal population and estimated
the standard error of the mean (σ_{M})
with s_{M}. What is the probability that
M would be within 1.96s_{M} of μ? Since
the sample size is 9, there are N - 1 = 8 df. From Table 1 you
can see that with 8 df the probability is 0.95 that the mean will
be within 2.306 s_{M} of μ. The probability that it will be within 1.96 s_{M} of μ is therefore lower than 0.95.

As shown
in Figure 2, the "t
distribution" calculator can be used to find that 0.086 of
the area of a t distribution is more than 1.96 standard deviations
from the mean, so the probability that
M would be less than 1.96sM from μ is 1 - 0.086 = 0.914.

Figure 2. Area more than 1.96 standard
deviations from the mean in a t distribution with 8 df.
Note that the two-tailed button is selected so that the area
in both tails will be included.

As expected, this probability is less than 0.95 that would have
been obtained if σM had been known
instead of estimated.