|
The
TI 82/3 can display the "gradient function" or derivative of a curve. In other
words, given the curve y=f(x), the curve dy/dx=f'(x) (where f'(x) is the
derivative of f(x) with respect to x) can be displayed on the same screen.
To set this up, go to the Y= screen. Against Y1= enter a simple curve,
3x²+1 for example. Against Y2= you should make the following selections:
MATH 8 VARS Y-VARS 1 1 then type in ,X,X)
The
result should be that the Y= screen shows Y1=3x²+1 and
Y2=nDerive(Y1,X,X). |
 |
|
Go to the GRAPH screen, to see the graphs of
y=3x²+1 and the gradient function. In this case the gradient function
displayed will be that of y=6x. The calculator cannot tell you this though. You
will have to work that out for yourself. If now you change the equation
against Y1, the calculator will automatically show its
derivative.
|
 |
|
|
The
special statistical functions on a TI-83 are most easily found by using the
CATALOG function (2nd 0).
Normal
probabilities Suppose that X ~ N(0,1), the standard normal
distribution: P(X < 0.2) is given by normalcdf(-1E99,0.2)
=.5792596878 P(-0.1 < X < 0.3) is given by normalcdf(-0.1,0.2)
=.119087584 If P(X < z) = 0.7, then z is given by invNorm(0.7) =
.5244005101
Suppose this question was
asked: IQ scores are assumed to be normally distributed with mean 100 and
standard deviation 15. Find the percentage of people expected to have an IQ
between 80 and 120. X ~ N(100,15²), so P(80<X<120) is given
by: normalcdf(80,120,100,15) = .8175774363.
ShadeNorm(80,120,100,15) displays the distribution and the
probability too
Binomial
probabilities Suppose that X ~ B(12,0.2): P(X=4) is given by
binompdf(12,0.2,4)=.1328755507 P(X
£ 4) is given by
binomcdf(12,0.2,4)=.9274445005 A list of all the probabilites
for the distribution B(12,0.2) is given by
binompdf(12,0.2)
Poisson
probabilities Suppose that X ~ P(5): P(X=2) is given by
poissonpdf(5,2)=.0842243375 P(X
£ 2) is given by
poissoncdf(5,2)=.1246520195
Using STAT
LIST
A complete set of
probabilities can be calculated at once and displayed in a list format. For
example: Suppose that X ~ B(5,0.5). Select the STAT LIST option and
enter the numbers 0,1,2,3,4,5 in list L1. Next enter binompdf(5,0.5,L1)
L2 and you will find all the probabilities P(X=0) to
P(X=5) displayed in list L2.
A
"Goodness of Fit" test As part of ASA2 module S3, you are
required to carry out a Goodness of fit, or c², test. The following data is thought to
from a Poisson distribution. Carry out a c² test to test this hypothesis.
| X |
0 |
1 |
2 |
3 |
4 |
5 |
6 |
| O |
17 |
31 |
64 |
43 |
16 |
11 |
10 |
 Enter
the above data into the first two lists, and select STAT CALC 1. Enter
L1, L2 so that the display shows 1-Var Stats L1,L2. The display will
tell you that the mean, l, of the distribution
is 2.432291667. The total frequency is 192. Now enter poissonpdf(2.43229,
L1)
L3 to get the first display shown opposite. Next enter
L3 x 192
L4 and (L2-L4)²/L4
L5 to get the second display. Finally, LIST MATH 5 will
put sum( in the display. sum(L5) gives 15.15279386, which is the
c² test statistic to compare with the
relevant critical value.
Contingency
tables As part of ASA2 module S3, you are required to carry out a
Goodness of fit, or c², test for a two-way
Contingency table. The whole process can be perfomed on the
calculator. The manager of a leasure centre collected data on the usage
of facilities in the centre by its members. Test if there is any association
between gender and type of facility used.
|
Pool |
Jacuzzi |
Gym |
TOTALS |
| Male |
35 |
31 |
54 |
120 |
| Female |
68 |
32 |
30 |
130 |
| TOTALS |
103 |
63 |
84 |
250 |
| 1.
Select MATRX EDIT, enter 2 x 3 and the data from the above table -
not including the totals. |
 |
| 2.
Select STAT TESTS C. this will bring up the c²-Test display showing the Observed data in
matrix A and Expected data in matrix B. Confirm by pressing enter and you will
get the results screen, showing the c²
test statistic and the degrees of freedom: |
 |
| 3. The
expected data is displayed by selecting MATRX EDIT 2. In this case the
5% critical value would be 5.991, meaning we have a significant result and
there is evidence of an association between gender and type of facility
used. |
 |
It is
easy to forget where these special functions are on you calculator. Check this
before your exam. |