Using the TI 83/84
Summary Statistics
• Press STAT, ENTER (for EDIT)
• If you do not see L1, scroll left. If you still don’t see it,
press INSERT, 2nd, L1 (this is the 1 key)
• If there are old data under L1:
– Press the up arrow, then CLEAR, ENTER
– DON’T use DELETE, ENTER. This causes L1 to disappear
instead. Use INSERT, 2nd, L1 to get it back.
• Enter data values in L1 one at a time, pressing ENTER
after each.
– If you make an error, use the up or down arrows to highlight
the error, then enter the correct value. Use the arrows to get
to the bottom of the list for the next value, if necessary.
– Be sure to press ENTER after the last data value.
Summary Statistics, Continued
• Press STAT, Right Arrow (for CALC),
ENTER
• Press ENTER (for 1-Var Stats)
• Press ENTER again
• Read results
– The Standard Deviation is labeled Sx
Normal Probablilities
• Find P(90 < x < 105) if x follows the
normal model with mean 100 and standard
deviation 15:
• P(90 < x < 105)
x
x
= normalcdf( 90 , 105 , 100 , 15)
= .378
1
2
Normal Quantiles
• We must find a so that P(x < a) = 2% when x has
a normal distribution with a mean of 100 and a
standard deviation of 15.
• With the TI 83/84:
a = invNorm( .02, 100 , 15) = 69.2
x
Calculating r and Regression
Coefficients
• The first time you do this:
– Press 2nd, CATALOG (above 0)
– Scroll down to DiagnosticOn
– Press ENTER, ENTER
– Read “Done”
– Your calculator will remember this setting
even when turned off
Calculating r and Regression Coefficients Continued
• Press STAT, ENTER
• If there are old values in L1:
– Highlight L1, press CLEAR, then ENTER
• If there are old values in L2:
– Highlight L2, press CLEAR, then ENTER
• Enter predictor (x) values in L1
• Enter response (y) values in L2
– Pairs must line up
• Press STAT, > (to CALC)
• Scroll down to LinReg(ax+b), press ENTER,
ENTER
• Read a, b, r and r2
Binomial Probabilities
• To find binomial probability distribution values –
probabilities for a particular number of successes, say 3
successes in 5 trials with the chance of success = .9:
–
–
–
–
Let n = 5, p = .9, and x = 3 (Note the order: n, p, x)
Press 2nd, DISTR [VARS]
Scroll down to 0 and press ENTER, or just press 0
“binompdf(“ appears. Press 5, .9, 3 ) ENTER
(Note that the commas are required, and note the order: n, p, x)
– .0729 appears
– If the values for x = 2, 3, and 4 are all needed
•
•
•
•
Press 2nd, DISTR [VARS]
Scroll down to 0 and press ENTER, or just press 0
“binompdf(“ appears. Press 5, .9, {2, 3, 4} ) ENTER
All 3 values appear
– In general, press 2nd, DISTR, 0, then enter n, p, x
Binomial Probabilities Continued
• To find the binomial probability for a range of
successes, say 3 or fewer successes in 5 trials
with the chance of success = .9 :
–
–
–
–
Let n = 5, p = .9, and x = 3
Press 2nd, DISTR [VARS]
Scroll down to A and press ENTER
“binomcdf(“ appears. Press 5, .9, 3 ) ENTER
(Note that the commas are required)
• .0815 appears
– If the probability for fewer than 3 successes is needed:
• Enter binomcdf(5, .9, 2 ); read .00856
– If the probability for more than 3 successes is needed:
• Enter 1 - binomcdf(5, .9, 3 ); read .919
Confidence Intervals for Proportions
• Press STAT and use the cursor
to highlight TESTS.
• Scroll down to A: 1-PropZInt…
and press ENTER.
• The TI remembers your previous
problem and shows the entries
for it. You will overwrite these
with the new entries. After x:
enter the number of successes in
the sample.
• After n: enter the sample size.
• After C-Level: enter the
confidence level as a decimal
fraction.
• When the cursor blinks on
Calculate, press ENTER again.
Hypothesis Test for Proportions
• Of 4276 households sampled, 4019 had telephones. Test, at the 1%
level, the claim that the percentage of households with telephones is
now greater than 93%.
 = .01
Claim: p > .93
H0: p < .93
.93
After p0 : enter .35
(the null hypothesis value)
After x : enter 4019
(number of successes in the sample
After n : enter 4276
(or p = .93)
(the sample size)
HA: p > .93
After prop highlight > p0
Press STAT,
TESTS,
1-PropZTest,
Enter
(this is H A )
Highlight CALCULATE
Press ENTER
11
Hypothesis Test for Proportions, Continued
• Of 4276 households sampled, 4019 had telephones. Test the claim
that the percentage of households with telephones is greater than
the 93%.
H0: p < .93
(or p = .93)
HA: p > .93
Read
1-PropZTest
.93
prop > .35
Test Performed
HA
2.54 z-score
z  80.87315928
p-Value
p  .00560
0
Sample Estimate
.940 for p
pˆ  .9398971001
n  4276 Sample Size
12
Hypothesis Test for Two Proportions
Example: For the racial profiling sample data,
use a 0.05 significance level to test the claim that the
proportion of black drivers stopped by the police is greater
than the proportion of white drivers who are stopped.
We have that:
nB= 200
xB = 24
nW = 1400
xW = 147
And,
H0: pB ≤ pW
HA: pB > pW
On the TI-83/84:
Press STAT, arrow to TESTS and scroll
down to 2-PropZTest. Press Enter.
Make
these
entries:
Highlight
Calculate and
press ENTER
13
Hypothesis Test for Two Proportions,
Continued
Your TI screen should look
like this:
This is HA
This is the p- value.
This is p
If you press the down arrow twice,
the TI will tell you the sample sizes,
as a check.
14
Confidence Interval for a Mean
n = 106
y = 98.20o
s = 0.62o
 = 0.05
15
Hypothesis Test for a Mean
Claim :   1017
H 0 :   1017
H1 :   1017
TI  83 / 84 entries:
STAT TESTS 2
Stats Enter
0 : 1017
x : 1040
sx : 207
n : 20
 :  0 Enter
TI  83 / 84 entries (continued):
Highlight Calculate
Enter
Read
 >1017 ( H1 )
t  .4969...
p  .312787...
x  1040
sx  207
n  20
16
Hypothesis Test for Two Means
Press STAT and arrow over to TESTS.
Scroll down to 2-SampTTest and press
ENTER.
Highlight HA as μ1: ≠ μ2
Your screen should look like this:
Highlight STATS and press ENTER
Set μ0: 0
Enter McGwire’s stats for sample 1
McGwire
n1
x1
s1
70
418.5
45.5
Enter Bonds’s stats for sample 2
Bonds
n2
x2
s2
73
403.7
30.6
Select Pooled: No
Select Calculate
Press ENTER
Hypothesis Test for Two Means,
Continued
Your screen should look like this:
HA
Test Statistic
P-Value
Scrolling down reveals some to the statistics
you entered earlier.
Because the P-value is
less than .05, we reject The
Null Hypothesis
The sample data support the
claim that there is a
difference between the
mean home run distances of
Mark McGwire and Barry
Bonds.
Confidence Interval for Two Means
Using the sample data given in the preceding example,
construct a 95%confidence interval estimate of the
difference between the mean home run distances of Mark
McGwire and Barry Bonds. We need E so that
(x1 – x2) – E < (µ1 – µ2) < (x1 – x2) + E
As before, Press STAT, go to TESTS, and
now scroll down to 2-SampTInt and press
ENTER.
The TI remembers the entries from the
last time 2-SampTTest or 2-SampTInt was
used. Since these are the same, we need
only scroll down to C-Level to make sure it
is set to .95 (NOT 95%!). Pooled is
always No for our work.
Highlight CALCULATE and press ENTER.
Matched Pairs Test
TI-83/84 users may test paired sample hypotheses
from raw data as follows:
Enter first half of each
STAT
pair in L2
EDIT
Enter second half of each
Highlight L1
pair in L3
CLEAR
2nd QUIT
ENTER
2nd L2 – 2nd L3 STO> 2nd L1
Highlight L2
CLEAR
STAT
ENTER
TESTS
Highlight L3
2: T-Test
CLEAR
Highlight Data
ENTER
ENTER
Proceed as for one-sample
t - Test
Matched Pairs Test, Continued
Claim: there is a difference between the actual
low temperatures and the low temperatures that
were forecast five days earlier
That is, μd ≠ 0
H 0:  d = 0
H 1:  d  0
Matched Pairs Confidence Interval
Follow the instructions given above for entering the 2 samples into
L2 and L3, then storing the difference in L1.
We now need only compute a one-sample confidence
interval using STAT
TESTS
TInterval
ENTER
Data
List: L1
Freq: 1
C-Level: .95
Calculate
ENTER
Goodness-of-Fit Test
1. STAT,
EDIT,
CLEAR
L1, L2
and L3
2. Enter
observed
counts in
L2.
3. Enter
expected
counts in L3.
4. 2nd Quit
Store
(Obs-Exp) 2
in L1
Exp
5. STAT,
CALC,
1-Var Stats,
ENTER.
Write down
value of Σx
6. 2nd, DISTR,
8:X2cdf,
ENTER, Σx,
1000, k-1)
Read P-Value
Test for Homogeneity
Claim: There is no difference in the distribution of
the continent of origin for student and staff cars.
H0: There is no difference in the distribution of the
continent of origin for student and staff cars.
HA: There is a difference in the distribution of the
continent of origin for student and staff cars.
Set α = .05