1:1-Var Stats – summary statistics for one set of raw data
Enter data : Press STAT, right arrow to CALC Press ENTER for l1
If the data to be used are in L2, press 2nd 2. Press ENTER
1:1-Var Stats – summary statistics for a frequency distribution.
Enter the class midpoints e.g., 10, 13, 16 19, 22, and 25 into L1.
Enter the frequencies e.g., 3, 2, 5, 3, 6, and 4 into L2.
Press STAT, right arrow to CALC
Press ENTER.
With class midpoints in L1 and corresponding frequencies in
L2,press 2nd 1 , 2nd 2
Press ENTER to get the results.
2-Var Stats
2:2-Var Stats – summary statistics for two sets of data.
Enter the numbers e.g, 10, 12, 13, 20, 18, and 17 into L1
Enter the numbers e.g, 23, 21, 25, 30, 32, and 40 into L2.
Enter the numbers e.g, 4, 1, 6, 5, 9, and 2 in L3.
Press STAT, right arrow to CALC and down arrow to
2:2-Var Stats
Press ENTER.
If the data to be used are in L1 and L2 (default columns), press
ENTER. If the data to be used are in other columns, press 2nd and
the appropriate numbers for the columns. To use L2 and L3 data,
press 2nd 2 , 2nd 3. Press ENTER for the results.
LinReg(ax+b)
4:LinReg(ax+b) – Linear Regression line for y = ax + b using x as the
predictor (input or independent) variable and y as the predicted
(output or dependent) variable.
Set up the calculator for a statistical plot with X-values in L1 and Yvalues in L2.
Press 2ND Y=. Press ENTER
While the cursor in over ON, press ENTER. This turns on the stat plot.
Press right arrow, down arrow, and ENTER to highlight the first graph.
Press down arrow to Xlist: and press 2ND 1 to put L1.
Press down arrow to Ylist: and press 2ND 2 to put L2.
Clear the home screen (2ND MODE).
Press 2ND 0 (for CATALOG), D, down arrow until you
reachDiasgnosticOn.
Press ENTER twice to get the screen at right.
Now, the calculator is ready to receive the data (X-values in L1 and Yvalues in L2).
Then we can find the linear regression equation, r, r2, and graph
the scatterplot and linear regression line.
Clear the home screen. Enter the numbers 10, 12, 13, 20, 18, and 17
into L1.
Enter the numbers 23, 21, 25, 30, 32, and 40 into L2.
Press 2ND MODE to save the data. Press STAT, right arrow to CALC,
and down arrow to 4:LinReg(ax+b). Press ENTER. Press 2ND 1,
comma (,), 2ND 2, comma (,) to use L1 for X-values and L2 for Yvalues. Press VARS and right arrow,Press ENTER Press ENTER
Press ENTER,Press ZOOM 9
MATH function
4:! – Calculate the Factorial of a number
to find 5! using the
TI-83 calculator,
enter 5, press MATH,
PRB, 4:!,
ENTER
Use of TI-83 Calculator to Compute
Binomial Probability Distribution Value
Find the probability that 2 successes will occur out of 5 trials if the probability of
success is 0.3. That is, find P(x = 2).
Press 2nd VARS [DISTR].
Scroll down to A:binompdf(
Press ENTER.
Enter 5,.3,2)
and press ENTER
to get the answer .3087.
The syntax is binompdf(n, p, x-value)
Find the probability that 2 or less successes will occur out of 5 trials if the
probability of success is 0.3. That is, find P(x ≤ 2)
Press 2nd VARS [DISTR]. Scroll down to B:binomcdf(
Press ENTER.
Enter 5,.3,2) and press ENTER
to get the answer .83692.
The syntax is binomcdf(n, p, largest x-value).
Use of TI-83 Calculator to
Compute Normal Probability Distribution Values;
Compute the Value(s) Given an Area
Calculate the Normal Probability that a ≤ x ≤ b (cumulative).
Since z-scores are for the Standard Normal Distribution,
the syntax is normalcdf(smaller z, larger z).
Example (TI-83): Find the probability that a z-score
is between -1.5 and 2. That is, find P(-1.5 ≤ z ≤ 2).
Find probability or area
Press 2nd VARS [DISTR]. Scroll down to 2:normalcdf(
Press ENTER.
Enter -1.5,2) and press ENTER to get the answer .91044.
Since z-scores are for the Standard Normal Distribution,
the syntax is invNorm(area to left of desired z).
Example (TI-83): Find the z-score for an area of
0.25 to the left of the z-score.
Press 2nd VARS [DISTR]. Scroll down to 3:invNorm
Press ENTER.
Enter .25) and press ENTER to get the answer -.67. The syntax
is invNorm(area to left of desired z).
EXAMPLE: 1 Part 1
Example (TI-83): Adult IQs are normally distributed with µ = 100 and σ = 15.
Find the probability that a randomly selected IQ is less than 112. That is, find P(x
< 112). Since this is a continuous function, we have P(x < 112) = P(x ≤ 112).
Press 2nd VARS [DISTR]. Scroll down to 2:normalcdf(
Press ENTER.
Enter -9999,112,100,15) and press ENTER
to get the answer .7881. The syntax is normalcdf(smaller, larger, µ,
σ).
Note: The -9999 is used as the smaller value to be at least 5 standard
deviations from the mean.
EXAMPLE 1 Part 2
Find the probability that a randomly selected IQ is at least 122. That is, find P(x
≥ 122).
Press 2nd VARS [DISTR]. Scroll down to 2:normalcdf(
Press ENTER.
Enter 122,9999,100,15)
and press ENTER
to get the answer .0712.
The syntax is normalcdf(smaller, larger, µ, σ).
Note: The 9999 is used as the larger value to be at least 5 standard
deviations from the mean.
EXAMPLE 1 Part 3
Find the probability that a randomly selected IQ is between 112 and 122.
That is, find P(112 ≤ x ≤ 122).
Press 2nd VARS [DISTR]. Scroll down to 2:normalcdf(
Press ENTER.T
Enter 112,122,100,15) and press ENTER
to get the answer .1406. syntax is normalcdf(smaller, larger, µ, σ).
TESTS – Allows hypothesis testing, develops confidence intervals,
Hypothesis Test of Mean (z Test)
for Normal Distribution - One Large Sample
Press STAT and the right arrow twice to select TESTS.
Use the down arrow to select
1:Z-Test…
Press ENTER
Use right arrow to select Stats
(summary values rather than raw data).
Enter hypothesized mean, standard deviation, mean, and sample
size. Select alternate hypothesis.
Press down arrow to select Calculate and press ENTER.
Results:
Since the p-value is 0.1, do not reject the null hypothesis with an
alpha value of 0.10 or smaller (10% level of significance or smaller).
Hypothesis Test of Mean (2 sample z test)
for Normal Distribution - Two Large Samples
Press STAT and the right arrow twice to select TESTS.
Use the down arrow to select
3:2-SampZTest…
Press ENTER.
Use right arrow to select Stats
(summary values rather than raw data).
Enter standard deviations, mean and sample size for samples 1 and
2.
Select alternate hypothesis.
Press down arrow to select Calculate and press ENTER
Results:
Since the p-value is 0.004, reject the null hypothesis with an alpha
value of 0.10 or smaller (10% level of significance or smaller).
Conclude that the two population means are not different.
Hypothesis Test of Proportion
for Normal Distribution - One Large Sample
Press STAT and the right arrow twice to select TESTS.
Use the down arrow to select
5:1-PropZTest…
Press ENTER.
Enter hypothesized proportion, number of favorable outcomes, X,
sample size, n, and select the alternate hypothesis.
Use down arrow to select Calculate and press ENTER.
Results:
Since the p-value is near zero, reject the null hypothesis. Conclude
that the sample proportion of 0.15 is significantly different from the
hypothesized proportion of 0.52.
Hypothesis Test of Proportion
for Normal Distribution - Two Large Samples
Press STAT and the right arrow twice to select TESTS.
Use the down arrow to select
6:2-PropZTest… Press ENTER.
Enter number of favorable outcomes and sample size of samples 1
and 2. Select the alternate hypothesis.
Use down arrow to select Calculate and press ENTER.
Results:
Since the p-value is near zero, reject the null hypothesis. Conclude
that the sample 1 proportion of 0.90 is significantly greater than
sample 2 proportion of 0.50.
Hypothesis Test of Mean
for Normal Distribution (Sigma, σ, is Known) - One Sample
Example: A sample of size 200 has a mean of 20. Assume the population
standard deviation is 6. Use the TI-83 calculator to test the hypothesis that the
population mean is not different from 19.2 with a level of significance of α =
5%.
Solution: “The population mean is not different from 19.2” means the same as
“the population mean is equal to 19.2.” Therefore, the null and alternate
hypotheses are H0: µ = 19.2 and H1: µ ≠ 19.2, respectively. Follow the steps
below to solve the problem using the TI-83. [NOTE: If the p-value < α, reject
the null hypothesis; otherwise, do not reject the null hypothesis.]
Press STAT and the right arrow
twice to select TESTS.
To select the highlighted
1:Z-Test…
Press ENTER.
Use right arrow to select Stats (summary values rather than raw data)
and Press ENTER.
Use the down arrow to Enter the hypothesized mean, population
standard deviation, sample mean,and sample size.
Select alternate hypothesis.
Press down arrow to select Calculate and press ENTER.
Results:
Since the p-value is 0.1, do not reject the null hypothesis
with an α(alpha) value of 0.10 or smaller (10% level of significance or
smaller). [In this example, α = 0.05.]
Hypothesis Test of Mean
for Normal Distribution (Sigma, σ, is Known) - Two Samples
Example: Two samples were taken, one from each of two populations. Use
the TI-83 calculator to test the hypothesis that the two population means are
not different with a level of significance of α = 5%.
Solution: For the two samples, we have the following summary data:
n1 = 38
σ1 = 5
n2 = 35
σ2 = 7
H0: µ1 = µ2
H1: µ1 ≠ µ2
Use α = 5%
“The two population means are not different” means the same as “the two
population means are equal.” Therefore, the null and alternate hypotheses are
H0: µ1 = µ2 and H1: µ1 ≠ µ2, respectively. Follow the steps below to solve the
problem using the TI-83. [NOTE: If the p-value < α , reject the null hypothesis;
otherwise, do not reject the null hypothesis.
Press STAT and the right arrow
twice to select TESTS.
Use the down arrow to select
3:2-SampZTest…
Press ENTER
Use right arrow to select Stats
(summary values rather than raw data).
Enter standard deviations, mean and sample size for samples 1 and
2.
Select alternate hypothesis.
Press down arrow to select Calculate and press ENTER
Results:
Since the p-value is 0.0186, reject the null hypothesis with an alpha
value of 0.05 or larger (5% level of significance or larger).
Conclude that the two population means are not different.
Hypothesis Test of Mean
for Student T-Test - One Small Sample
Example: A sample of size 20 has a mean of 110 and a standard deviation of
16. Use the TI-83 calculator to test the hypothesis that the population mean is
greater than 100 with a level of significance of  = 5%.
Solution: “The population mean is greater than 100” means the
alternate hypothesis is H1:  > 100, and the null hypothesis is
H0:   100. Follow the steps below to solve the problem using the TI83. [NOTE: If the p-value < , reject the null hypothesis; otherwise, do
not reject the null hypothesis.
Press STAT and the right arrow twice to select TESTS.
To select the highlighted
2:T-Test…Press ENTER.
Use right arrow to select Stats (summary values rather than raw data)
and Press ENTER.
Use the down arrow to Enter the hypothesized mean, sample mean,
standard deviation, and sample size.
Select alternate hypothesis.
Press down arrow to select Calculate and press ENTER.
Results:
Since the p-value is 0.0058, reject the null hypothesis with an alpha
value larger than 0.0058 (0.58% level of significance or larger).
Hypothesis Test of Mean
for Student T-Test - Two Small Samples
Example: Two samples were taken, one from each of two populations. Use the
TI-83 calculator to test the hypothesis that the two population means are equal
with a level of significance of  = 2%.
Solution: For the two samples, we have the following summary data:
sx1 = 5
sx2 = 7
n1 = 8
n2 = 5
H0: 1 = 2
H1: 1  2
Use  = 2%
“The two population means are equal” means the null and alternate hypotheses
are H0: 1 = 2 and H1: 1  2, respectively. Follow the steps below to solve the
problem using the TI-83. [NOTE: If the p-value < , reject the null hypothesis;
otherwise, do not reject the null hypothesis.
Press STAT and the right arrow twice to select TESTS.
Use the down arrow to select
4:2-SampTTest…
Press ENTER.
Use right arrow to select Stats
(summary values rather than raw data).
Enter sample mean, standard deviation, and sample size for samples
1 and 2.
Select alternate hypothesis.
Select pooled: Yes
Press down arrow to select Calculate and press ENTER.
Results:
Since the p-value is 0.384, do not reject the null hypothesis with an
alpha value of 0.02 (because 0.384 is not less than 0.02).
Conclude that the two population means are not different.
Hypothesis Test of Proportion
for Normal Distribution - One Sample
Example: In sampling 200 people, we found that 30% of them favored a certain
candy. Use α = 10% to test the hypothesis that the proportion of people who
favored that candy is less than 35%.
Solution: This represents a one-sample test of proportion. So we use the "1PropZTest" function. The sample proportion is 30% or p = 0.30, and the
hypotheses are H0: p ≥ 0.35 and H1: p < 0.35 (claim). Hypothesized value is 0.35.
Press STAT and the right arrow twice to select TESTS.
Use the down arrow to select
5:1-PropZTest…
Press ENTER.
Enter hypothesized proportion, number of favorable outcomes, x,
sample size, n, and select the alternate hypothesis.
Use down arrow to select Calculate and press ENTER
Results:
Since the p = 0.069 is less than α = 0.10, reject the
null ypothesis. Conclude that the sample proportion of 0.30 is
significantly less than the hypothesized proportion of 0.35.
Confidence Interval of Mean
for Normal Distribution - One Large Sample
Confidence Interval of Proportion
for Normal Distribution - One Large Sample
Confidence Interval of Mean
for T-Test - One Small Sample