# MATH200B Program —

Extra Statistics Utilities for TI-83/84

Copyright © 2008–2020 by Stan Brown

Extra Statistics Utilities for TI-83/84

Copyright © 2008–2020 by Stan Brown

Your first course in statistics probably won’t use these features, but they’re offered here for advanced students and those who are studying on their own.

**See also:**
Troubles? See TI-83/84 Troubleshooting.

`MATH200B`

Program OverviewBecause this program helps you,

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please click to donate!Because this program helps you,

please donate at

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The program is in two parts, MATH200B and
MATH200Z. You need both on your calculator, even though
**you won’t run MATH200Z directly.** It works with
all TI-83 Plus calculators and all TI-84 calculators, including the
color models.

**If you have a “classic” TI-83**, not a Plus or Silver,
follow the directions below but put M20083B and
M20083Z on your calculator, not MATH200B and
MATH200Z. (M20083B and M20083Z aren’t
being updated after version 4.2,
which was released in August 2012, so you will see some differences
from the screen shots in this document.)

There are three methods to get the programs into your calculator:

- If you have a TI-84, download MATH200B.zip (110 KB, updated 18 July 2019), and unzip it. Use the USB cable that came with your calculator, and the free TI Connect CE software from Texas Instruments, to transfer the MATH200B.8XP and MATH200Z.8XP programs to your calculator.
- If a classmate has the programs on her calculator
(any model TI-83/84),
she can transfer them to yours, provided you both have a USB port
or you both have a round I/O port. Connect the appropriate cable to
both calculators, inserting each end firmly.
On your calculator, press [
`2nd`

`x,T,θ,n`

*makes*`LINK`

] [`►`

] [`ENTER`

]. Then on hers press [`2nd`

`x,T,θ,n`

*makes*`LINK`

] [`3`

], select MATH200B (or M20083B; see above), and finally press [`►`

] [`ENTER`

]. If you get a prompt about a duplicate program, choose Overwrite.Repeat for MATH200Z (or M20083Z; see above).

- Or, as a last resort, key in the programs. See MATH200B.pdf, MATH200Z.pdf, and MATH200B_hints.htm in the MATH200B.zip file.

Press the [`PRGM`

] key. If you can see `MATH200B`

in the menu, press its number; otherwise, scroll to it and press
[`ENTER`

]. When the program name appears on your home screen,
press [`ENTER`

] a second time to run it. Check the splash screen to make sure you have the
latest version (v4.4a), then press
[`ENTER`

].

The menu at right shows what the program can do:

`Skew/kurtosis`

: compute skewness and kurtosis, which are numerical measures of the shape of a distribution`Time series`

: plot time-series data`Critical t`

: find the t value that cuts the distribution with a given probability in the right-hand tail`Critical χ˛`

: find the χ˛ value that cuts the distribution with a given probability in the right-hand tail`Infer about σ`

: hypothesis tests and confidence intervals for population standard deviation and variance`Correlatn inf`

: hypothesis tests and confidence intervals for the linear correlation of a population; the hypothesis test for correlation doubles as a hypothesis test for slope of the regression line`Regression inf`

: confidence intervals for slope of the regression line,*y*intercept, and*ŷ*for a particular*x*, plus prediction intervals for*ŷ*for a particular*x*

If you ever need to break out of the program
before finishing the prompts, press [`ON`

] [`1`

].

If you run the program on a TI-84 with a higher-resolution screen, some displays will look slightly different, but all keystrokes will be the same.

The program is protected so that you can’t edit it accidentally. If you want to look at the program source code, see MATH200B.PDF and MATH200Z.PDF in the downloadable MATH200B.ZIP file.

Each procedure leaves its results in variables in case you want to use them for further computations. For details, please see the separate document MATH200B Program — Technical Notes.

A histogram gives
you a general idea of the shape of a data set, but two
**numeric measures of shape**
are also available. **Skewness** measures how far a
distribution departs from symmetry, and in which direction.
**Kurtosis** measures the height or shallowness of the central
peak, using the normal distribution (bell curve) as a reference.

The `1:Skew/kurtosis`

part of the `MATH200B`

program computes these statistics
for a list of numbers or a grouped or ungrouped frequency
distribution. This section of the document explains how to use the
program and how to interpret the numbers.

**See also:**
For interpretation of skewness and kurtosis, and
technical details of how they are
calculated, see Measures of Shape: Skewness and Kurtosis.

If you have a frequency or probability distribution, put the data points or class midpoints (class marks) in one statistics list and the frequencies or probabilities in another. If you have a simple list of numbers, put them in a statistics list.

Then press [`PRGM`

], scroll if necessary and select
`MATH200B`

, and in the program menu select `1:Skew/kurtosis`

. Specify your
data arrangement, enter your data list, and if appropriate enter your
frequency or probability list. The program will produce a great many
statistics.

Here are grouped data for heights of 100 randomly selected male students:

Class boundaries | 59.5–62.5 | 62.5–65.5 | 65.5–68.5 | 68.5–71.5 | 71.5–74.5 |
---|---|---|---|---|---|

Class midpoints, x |
61 | 64 | 67 | 70 | 73 |

Frequency, f |
5 | 18 | 42 | 27 | 8 |

Data are adapted from Spiegel 1999 [full citation in “References”, below], page 68. |

A histogram, prepared with the MATH200A
program, shows the data are skewed
left, not symmetric.
But **how highly skewed** are they? And
**how does the central peak compare** to the normal distribution
for height and sharpness? To answer these questions, you have to
compute the skewness and kurtosis.

Enter the *x*’s in one statistics list and the *f*’s in
another. If you’re not sure how to create statistics lists,
please see Sample Statistics on TI-83/84.

Then run the `MATH200B`

program and select `1:Skew/kurtosis`

.
Your data arrangement is `3:Grouped dist`

.
When prompted, enter the list that contains the
*x*’s and then the list that contains
the *f*’s. I’ve used L5 and L6, but you could use any lists.

The program gives its results on three screens of data.

The first screen shows some basic statistics: the sample size,
the mean, the standard deviation, and the variance.
As usual, you have to consider whether the data are a sample or the
whole population; the program gives you both σ and *s*,
σ˛ and *s*˛.

The program stores key results in variables in case you want to do any further computations with them. See MATH200B Program — Technical Notes for a complete list of variables computed by the program.

The second screen shows results for **skewness**. The third moment divided
by the 1.5 power of the variance is the skewness, which is about
−0.11 for this data set. Again, you are given the values to
use if this is the whole population and if it is a sample.

If this is the whole population, then you stop with the first skewness figure and can state that the population is negatively skewed (skewed left).

But this is just a sample, so you use the “as
sample” figure for your skewness. (This is also the figure that
Excel reports.)
The sample is negatively skewed (skewed left),
but can you say anything about the skew of the population?
To answer that question, use the standard
error of skewness, which is also shown on the screen.
As a **rule of thumb**, if sample
skewness is more than about two standard errors either side of zero,
you can say that the population is skewed in that direction.
In this example, the standard error of skewness is 0.24, and
the statistic of −0.45 tells you that
the skewness is only 0.45 standard errors below zero. This is not enough to
let you say anything about whether the population is skewed in either
direction or symmetric.

The last screen shows results for **kurtosis**. The fourth moment divided
by the square of the variance gives the kurtosis, which is 2.74.
Some authors, and Microsoft Excel, prefer to subtract 3 and consider
the **excess kurtosis**: 2.74−3 is −0.26.

A bell curve (normal distribution) has kurtosis of 3 and
excess kurtosis of 0. If excess kurtosis is negative, as it is here,
then the distribution has a lower peak and higher
“shoulders” than a normal distribution, and it is called
**platykurtic**.
(An excess kurtosis greater than 0 would mean that the distribution
was **leptokurtic**, with a narrower and higher peak than a bell
curve.)

Since this is just a sample, and not the whole population, use
the “as sample” excess kurtosis of −0.21. (This is
the figure Excel reports.)
Can you say anything about the kurtosis of the population from
which this sample was taken? Yes, just as you did for skewness.
The **rule of thumb**
is that an excess kurtosis of at least two standard errors
is significant. For this sample,
the standard error of kurtosis is 0.48, and −0.21/0.48 =
−0.44, so the excess
kurtosis is only 0.44 standard errors below zero. (Or,
the kurtosis is only 0.44 standard errors below 3.) Therefore you
can’t say whether the population is peaked like a normal
distribution, more than normal, or less than normal.

On high-resolution screens (the TI-84 Plus C and TI-84 Plus CE), there’s enough room to show skewness and kurtosis on the same screen, as shown at right.

You can also use this part of the program to compute the shape of a probability distribution. For instance, here’s the probability distribution for the number of spots showing when you throw two dice:

Probability Distribution for Throwing Two Dice | |||||||||||
---|---|---|---|---|---|---|---|---|---|---|---|

Spots, x |
2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 |

Probability, P(x) |
1/36 | 2/36 | 3/36 | 4/36 | 5/36 | 6/36 | 5/36 | 4/36 | 3/36 | 2/36 | 1/36 |

The *x*’s go in one list and the *P*’s in another.
(Enter the probabilities as
fractions, not decimals, to ensure that they
add to exactly 1. The calculator displays rounded decimals but
keeps full precision internally, and the program will tell you if your
probabilities don’t add to 1.)
Now run the `MATH200B`

program and select `1:Skew/kurtosis`

.
Your data arrangement is
`4:Discrete PD`

,
and you’ll see the following results:

On the first screen, no sample size is shown because a probability distribution is a population.

On the second screen, the skewness is essentially zero. This confirms what you can see in the histogram: the distribution is symmetric. Standard error and test statistic don’t apply because you have a probability distribution (population) rather than a sample.

On the same screen, the kurtosis is 2.37 (not shown for reasons of space), and the excess kurtosis is −0.63; the dice make a platykurtic distribution. Compared to a normal distribution, this distribution of dice throwing has a lower, less distinct peak and shorter tails.

On high-resolution screens, namely the TI-83 Plus C Silver Edition and TI-84 Plus CE, complete information about a probability distribution fits on one screen.

You may notice that, although the skewness is still essentially zero, it’s a different very small number from the very small number the older TI-84s gave us, on the screen shot above. I can’t account for this in detail, but I think it’s likely that the newer calculator’s chip processes floating point with very slightly different precision than the old one. Don’t obsess about it — for all practical purposes, both numbers are zero.

**Summary:**
To plot a time series or trend line,
put the numbers in a statistics list and use the `2:Time series`

part of
the `MATH200B`

program.

**Example**:
Let’s plot the closing prices of Cisco Systems stock
over a two-year period. The following table is adapted from
Sullivan 2008 [full citation in “References”, below],
page 82, which credits NASDAQ as the source.

Month | 3/03 | 4/03 | 5/03 | 6/03 | 7/03 | 8/03 | 9/03 | 10/03 |
---|---|---|---|---|---|---|---|---|

Closing | 12.98 | 15.00 | 16.41 | 16.79 | 19.49 | 19.14 | 19.59 | 20.93 |

Month | 11/03 | 12/03 | 1/04 | 2/04 | 3/04 | 4/04 | 5/04 | 6/04 |

Closing | 22.70 | 24.23 | 25.71 | 23.16 | 23.57 | 20.91 | 22.37 | 23.70 |

Month | 7/04 | 8/04 | 9/04 | 10/04 | 11/04 | 12/04 | 1/05 | 2/05 |

Closing | 20.92 | 18.76 | 18.10 | 19.21 | 18.75 | 19.32 | 18.04 | 17.42 |

Enter the closing prices in a statistics list such as L1, ignoring the dates.

Now run the `MATH200B`

program and select `2:Time series`

. The program prompts you
for the data list. (**Caution**: The program assumes the
time intervals are all equal. If they aren’t, the horizontal
scale will not be uniform and the graph will not be correct.)

It’s usually good practice to start the vertical scale at
zero, or in other words to show the *x* axis at its proper level on
the graph. But the program gives you the choice. If you have good
reason, you can let the program scale the data to take up the entire screen.
This exaggerates the amount of change from one time period to the
next.
(If the data include any negative or zero values, the
*x* axis will naturally appear in the graph, and program skips the
yes/no prompt.)

Below you see the effect of a “yes” at left and the effect of a “no” at right.

As you can see, the graph that doesn’t include the zero
looks a lot more dramatic, with bigger changes. But that can be
deceptive. A more accurate picture is shown in the first graph,
the one that does include the *x* axis.

If you wish, you can press the [`TRACE`

] key and display
the closing prices, scrolling back and forth with the
[`◄`

] and [`►`

] keys. If you want
to jump to a particular month, say June 2004, the 16th month, type
`16`

and then press [`ENTER`

].

The TI-83 doesn’t have an `invT`

function as
the TI-84 does, but if you need to find critical *t* or inverse *t* on
either calculator you can use this part of the `MATH200B`

program.

Caution: our notation of *t*(*df*,*rtail*) matches most books
in specifying the area of the right-hand tail for critical *t*. But the
TI calculator’s built-in menus specify the area of the left-hand tail. Make sure
you know whether you expect a positive or negative *t* value.

Some textbooks interchange the arguments: *t*(*rtail*,*df*).
Since degrees of freedom must always be a whole number and the tail
area must always be less than 1, you’ll always know which
argument is which.

**Example**: find *t*(27,0.025), the *t* statistic with 27 degrees of
freedom (sample size 28) for a one-tailed significance test with
α = 0.025, a two-tailed test with
α = 0.05, or a confidence interval with
1−α = 95%.

**Solution**: run the `MATH200B`

program and select `3:Critical t`

. When
prompted, enter 27 for degrees of freedom and 0.025 for the area of
the right-hand tail, as shown in the first screen. After a short
pause, the calculator gives you the answer: *t*(27,0.025) =
2.05.

Interpretation: with a sample of 28 items (*df*=27), a *t* score
of 2.05 cuts the *t* distribution with 97.5% of the area to the left and
2.5% to the right.

χ˛(*df*,*rtail*) is the critical value for the χ˛
distribution with *df* degrees of freedom and probability
*rtail*. (In the context of a hypothesis test, *rtail* is
α, the significance level of the test.)

In the illustration, *rtail* is the area of the
right-hand tail, and the asterisk * marks the critical value
χ˛(*df*,*rtail*). The critical value or inverse χ˛ is
the χ˛ value such that a higher value of χ˛ has only an
*rtail* probability of occurring by chance.

You can compute critical χ˛ only for the right-hand tail, because the χ˛ distribution has no left-hand tail.

**Caution:**
Some textbooks write the function the
other way, χ˛(*rtail*,*df*). Since *df* is a whole number
and *rtail* is a decimal between 0 and 1, you will be
able to adapt.

**Example**:
What is the critical χ˛ for a 0.05 significance test
with 13 degrees of freedom?

Run the `MATH200B`

program and select
`4:Critical χ˛`

. Enter the number of degrees of
freedom and the area of the right-hand tail. Be patient: the
computation is slow. But the program gives you the critical χ˛
value of 22.36, as shown in the second screen.

Interpretation: For a χ˛ distribution with 13 degrees of freedom, the value χ˛ = 22.36 divides the distribution such that the area of the right-hand tail is 0.05.

**Summary:**
This part performs hypothesis tests
and computes confidence intervals for the standard deviation of a
population. Since variance is the square of standard deviation, it can
also do those calculations for the variance of a population.

The tests on standard deviation or variance of a population require that
**the underlying population must be normal**.
They are not robust, meaning that even moderate departures from
normality can invalidate your analysis.
See MATH200A Program part 4
for procedures to test whether a
population is normal by testing the sample.

**Outliers are also unacceptable** and must be ruled out.
See MATH200A Program part 2
for an easy way to test for outliers.

**See also:**
Inferences about One-Population Standard Deviation gives the statistical
concepts with examples of calculation “by hand” and
in an Excel workbook.

You already know how to test the mean of a population with a t test, or estimate a population mean using a t interval. Why would you want to do that for the standard deviation of a population?

The standard deviation measures **variability**. In many
situations not just the average is important, but
also the variability. Another way to look at it is that
**consistency** is important: the variability must not be too
great.

For example, suppose you are thinking about investing in one of two mutual funds. Both show an average annual growth of 3.8% in the past 20 years, but one has a standard deviation of 8.6% and the other has a standard deviation of 1.2%. Obviously you prefer the second one, because with the first one there’s quite a good chance that you’d have to take a loss if you need money suddenly.

Industrial processes, too, are monitored not only for average output but for variability within a specified tolerance. If the diameter of ball bearings produced varies too much, many of them won’t fit in their intended application. On the other hand, it costs more money to reduce variability, so you may want to make sure that the variability is not too low either.

To use the program, first check the requirements for your
sample; see Cautions above.
Then run the `MATH200B`

program and select `5:Infer about σ`

.
When prompted, enter the standard deviation and size of the sample,
pressing [`ENTER`

] after each one. If you know the variance of
the sample rather than the standard deviation, use the square root
operation since *s* is the square root of the variance *s*˛ (see
example below).

The program then presents you with a five-item menu: confidence interval for the population standard deviation σ, confidence interval for the population variance σ˛, and three hypothesis tests for σ or σ˛ less than, different from, or greater than a number. Make your selection by pressing the appropriate number.

If you select one of the confidence intervals, the program will prompt you for the confidence level and then compute the interval. Because this involves a process of successive approximations, it can take some time, so please be patient.

The program displays the endpoints of the interval on screen
and also leaves them in variables `L`

and `H`

in
case you want to use them in further calculations. You can include
them in any formula by pressing [`ALPHA`

`)`

*makes* `L`

] and [`ALPHA`

`^`

*makes* `H`

].

By the way, confidence intervals about a population standard deviation are not symmetric around the sample standard deviation. That’s different from the simpler cases of means and proportions. In this example, the 95% interval for σ extends 2.7 units below the sample standard deviation, but 4.3 units above it.

If you select one of the hypothesis tests, the program will
prompt you for σ, the population standard deviation in the
null hypothesis. If your H_{0} is about population variance
σ˛ rather than σ, use the square root symbol to
convert the hypothetical variance to standard deviation.

The program then displays the χ˛ test statistic, the
degrees of freedom, and the p-value. These are also left in variables
`X`

, `D`

, and `P`

in case you wish to use
them in further calculations. You can include them in any formula with
[x,T,θ,n], [`ALPHA`

`x`

^{-1}*makes* `D`

], and [`ALPHA`

`8`

*makes* `P`

].

**Example 1**: A machine packs cereal into
boxes, and you don’t want too much variation from box to box.
You decide that a standard deviation of no more than five
grams (about 1/6 ounce) is acceptable. To determine whether the machine is
operating within specification, you randomly select 45 boxes. Here are
the weights of the boxes, in grams:

386 | 388 | 381 | 395 | 392 | 383 | 389 | 383 | 370 |

379 | 382 | 388 | 390 | 386 | 393 | 374 | 381 | 386 |

391 | 384 | 390 | 374 | 386 | 393 | 384 | 381 | 386 |

386 | 374 | 393 | 385 | 388 | 384 | 385 | 388 | 392 |

400 | 377 | 378 | 392 | 380 | 380 | 395 | 393 | 387 |

**Solution**: First, use `1-Var`

`Stats`

to find the sample standard deviation,
which is 6.42 g. Obviously this is greater than the target
standard deviation of 5 g, but is it enough greater that you can
say the machine is not operating correctly, or could it have come from
a population with standard deviation no more than 5 g?
Your hypotheses are

H_{0}: σ = 5, the machine is within spec
(some books would say H_{0}: σ ≤ 5)

H_{1}: σ > 5, the machine is not working right

No α was specified, but for an industrial process with no possibility of human injury α = 0.05 seems appropriate.

Next, check the requirements: is the sample
**normally distributed and free of outliers**?
Use MATH200A part 2 to make a box-whisker plot to rule out outliers, and
MATH200A part 4 to check normality. The outputs are shown at right. You can see
that the sample has no outliers and
that it is extremely close to normal,
so **requirements are met** and you can proceed with the
hypothesis test.

Now, run the `MATH200B`

program and select
`5:Infer about σ`

. Enter *s*:6.42 and
*n*:45, and select `5:Test σ>const`

. Enter 5
for σ in H_{0}.

The results are shown at far right. The test statistic is χ˛ = 72.54 with 44 degrees of freedom, and the p-value is 0.0043.

Since *p*<α, you
**reject H _{0} and accept H_{1}.**
At the 0.05 level of significance, the population standard deviation
σ is greater than 5, and the machine is not operating within
specificaton.

**Example 2**:
You have a random sample of size 20, with a standard deviation of 125. You
have good reason to believe that the underlying population is normal,
and you’ve checked the sample and found no outliers.
Is the population standard deviation different from 100, at the 0.05
significance level?

**Solution**:
*n* = 20, *s* = 125, σ_{o} = 100,
α = 0.05. Your hypotheses are

H_{0}: σ = 100

H_{1}: σ ≠ 100

This time in the `INFER ABOUT σ`

menu you select
`4:Test σ≠const`

.

Results are shown at right. χ˛ = 29.69 with 19 degrees of freedom, and the p-value is 0.1118.

*p*>α; fail to reject H_{0}. At the 0.05 significance level,
you can’t say whether the population standard deviation
σ is different from 100 or not.

**Example 3**: Of several thousand students
who took the same exam, 40 papers were selected randomly and
statistics were computed. The standard deviation of the sample was 17
points. Estimate the standard deviation of the population, with 95%
confidence. (Recall that test scores are normally distributed.)

**Solution**:
Check the data and make sure there are no outliers.
Run `MATH200B`

and select [`2`

] in the first
menu. Enter *s* and *n*, and in the second menu select
`1:σ interval`

with a C-level of 95 or .95.
The results screen is shown at right.

Conclusion: You’re 95% confident that the standard deviation of test scores for all students is between 13.9 and 21.8.

**Remark**: The center of the confidence
interval is about
17.9, which is different from the point estimate *s*=17. This is a
feature of confidence intervals for σ or σ˛:
they are **asymmetric** because the χ˛ distribution used to
compute them is asymmetric.

**Example 4**:
Heights of US males aged 18–25 are normally distributed. You
take a random sample of 100 from that population and find a mean of
65.3 in and a variance of 7.3 in˛. (Remember that
the units of variance are the square of the units of the original
measurement.)

Estimate the mean and variance of the height of US males aged 18–25, with 95% confidence.

**Solution for mean**:
Computing a confidence interval for the mean is a straightforward
`TInterval`

. Just remember that for `Sx`

the
calculator wants the sample standard deviation, but you have the
sample variance, which is *s*˛. Therefore you take the square root
of sample variance to get sample standard deviation, as shown in the
input screen at near right.

The output screen at far right shows the confidence interval. You’re 95% confident that the mean height of US males aged 18–25 is between 64.8 and 65.8 in.

**Solution for variance**:
Run the `MATH200B`

program and select
`5:Infer about σ`

. Enter *s*:√7.3 and *n*:100.
Select `2:σ˛ interval`

and enter C-Level:.95 (or
95). The program computes the confidence interval for population
variance as 5.6 ≤ σ˛ ≤ 9.9.
Notice that the output screen shows the point estimate for variance, *s*˛,
and that as expected the confidence interval is not symmetric.

You’re 95% confident that the variance in heights of US males aged 18–25 is between 5.6 and 9.9 in˛.

**Complete answer**:
You’re 95% confident that the heights of US males aged
18–25 have mean 64.8 to 65.9 in and variance
5.6 to 9.9 in˛.

**Summary:**
With linear correlation, you compute
a sample correlation coefficient *r*. But what can you say about the
correlation in the population, ρ? The `MATH200B`

program computes a
confidence interval about ρ or performs a hypothesis test to
tell whether there is correlation in the population.

**See also:**
Inferences about Linear Correlation gives the statistical
concepts with examples of calculation “by hand” and
in an Excel workbook.

To perform inferences about linear regression, first load your
*x*’s and *y*’s in any two statistics lists. Then run
the `MATH200B`

program and select `7:Regression inf`

.

**Example**: The
following sample of commuting distances and times for fifteen randomly
selected co-workers is adapted from
Johnson & Kuby 2004 [full citation at https://BrownMath.com/swt/sources.htm#so_Johnson2004], page 623.

Commuting Distances and Times | |||||||||||||||
---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|

Person | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | 13 | 14 | 15 |

Miles, x |
3 | 5 | 7 | 8 | 10 | 11 | 12 | 12 | 13 | 15 | 15 | 16 | 18 | 19 | 20 |

Minutes, y |
7 | 20 | 20 | 15 | 25 | 17 | 20 | 35 | 26 | 25 | 35 | 32 | 44 | 37 | 45 |

The TI’s `LinReg(ax+b)`

command can tell you
that the correlation of the sample is 0.88. But what can you infer
about ρ, the correlation of the population? You can get a confidence
interval estimate for ρ, or you can perform a hypothesis test
for ρ≠0.

Before you can make any inference (hypothesis test or confidence interval) about correlation or regression in the population, check these requirements:

- The data are a
**simple random sample**. - The
**plot of residuals versus**— no bending, no thickening or thinning trend from left to right, and no outliers.*x*is featureless - The
**residuals are normally distributed**. You can check this with a normal probability plot, available in most statistics packages and in MATH200A part 4. Since the test statistic is a t, and the t test is robust, moderate departures from normality are okay.

To make a scatterplot of residuals, perform a regression
with `LinReg(ax+b) L1,L2`

(or whichever lists contain
your data). This computes the residuals automatically. You can then
plot them by following the procedure in
Display the Residuals, part of
Linked Variables. As you see from the graph at right, the
residuals don’t show any problem features.

To check normality of the residuals, run MATH200A part 4 and when
prompted for the data list press [`2nd`

`STAT`

*makes* `LIST`

]
[`▲`

], scroll to `RESID`

if necessary, and
press [`ENTER`

] [`ENTER`

]. The graph at right shows that the
residuals are approximately normally distributed.

It can be hard to tell whether a normal probability
plot is close enough to a straight line. But MATH200A part 4 shows the *r*
and critical values from the
Ryan-Joiner
test. When *r* > the critical value, the points are near
enough to a normal distribution. Here *r*=0.9772 > crit=0.9383,
so the residuals are close enough to normal.

Enter your *x*’s and *y*’s in two statistics lists,
such as L1 and L2. Run the `MATH200B`

program and select `6:Correlatn inf`

. When
prompted, enter your *x* list and *y* list, select
`1:Conf interval`

, and enter your desired confidence
level, such as .95 or 95 for 95%.

The output screen is shown at right. For this sample of
*n* = 15 points, the sample correlation coefficient is
*r* = 0.88. For the correlation of the population (distances
and times for all commuters at this company),
you’re 95% confident that
0.67 ≤ ρ ≤ 0.96.

(Just like confidence intervals about σ, confidence intervals about ρ extend different amounts above and below the sample statistic.)

You can also do a hypothesis test to see whether there is any
correlation in the population. The null hypothesis H_{0} is that
there is no correlation in the population, ρ = 0; the
alternative H_{1} is that there is correlation in the population,
ρ ≠ 0.

Select your α; 0.05 is a common choice.
Run the `MATH200B`

program and select `6:Correlatn inf`

. Enter your *x* and *y*
lists and select `2:Test ρ≠0`

.

The output screen is shown at right. Sample size *n* = 15,
and sample correlation is *r* = 0.88. The *t* statistic for this
hypothesis test is 6.64, and with 13 (*n*−2) degrees of freedom
that yields a p-value of <0.0001.

p<α; reject H_{0} and accept H_{1}. At the 0.05
level of significance, ρ ≠ 0: there is some
correlation in the population. Furthermore, the population
correlation is positive. (See *p* < α in Two-Tailed Test: What Does It Tell You? for interpreting the
result of a two-tailed test in a one-tailed manner like this.)

**Remark**: When *p* is greater than α, you
fail to reject H_{0}. In that case, you conclude that
it is impossible to say, at the 0.05 level of significance,
whether there is correlation in the population or not.

The `MATH200B`

program finds
**confidence intervals for the slope β _{1} and intercept β_{0}**
of the line that best fits the entire population of points, not just a
particular sample. It can also find a

The program doesn’t do any hypothesis tests on the
regression line. The standard test is to
**test whether the regression line has a nonzero slope, β _{1} ≠ 0.**
But that
test is identical to the test for a nonzero correlation coefficient,
ρ ≠ 0, which the

`MATH200B`

program performs as part of
the `6:Correlatn inf`

menu selection.**See also:**
Inferences about Linear Regression explains the principles and
calculations behind inferences about linear regression; there’s
even an Excel workbook.

Let’s use the same data on commuting
distances and times from Inferences about Linear Correlation.
The TI-83/84 command `LinReg(ax+b)`

will show the best
fitting regression line for this particular sample, but what can you say about
the regression for all commuters at that company?

The requirements for inference about regression are the same as the requireemnts for inference about correlation, listed above.

**Solution**: Enter the *x*’s and *y*’s in any two
statistics lists, such as L1 and L2. Run the `MATH200B`

program and select
`7:Regression inf`

. Specify the two lists and your desired confdence level,
such as .95 or 95 for 95%.

**Results**: Always look first at the sample size (bottom
of the screen) to make sure you haven’t left out any points.
The slope of the sample regression line is
1.89, meaning that on average each extra mile of commute takes 1.89
minutes (a speed of about 32 mph). But the 95% confidence
interval for the slope is 1.28 to 2.51: you’re 95% confident
that the slope of commuting time per distance, for all commuters at
this company, is between 1.28 and 2.51 minutes per mile.

The second section of the screen shows that the
*y* intercept of the sample is 3.6: this represents the “fixed
cost” of the commute, as opposed to the “variable cost”
per mile represented by the slope. But the 95% confidence interval is
−4.5 to +11.8 minutes.

**Interpretation**: the line that best fits the sample data is

*ŷ* = 1.89*x* + 3.6

and the regression line for the whole population is

*ŷ* = β_{1}*x* + β_{0}

where you’re 95% confident that

1.28 ≤ β_{1} ≤ 2.51
and
−4.5 ≤ β_{0} ≤ +11.8

Let’s think a bit more about that intercept, with a 95%
confidence interval of −4.5 to +11.8 minutes. This is a good
illustration that
**it’s a mistake to use a regression line too far outside your actual data.**
Here, the *x*’s run
from 3 to 20. The *y* intercept corresponds to *x* = 0,
and a commute of zero miles is not a commute at all. (Yes, there are
people who work from home, but they don’t get in their cars and
drive to work.) While the *y* intercept can be discussed as a
mathematical concept, it really has no relevance to this particular
problem.

The first output screen was about the line as a whole; now the
program turns to **predictions for a specific x value**.
First it asks for the

**Caution**: You should only use *x* values that are
within the domain of *x* values in your data, or close to it. No
matter how good the straight-line relationship of your data, you
don’t really know whether that relationship continues for lower
or higher *x* values.

The program arbitrarily limits you to the domain plus or
minus 15% of the domain width, but even that may be too much in some
problems. In this problem, commuting distances range from 3 mi to
20 mi, a width of 17 mi. The program will let you make
predictions about any *x* value from
3−.15*12 = 0.45 mi to
15+.15*12 = 22.55 mi, but you have to decide how far
you’re justified in extrapolating.

The input and output screens are shown at right.
*ŷ* (“y-hat”) is simply the *y* value on the regression
line for the given *x* value, found by
*ŷ* = (slope)×10+(intercept) = 22.6. That is a
prediction for μ_{y|x=10}, the average time for many
10-mile commutes. The screen shows a 95% confidence interval for that
mean: you’re
**95% confident that the average commute time for all 10-mile commutes**
(not just in the sample) is between 19.3 and 25.9
minutes.

But that is an estimate of the mean. Can we say anything about
individual commutes? Yes, that is the prediction interval at the
bottom of the
screen. It says that **95% of all 10-mile commutes** take between
10.4 and 34.7 minutes.

- Spiegel, Murray R., and Larry J. Stephens. 1999.
- Theory and Problems of Statistics. 3d ed. McGraw-Hill.
- Sullivan, Michael. 2008.
- Fundamentals of Statistics. 2d ed. Pearson Prentice Hall.

**17 Nov 2020**: Converted page from HTML 4.01 to HTML5, and italicized variable names.**18 July 2019**, program version 4.4a:With some very unlikely data sets, the program could falsely tell you that class widths were unequal or that discrete probabilities didn’t add to 1, and refuse to compute any results. I changed the floating-point tests involved, eliminating that possibility. My thanks to Ernest Brock for drawing this potential problem to my attention.

Don’t worry that you might have been getting incorrect computations!

**If MATH200B computed results for you, they were correct.**The only problem could have been in the program saying your data set was invalid when it was actually valid — and I don’t know of any actual case where even that has happened.**15 May 2016**: Moved citations to the new References section.**27 Dec 2015**:Program version 4.4:

- MATH200B now senses whether it’s being run on a hi-resolution color screen or low-resolution b&w screen and adjusts itself acordingly, so there’s no more need for a separate MATH200C. I learned the technique in the forums at Cemetech.
- On high-resolution TI-84s, skewness and kurtosis fit on a single screen.
- Round correation coefficients to four decimal places.
- On screens for regression confidence intervals, show the confidence leval.

Mention the newer TI-84 models’ different floating-point processing as compared to the older models.

Add a paragraph explaining the normality test for residuals.

Show the results screen for a confidence interval about the standard deviation of a population, and note that the interval is not symmetric.

Lose the comparisons to Sullivan’s book.

**17 Dec 2015**:Program version 4.3:

- Fix error in correlation inference:
*R*˛ is now a reserved symbol, so use R*R instead. - Adjust spacing for outputs of correlation inference.
- Update Web address, show compatibility on splash screen, and shorten menu header.
- There is now a separate MATH200C program for the TI-84+ C and CE calculators. (MATH200C was merged back into MATH200B ten days later.)

Shrink most of the screens with larger print by 25% to make the article a little shorter, especially for printing.

- Fix error in correlation inference:
**17 Oct 2015**: Add TI Connect CE in Getting the Program.- intervening changes suppressed)
**Dec 2008**: program version 1

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