# MATH200B Program —

Extra Statistics Utilities for TI-83/84

Copyright © 2008–2015 by Stan Brown, Oak Road Systems

Extra Statistics Utilities for TI-83/84

Copyright © 2008–2015 by Stan Brown, Oak Road Systems

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.

If you’re in Stan Brown’s MATH200 classes at TC3, this is the optional extra program; see MATH200A Program — Statistics Utilities for TI-83/84 for the required program.

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

`MATH200B`

Program OverviewThere are three methods to get the programs into your calculator:

- If a classmate has the programs on her calculator
(any model TI-83/84), she
can transfer them to yours using the short cable with headphone-style
plugs that comes with all TI
calculators. On your calculator, press [
`2nd`

`x,T,θ,n`

*makes*`LINK`

] [`►`

] [`ENTER`

], and then on hers press [`2nd`

`x,T,θ,n`

*makes*`LINK`

] [`3`

], select MATH200B, then press [`►`

] [`ENTER`

]. If you get a prompt about a duplicate program, choose Overwrite.Repeat for MATH200Z.

**If you have the original TI-83**(not a Plus or Silver), instead of the above you need M20083B and M20083Z. (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.) - Or, download MATH200B.ZIP
(107 KB,
updated 2 Aug 2012), unzip it, and transfer
files MATH200B.8XP and MATH200Z.8XP to your calculator. (This
requires TI-Connect or TI-Graph Link software and a cable.)
If you get a prompt about a duplicate program, choose Replace.
**If you have the original TI-83**(not a Plus or Silver), instead of the above you need M20083B.83P and M20083Z.83P, located in the For_Original_TI83 folder in MATH200B.ZIP. (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.) - 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`

] to run it. Check to make sure you have the
latest version, as shown on the splash screen, 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 χ²`

: find the χ² 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

When running the program, be careful to select
`EXEC`

, not `EDIT`

, on the screen: you
don’t want to change the code accidentally.
To look at the program source code, see MATH200B.PDF and
MATH200Z.PDF in the downloadable
MATH200B.ZIP file.

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

] [`1`

].

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 & Stephens, Theory and Problems of Statistics 3/e (McGraw-Hill, 1999), 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, σ² and s².

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.

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.

**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.

**MATH200 students:**
Time series are discussed on
page 90 of Sullivan, Michael, Fundamentals of Statistics 3/e (Pearson Prentice Hall, 2011), but are not part of the MATH200
syllabus.

**Example**:
Let’s plot the closing prices of Cisco Systems stock
over a two-year period. The following table is adapted from Sullivan, Michael, Fundamentals of Statistics 2/e (Pearson Prentice Hall, 2008),
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.

χ²(*df,rtail*) is the critical value for the χ²
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
χ²(*df,rtail*). The critical value or inverse χ² is
the χ² value such that a higher value of χ² has only an
*rtail* probability of occurring by chance.

You can compute critical χ² only for the right-hand tail, because the χ² distribution has no left-hand tail.

**Caution:**
Some textbooks write the function the
other way, χ²(*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 χ² for a 0.05 significance test
with 13 degrees of freedom?

Run the `MATH200B`

program and select
`4:Critical χ²`

. 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 χ²
value of 22.36, as shown in the second screen.

Interpretation: For a χ² distribution with 13 degrees of freedom, the value χ² = 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.

**MATH200 students:**
Sullivan, Michael, Fundamentals of Statistics 3/e (Pearson Prentice Hall, 2011)
**sections C.2 and C.3** discuss confidence intervals and
hypothesis tests about σ, but this is not part of the MATH200
syllabus.

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 s² (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 σ², and three hypothesis tests for σ or σ² 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`

].

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
σ² rather than σ, use the square root symbol to
convert the hypothetical variance to standard deviation.

The program then displays the χ² 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 on a standard deviation of no more than five
grams (about 1/6 ounce). 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, 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 4 to check normality and MATH200A part 2 to make a box-whisker
plot to rule out outliers. The outputs are shown at right. You can see
that the sample is extremely close to normal and that it has no
outliers, 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 χ² = 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. χ² = 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 σ²:
they are **asymmetric** because the χ² 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 in². (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 s². 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:σ² interval`

and enter C-Level:.95 (or
95). The program computes the confidence interval for population
variance as 5.6 ≤ σ² ≤ 9.9.
Notice that the output screen shows the point estimate for variance, s²,
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 in².

**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 in².

**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, Elementary Statistics 9/e (Thomson, 2004), 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 x is featureless**— no bending, no thickening or thinning trend from left to right, and no outliers. - 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.

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.**MATH200 students:**
Sullivan, Michael, Fundamentals of Statistics 3/e (Pearson Prentice Hall, 2011)
**sections 12.3 and 12.4** discuss confidence and prediction intervals,
but this is not part of the MATH200 syllabus.

**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 = 1.89x + 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 x value you’re interested in. This time,
let’s make predictions about a commute of 10 miles.

**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.

**2 Aug 2012**—program version 4.2:

- Skewness and Kurtosis now includes binomial PD.
- Clarify the data arrangement in the example of throwing dice.
- Clarify the applicability of the
`MATH200B`

program for my classes.

**9 Jun 2012**: Protection was causing problems with some
calculators, so the programs are now unprotected and there’s a
warning about editing.

**9 Mar 2012**: Update one
missed reference to the second editionn of
Sullivan

(intervening changes suppressed)

**Dec 2008**: program version 1

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For updates and new info, go to http://www.tc3.edu/instruct/sbrown/ti83/