TI-83/84 PLUS BASIC MATH PROGRAMS (GEOMETRY) Archive Statistics Number of files. Volume, and circumferencen of 2D and 3D figures. If the shape you are looking for is not in this program, feel free to e-mail me and I will add it, or feel free to add it yourself. Distance Formula A program that calculates the distance between two points!
BrownMath.com →TI-83/84/89 →Extra Statistics Utilities
Copyright © 2008–2019 by Stan Brown
Summary:This page presents a downloadable TI-83/84 program with easierversions of some calculator procedures plus new capabilities likecomputing skewness and kurtosis and making statistical inferencesabout standard deviation, correlation, and regression.See Using the Program below for afull list of features.
Your first course in statistics probably won’t use thesefeatures, but they’re offered here for advanced students andthose who are studying on their own.
If you’re in Stan Brown’s MATH200classes at TC3, this is the optional extra program; seeMATH200A Program — Basic Statistics Utilities for TI-83/84 for the required program.
Contents:
See also:Troubles? See TI-83/84 Troubleshooting.
|
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 fromSpiegel 1999 [full citation in “References”, below], page 68. |
A histogram, prepared with the MATH200Aprogram, shows the data are skewedleft, not symmetric.But how highly skewed are they? Andhow does the central peak compare to the normal distributionfor height and sharpness? To answer these questions, you have tocompute the skewness and kurtosis.
Enter the x’s in one statistics list and the f’s inanother. 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 thex’s and then the list that containsthe 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 thewhole population; the program gives you both σ and s,σ² and s².
The program stores keyresults in variables in case you want to do any further computationswith them. See MATH200B Program — Technical Notes for a complete listof variables computed by the program.
The second screen shows results for skewness. The third moment dividedby 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 touse if this is the whole population and if it is a sample.
If this is the whole population, then you stop with the firstskewness figure and can state that the population is negatively skewed(skewed left).
But this is just a sample, so you use the “assample” figure for your skewness. (This is also the figure thatExcel 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 standarderror of skewness, which is also shown on the screen.As a rule of thumb, if sampleskewness 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, andthe statistic of −0.45 tells you thatthe skewness is only 0.45 standard errors below zero. This is not enough tolet you say anything about whether the population is skewed in eitherdirection or symmetric.
The last screen shows results for kurtosis. The fourth moment dividedby the square of the variance gives the kurtosis, which is 2.74.Some authors, and Microsoft Excel, prefer to subtract 3 and considerthe excess kurtosis: 2.74−3 is −0.26.
A bell curve (normal distribution) has kurtosis of 3 andexcess 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 calledplatykurtic.(An excess kurtosis greater than 0 would mean that the distributionwas leptokurtic, with a narrower and higher peak than a bellcurve.)
Since this is just a sample, and not the whole population, usethe “as sample” excess kurtosis of −0.21. (This isthe figure Excel reports.)Can you say anything about the kurtosis of the population fromwhich this sample was taken? Yes, just as you did for skewness.The rule of thumbis that an excess kurtosis of at least two standard errorsis significant. For this sample, the standard error of kurtosis is 0.48, and −0.21/0.48 =−0.44, so the excesskurtosis is only 0.44 standard errors below zero. (Or,the kurtosis is only 0.44 standard errors below 3.) Therefore youcan’t say whether the population is peaked like a normaldistribution, more than normal, or less than normal.
On high-resolution screens (the TI-84 Plus C and TI-84Plus CE), there’s enough room to show skewness and kurtosis onthe same screen, as shown at right.
Example: Throwing Dice
You can also use this part of the program to compute the shapeof a probability distribution. For instance, here’s the probabilitydistribution for the number of spots showing when you throw twodice:
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 asfractions, not decimals, to ensure that theyadd to exactly 1. The calculator displays rounded decimals butkeeps full precision internally, and the program will tell you if yourprobabilities don’t add to 1.)Now run the
MATH200B
program and select 1:Skew/kurtosis
. Your data arrangement is4:Discrete PD
,and you’ll see the following results:
Ti 84 Free Calculator Download
On the first screen, no sample size is shown becausea probability distribution is a population.
![Distance Distance](/uploads/1/2/6/2/126227138/358595595.jpg)
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 statisticdon’t apply because you have a probability distribution(population) rather than a sample.
On the same screen, the kurtosis is 2.37 (not shown forreasons of space), and the excesskurtosis is −0.63; the dice make a platykurtic distribution.Compared to a normal distribution, this distribution of dice throwinghas a lower, less distinct peak and shorter tails.
On high-resolution screens, namely the TI-83 Plus C Silver Edition andTI-84 Plus CE, complete information about a probability distributionfits on one screen.
You may notice that, although the skewness is stillessentially zero, it’s a different very small number from thevery 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’slikely that the newer calculator’s chip processes floating pointwith very slightly different precision than the old one. Don’t obsess about it — for all practical purposes,both numbers are zero.
2. Time Series Plots
Summary: To plot a time series or trend line,put the numbers in a statistics list and use the
2:Time series
part ofthe MATH200B
program.
Example:Let’s plot the closing prices of Cisco Systems stockover a two-year period. The following table is adapted fromSullivan 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 youfor the data list. (Caution: The program assumes thetime intervals are all equal. If they aren’t, the horizontalscale will not be uniform and the graph will not be correct.)
It’s usually good practice to start the vertical scale atzero, or in other words to show the x axis at its proper level onthe graph. But the program gives you the choice. If you have goodreason, 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 thenext.(If the data include any negative or zero values, thex axis will naturally appear in the graph, and program skips theyes/no prompt.)
Below you see the effect of a “yes” at left and theeffect of a “no” at right.
As you can see, the graph that doesn’t include the zerolooks a lot more dramatic, with bigger changes. But that can bedeceptive. A more accurate picture is shown in the first graph,the one that does include the x axis.
optional extra: Tracing the Graph
If you wish, you can press the [
TRACE
] key and displaythe closing prices, scrolling back and forth with the[◄
] and [►
] keys. If you wantto jump to a particular month, say June 2004, the 16th month, type16
and then press [ENTER
].
3. Critical t or Inverse t
The TI-83 doesn’t have an
invT
function asthe TI-84 does, but if you need to find critical t or inverse t oneither calculator you can use this part of the MATH200B
program.
Caution: our notation of t(df,rtail) matches most booksin specifying the area of the right-hand tail for critical t. But theTI calculator’s built-in menus specify the area of the left-hand tail. Make sureyou 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 tailarea must always be less than 1, you’ll always know whichargument is which.
Example: find t(27,0.025), the t statistic with 27 degrees offreedom (sample size 28) for a one-tailed significance test withα = 0.025, a two-tailed test withα = 0.05, or a confidence interval with1−α = 95%.
Solution: run the
MATH200B
program and select 3:Critical t
. Whenprompted, enter 27 for degrees of freedom and 0.025 for the area ofthe right-hand tail, as shown in the first screen. After a shortpause, the calculator gives you the answer: t(27,0.025) =2.05.
Interpretation: with a sample of 28 items (df=27), a t scoreof 2.05 cuts the t distribution with 97.5% of the area to the left and2.5% to the right.
4. Critical χ² or Inverse χ²
χ²(df,rtail) is the critical value for the χ²distribution with df degrees of freedom and probabilityrtail. (In the context of a hypothesis test, rtail isα, the significance level of the test.)
In the illustration, rtail is the area of theright-hand tail, and the asterisk * marks the critical valueχ²(df,rtail). The critical value or inverse χ² isthe χ² value such that a higher value of χ² has only anrtail probability of occurring by chance.
You can compute critical χ² only for the right-handtail, because the χ² distribution has no left-hand tail.
Caution: Some textbooks write the function theother way, χ²(rtail,df). Since df is a whole numberand rtail is a decimal between 0 and 1, you will beable to adapt.
Example:What is the critical χ² for a 0.05 significance testwith 13 degrees of freedom?
Run the
MATH200B
program and select4:Critical χ²
. Enter the number of degrees offreedom and the area of the right-hand tail. Be patient: thecomputation is slow. But the program gives you the critical χ²value of 22.36, as shown in the second screen.
Interpretation: For a χ² distribution with 13 degreesof freedom, the value χ² = 22.36 divides thedistribution such that the area of the right-hand tail is 0.05.
5. Inferences about σ, the Standard Deviation of a Population
Summary:This part performs hypothesis testsand computes confidence intervals for the standard deviation of apopulation. Since variance is the square of standard deviation, it canalso do those calculations for the variance of a population.
Cautions:
The tests on standard deviation or variance of a population require thatthe underlying population must be normal.They are not robust, meaning that even moderate departures fromnormality can invalidate your analysis.See MATH200A Program part 4for procedures to test whether apopulation is normal by testing the sample.
Outliers are also unacceptable and must be ruled out.See MATH200A Program part 2for an easy way to test for outliers.
See also:Inferences about One Population Standard Deviation gives the statisticalconcepts with examples of calculation “by hand” andin an Excel workbook.
You already know how to test the mean of apopulation with a t test, or estimate a population mean using at interval. Why would you want to do that for the standarddeviation of a population?
The standard deviation measures variability. In manysituations not just the average is important, butalso the variability. Another way to look at it is thatconsistency is important: the variability must not be toogreat.
For example, suppose you are thinking aboutinvesting in one of two mutual funds. Both show an average annualgrowth of 3.8% in the past 20 years, but one has a standard deviationof 8.6% and the other has a standard deviation of 1.2%. Obviously youprefer the second one, because with the first one there’squite a good chance that you’d have to take a loss if you needmoney suddenly.
Industrial processes, too, are monitored not only for averageoutput but for variability within a specified tolerance. If thediameter of ball bearings produced varies too much, many of themwon’t fit in their intended application. On the other hand, itcosts more money to reduce variability, so you may want to make surethat the variability is not too low either.
To use the program, first check the requirements for yoursample; 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 ofthe sample rather than the standard deviation, use the square rootoperation since s is the square root of the variance s² (seeexample 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, andthree hypothesis tests for σ or σ² less than,different from, or greater than a number. Make your selection bypressing the appropriate number.
Confidence Intervals
If you select one of the confidence intervals, the program willprompt you for the confidence level and then compute the interval.Because this involves a process of successive approximations, it cantake some time, so please be patient.
The program displays the endpoints of the interval on screenand also leaves them in variables
L
and H
incase you want to use them in further calculations. You can includethem in any formula by pressing [ALPHA
)
makesL
] and [ALPHA
^
makesH
].
By the way, confidence intervals about a populationstandard deviation are not symmetric around the sample standarddeviation. That’s different from the simpler cases of means andproportions. In this example, the 95% interval for σ extends2.7 units below the sample standard deviation, but 4.3 units aboveit.
Hypothesis Tests
If you select one of the hypothesis tests, the program willprompt you for σ, the population standard deviation in thenull hypothesis. If your H0 is about population varianceσ² rather than σ, use the square root symbol toconvert the hypothetical variance to standard deviation.
The program then displays the χ² test statistic, thedegrees of freedom, and the p-value. These are also left in variables
X
, D
, and P
in case you wish to usethem in further calculations. You can include them in any formula with[x,T,θ,n], [ALPHA
x-1
makesD
], and [ALPHA
8
makesP
].
Examples
Example 1: A machine packs cereal intoboxes, and you don’t want too much variation from box to box. You decide that a standard deviation of no more than fivegrams (about 1/6 ounce) is acceptable. To determine whether the machine isoperating within specification, you randomly select 45 boxes. Here arethe 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 targetstandard deviation of 5 g, but is it enough greater that you cansay the machine is not operating correctly, or could it have come froma population with standard deviation no more than 5 g?Your hypotheses are
H0: σ = 5, the machine is within spec(some books would say H0: σ ≤ 5)
H1: σ > 5, the machine is not working right
No α was specified, but for an industrial process with nopossibility of human injury α = 0.05 seemsappropriate.
Next, check the requirements: is the samplenormally distributed and free of outliers?Use MATH200A part 2 to make a box-whisker plot to rule out outliers, andMATH200A part 4 to check normality. The outputs are shown at right. You can seethat the sample has no outliers andthat it is extremely close to normal,so requirements are met and you can proceed with thehypothesis test.
Now, run the
MATH200B
program and select5:Infer about σ
. Enter s:6.42 andn:45, and select 5:Test σ>const
. Enter 5for σ in H0.
The results are shown at far right. The test statistic isχ² = 72.54 with 44 degrees of freedom, and thep-value is 0.0043.
Since p<α, youreject H0 and accept H1.At the 0.05 level of significance, the population standard deviationσ is greater than 5, and the machine is not operating withinspecificaton.
Example 2:You have a random sample of size 20, with a standard deviation of 125. Youhave 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 0.05significance level?
Solution:n = 20, s = 125, σo = 100,α = 0.05. Your hypotheses are
H0: σ = 100
H1: σ ≠ 100
This time in the
INFER ABOUT σ
menu you select4:Test σ≠const
.
Results are shown at right. χ² = 29.69 with19 degrees of freedom, and the p-value is 0.1118.
p>α; fail to reject H0. At the 0.05 significance level,you can’t say whether the population standard deviationσ is different from 100 or not.
Example 3: Of several thousand studentswho took the same exam, 40 papers were selected randomly andstatistics were computed. The standard deviation of the sample was 17points. 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 firstmenu. Enter s and n, and in the second menu select1:σ interval
with a C-level of 95 or .95.The results screen is shown at right.
Conclusion: You’re 95% confident that the standarddeviation of test scores for all students is between 13.9 and21.8.
Remark: The center of the confidenceinterval is about17.9, which is different from the point estimate s=17. This is afeature of confidence intervals for σ or σ²:they are asymmetric because the χ² distribution used tocompute them is asymmetric.
Example 4:Heights of US males aged 18–25 are normally distributed. Youtake a random sample of 100 from that population and find a mean of65.3 in and a variance of 7.3 in². (Remember thatthe units of variance are the square of the units of the originalmeasurement.)
Estimate the mean and variance of the height of US malesaged 18–25, with 95% confidence.
Solution for mean:Computing a confidence interval for the mean is a straightforward
TInterval
. Just remember that for Sx
thecalculator wants the sample standard deviation, but you have thesample variance, which is s². Therefore you take the square rootof sample variance to get sample standard deviation, as shown in theinput 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 aged18–25 is between 64.8 and 65.8 in.
Solution for variance:Run the
MATH200B
program and select5:Infer about σ
. Enter s:√7.3 and n:100.Select 2:σ² interval
and enter C-Level:.95 (or95). The program computes the confidence interval for populationvariance 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 malesaged 18–25 is between 5.6 and 9.9 in².
Complete answer:You’re 95% confident that the heights of US males aged18–25 have mean 64.8 to 65.9 in and variance5.6 to 9.9 in².
6. Inferences about Linear Correlation
Summary:With linear correlation, you computea sample correlation coefficient r. But what can you say about thecorrelation in the population, ρ? The
MATH200B
program computes aconfidence interval about ρ or performs a hypothesis test totell whether there is correlation in the population.
See also:Inferences about Linear Correlation gives the statisticalconcepts with examples of calculation “by hand” andin an Excel workbook.
To perform inferences about linear regression, first load yourx’s and y’s in any two statistics lists. Then runthe
MATH200B
program and select 7:Regression inf
.
Example: Thefollowing sample of commuting distances and times for fifteen randomlyselected co-workers is adapted fromJohnson & 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 youthat the correlation of the sample is 0.88. But what can you inferabout ρ, the correlation of the population? You can get a confidenceinterval estimate for ρ, or you can perform a hypothesis testfor ρ≠0.
Requirements
Before you can make any inference (hypothesis test orconfidence interval) about correlation or regression in thepopulation, 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 nooutliers.
- The residuals are normally distributed. You can checkthis with a normal probability plot, available in most statisticspackages and in MATH200A part 4. Since the test statistic is a t, and thet test is robust, moderate departures from normality are okay.
To make a scatterplot of residuals, perform a regressionwith
LinReg(ax_b) L1,L2
(or whichever lists containyour data). This computes the residuals automatically. You can thenplot them by following the procedure in Display the Residuals, part ofLinked Variables. As you see from the graph at right, theresiduals don’t show any problem features.
To check normality of the residuals, run MATH200A part 4 and whenprompted for the data list press [
2nd
STAT
makesLIST
][▲
], scroll to RESID
if necessary, andpress [ENTER
] [ENTER
]. The graph at right shows that theresiduals are approximately normally distributed.
It can be hard to tell whether a normal probabilityplot is close enough to a straight line. But MATH200A part 4 shows the rand critical values from theRyan-Joinertest. When r > the critical value, the points are nearenough to a normal distribution. Here r=0.9772 > crit=0.9383,so the residuals are close enough to normal.
Confidence Interval about ρ
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
. Whenprompted, enter your x list and y list, select1:Conf interval
, and enter your desired confidencelevel, such as .95 or 95 for 95%.
The output screen is shown at right. For this sample ofn = 15 points, the sample correlation coefficient isr = 0.88. For the correlation of the population (distancesand times for all commuters at this company),you’re 95% confident that0.67 ≤ ρ ≤ 0.96.
(Just like confidence intervals about σ, confidenceintervals about ρ extend different amounts above and below thesample statistic.)
Hypothesis Test about ρ
You can also do a hypothesis test to see whether there is anycorrelation in the population. The null hypothesis H0 is thatthere is no correlation in the population, ρ = 0; thealternative H1 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 ylists 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 thishypothesis test is 6.64, and with 13 (n−2) degrees of freedomthat yields a p-value of <0.0001.
p<α; reject H0 and accept H1. At the 0.05level of significance, ρ ≠ 0: there is somecorrelation in the population. Furthermore, the populationcorrelation is positive. (See p < α in Two-Tailed Test: What Does It Tell You? for interpreting theresult of a two-tailed test in a one-tailed manner like this.)
Remark: When p is greater than α, youfail to reject H0. In that case, you conclude thatit is impossible to say, at the 0.05 level of significance,whether there is correlation in the population or not.
7. Inferences about Linear Regression
Summary:A linear regression fits anequation of the formŷ = b1x + b0 to thesample data, but the slope b1 and the y interceptb0 are just sample statistics. If you took a differentsample you would likely get a different regression line.
The
MATH200B
program findsconfidence intervals for the slope β1 and intercept β0of the line that best fits the entire population of points, not just aparticular sample. It can also find aconfidence interval about the mean ŷ for a particular xand aprediction interval about all ŷ’s for a particular x.
The program doesn’t do any hypothesis tests on theregression line. The standard test is totest whether the regression line has a nonzero slope, β1 ≠ 0.But thattest is identical to the test for a nonzero correlation coefficient,ρ ≠ 0, which the
MATH200B
program performs as part ofthe 6:Correlatn inf
menu selection.
See also:Inferences about Linear Regression explains the principles andcalculations behind inferences about linear regression; there’seven an Excel workbook.
The Example
Let’s use the same data on commutingdistances and times from Inferences about Linear Correlation.The TI-83/84 command
LinReg(ax+b)
will show the bestfitting regression line for this particular sample, but what can you say aboutthe regression for all commuters at that company?
Requirements
The requirements for inference about regression are the sameas the requireemnts for inference about correlation, listed above.
Regression Coefficients for the Population
Solution: Enter the x’s and y’s in any twostatistics lists, such as L1 and L2. Run the
MATH200B
program and select7: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 (bottomof the screen) to make sure you haven’t left out any points.The slope of the sample regression line is1.89, meaning that on average each extra mile of commute takes 1.89minutes (a speed of about 32 mph). But the 95% confidenceinterval for the slope is 1.28 to 2.51: you’re 95% confidentthat the slope of commuting time per distance, for all commuters atthis company, is between 1.28 and 2.51 minutes per mile.
The second section of the screen shows that they intercept of the sample is 3.6: this represents the “fixedcost” 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 = β1x + β0
where you’re 95% confident that
1.28 ≤ β1 ≤ 2.51 and −4.5 ≤ β0 ≤ +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 goodillustration thatit’s a mistake to use a regression line too far outside your actual data.Here, the x’s runfrom 3 to 20. The y intercept corresponds to x = 0,and a commute of zero miles is not a commute at all. (Yes, there arepeople who work from home, but they don’t get in their cars anddrive to work.) While the y intercept can be discussed as amathematical concept, it really has no relevance to this particularproblem.
Inferences about a Particular x Value
The first output screen was about the line as a whole; now theprogram 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 arewithin the domain of x values in your data, or close to it. Nomatter how good the straight-line relationship of your data, youdon’t really know whether that relationship continues for loweror higher x values.
The program arbitrarily limits you to the domain plus orminus 15% of the domain width, but even that may be too much in someproblems. In this problem, commuting distances range from 3 mi to20 mi, a width of 17 mi. The program will let you makepredictions about any x value from3−.15*12 = 0.45 mi to15+.15*12 = 22.55 mi, but you have to decide how faryou’re justified in extrapolating.
The input and output screens are shown at right.ŷ (y-hat) is simply the y value on the regressionline for the given x value, found byŷ = (slope)×10+(intercept) = 22.6. That is aprediction for μy|x=10, the average time for many10-mile commutes. The screen shows a 95% confidence interval for thatmean: you’re95% confident that the average commute time for all 10-mile commutes(not just in the sample) is between 19.3 and 25.9minutes.
But that is an estimate of the mean. Can we say anything aboutindividual commutes? Yes, that is the prediction interval at thebottom of thescreen. It says that 95% of all 10-mile commutes take between10.4 and 34.7 minutes.
References
- 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.
What’s New
18 July 2019, program version 4.4a:With some very unlikely data sets,the program could falsely tell you that class widths were unequal orthat discrete probabilities didn’t add to 1, and refuse tocompute any results. I changed the floating-point tests involved,eliminating that possibility. My thanks to Ernest Brock for drawingthis potential problem to my attention.
Don’t worry that you might have been getting incorrectcomputations! If MATH200B computed results for you, they were correct.The only problem could have been in the program saying your data setwas invalid when it was actually valid — and Idon’t know of any actual case where even that has happened.
15 May 2016: Move citations to the newReferences section.
27 Dec 2015:
- Program version 4.4:
- MATH200B now senses whether it’s being run on ahi-resolution color screen or low-resolution b&w screen andadjusts itself acordingly, so there’s no more need for aseparate MATH200C. I learned the technique in the forums atCemetech.
- On high-resolution TI-84s, skewness and kurtosis fiton a single screen.
- Round correation coefficients to four decimalplaces.
- On screens for regression confidence intervals, show theconfidence leval.
- Mention the newer TI-84 models’ different floating-point processing ascompared to the older models.
- Add a paragraph explaining the normalitytest for residuals.
- Show the results screen for a confidenceinterval about the standard deviation of a population, and note thatthe interval is not symmetric.
- Lose the comparisons to Sullivan’s book.
17 Dec 2015:
- Program version 4.3:
- Fix error in correlation inference: R² is now a reservedsymbol, 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 andCE calculators. (MATH200C was merged back into MATH200B ten dayslater.)
- Shrink most of the screens with larger print by 25% tomake the article a little shorter, especially for printing.
17 Oct 2015: Add TI Connect CE inGetting the Program.
(intervening changes suppressed)
Dec 2008: program version 1
Because this program helps you,
please click to donate!Because this program helps you,
please donate at
BrownMath.com/donate.
please click to donate!Because this program helps you,
please donate at
BrownMath.com/donate.
Quadratic Formula Ti 84 Program
Updates and new info: https://BrownMath.com/ti83/
Site Map |Home Page| Contact
The programming language used by the TI-84 Plus calculator is similar to the Basic programming language. It uses the standard commands (such as the “If …, then …, else …” command) that are familiar to anyone who has ever written a program. And, of course, it also makes use of commands that are unique to the calculator (such as ClrHome, which clears the Home screen).
These are the basic steps for creating a program on the calculator:
-
To create a new program using the Program editor, pressThis is illustrated in the first screen.
-
Give your program a name and then press [ENTER].The name of your program can consist of one to eight characters that must be letters, numbers, or the Greek letter theta. The first character in the name must be a letter or theta, as in the second screen.The A after Name = indicates that the calculator is in Alpha mode. In this mode, when you press a key you enter the green letter above that key. To enter a number, press [ALPHA] to take the calculator out of Alpha mode and then enter the number.To enter a letter after entering a number, you must press [ALPHA] to put the calculator back in Alpha mode. Press [2nd][ALPHA] to place your calculator in Alpha-lock mode, which allows you to enter multiple letters without having to press a between them.When you press [ENTER] after naming your program, the calculator puts you in the Program editor, as in the third screen. The program appearing in this screen is entered in the next step.
-
Enter your program in the Program editor.Your program consists of a series of commands, each of which must be preceded by a colon, as shown in the third screen. After entering a command, press [ENTER] so the calculator supplies the colon preceding the next command you enter. When you finish writing your final command, press [ENTER] and ignore the colon that is waiting for a command to be entered.An example of entering a program appears in the third screen. The program in this screen writes HI on the Home screen. The Disp command is entered by pressing
-
Press [2nd][MODE] when you’re finished writing your program.This saves your program in the memory of the calculator and returns you to the Home screen. The name under which the program is stored in the calculator is the same name you gave the program in Step 2.
Editing a program on the TI-84 Plus
To edit a program stored on the calculator, follow these steps:
-
Press [PRGM] and the right-arrow key and press the number of the program or use the up- and down-arrow keys to highlight the program you want to edit.
-
Edit the program.Pressing [CLEAR] deletes the line containing the cursor.
-
Press [2nd][MODE] to save the program and return to the Home screen.
Executing a TI-84 Plus calculator program
Ti 83 Distance Formula Program
After creating your program and saving it on the calculator, you can run the program on the calculator by performing the following steps:
-
Press [PRGM] to enter the Program Execute menu, and use the down-arrow key to move the indicator to your program.This is illustrated in the first screen.
-
Press [ENTER] to place the program on the Home screen.This is illustrated in the second screen.
-
Press [ENTER] to execute the program.This operation is shown in the third screen. When the calculator is finished executing the program, it writes Done on the Home screen.
Deleting a program from the TI-84 Plus calculator
To delete a program from the calculator:
Ti 84 Download Free
-
Press [2nd][+] to access the Memory menu.
-
Press [2] to access the Mem Mgt/Del menu.
-
Press [7] to access the Program files stored in the calculator.
-
If necessary, repeatedly press the down-arrow key to move the indicator to the program you want to delete.
-
Press [DEL] to delete the program.You are asked whether you really want to delete this program. Press [2] if you want it deleted or press [1] if you’ve changed your mind.
-
Press [2nd][MODE] to exit this menu and return to the Home screen.
Comments are closed.
Author
Write something about yourself. No need to be fancy, just an overview.