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TI-89 BASIC Math Programs - ticalc.org:::::: TI-89 BASIC Math Programs File Archives TI-89 BASIC MATH PROGRAMS Click a filename to download that file. Click a folder name to view files in that folder. Click for file information.

Because this program helps you, please donate at BrownMath.com/donate. Getting the Program The program is in two parts, MATH200A 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.

TI-89 BASIC MATH PROGRAMS (STATISTICS) Archive Statistics Number of files 119 Last updated Thursday, 12 December 2013 Total downloads 385,674 Most popular file.

If you have a “classic” TI-83, not a Plus or Silver, follow the directions below but put M20083A and M20083Z on your calculator, not MATH200A and MATH200Z. (M20083A and M20083Z aren’t being updated after version 5, 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, (111 KB, updated 28 Dec 2016), and unzip it. Use the USB cable that came with your calculator, and the free software from Texas Instruments, to transfer the MATH200A.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 MATH200A (or M20083A; 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 MATH200A.pdf, MATH200Z.pdf, and MATH200Ahints.htm in the. Using the Program. Press the PRGM key. If you can see MATH200A 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 (v7.2), then press ENTER. The menu at right shows what the program can do:. Histograms etc:, or overlay both, for a frequency distribution, a relative frequency distribution, a simple list of numbers, a binomial probability distribution, or a generic discrete probability distribution. Box-whisker: plot a.

Binomial prob: compute. Normality chk: test whether your sample is.

Sample size: for a given confidence level and margin of error for binomial data (one or two populations) or numeric data (with population standard deviation σ known or unknown). GoF test: for categorical data in one population. Two-way table: for categorical data in a two-way table 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 MATH200A.PDF and MATH200Z.PDF in the. Each procedure leaves its results in variables in case you want to use them for further computations. For details, please see the separate document. Histograms and Frequency Polygons. Summary: Making a histogram or frequency polygon using native TI-83/84 commands is kind of tedious, especially setting up the WINDOW screen. This part of the MATH200A program automates the process. The program can create a histogram or polygon, or both in overlay, for these distributions:.

(discrete or continuous) with frequencies, relative frequencies, or probabilities. of numbers (discrete or continuous). (discrete data) with frequencies or relative frequencies. general. To use the program, put the class midpoints or ungrouped data in a statistics list and the frequencies (if applicable) in another. (Some people use the term class marks instead of class midpoints; they mean the same thing.) Then press PRGM, select MATH200A, and press 1:Histograms etc. The program will prompt you for the necessary information and will check silently to make sure your inputs are in valid form.

Then it will ask whether you want a histogram, polygon, or both, and will produce your desired graph. (The program uses an algorithm to ensure that there are an appropriate number of dots vertically on the screen.) Restrictions: If you have frequency data, the data list and frequency list must be the same length and the class widths in the data list must all be the same; the program checks this. The TI-83 and TI-84 won’t let you make a histogram with more than 47 classes, and the program also checks this. Finally, the program also won’t let you make a grouped frequency histogram with just one or two classes, because that’s silly. Grouped Frequency Data.

Class Boundaries Class Midpoints Frequency 20 ≤ x. 11 15 14 12 9 8 7 5 6 11 10 10.5 12 11 13 2 6 4 13.5 Suppose you want to make a histogram of the class performance on a 15-point quiz, where the scores are shown at left. You don’t want to bother to group these 19 scores by hand, so you enter them in a statistics list such as L1 and let the program do the grouping for you.

(An alternative graphical display is the.) Run the MATH200A program and select 1:Histograms etc. When the program asks your data arrangement, select 1 for a plain list of numbers. Then enter the name of the list that contains your numbers. For a plain list of numbers, the program needs to know how you want to group your data, so it asks you to specify the lower bound of the first class as well as the class width. Since 0 is the lowest possible grade, that’s the obvious lower bound for this example. The quiz has a possible maximum score of 15, and 10% of that is 1.5 points.

This is a good class width because grades of D, C, B, and A will each be one bar of the histogram. What if you’re not sure which class width to use? Try several class widths, one at a time, and see what the histograms look like.

You can pick the class width that seems to display the shape of the data most clearly. At this point you may see a pause as the program computes the number of classes and places each data point into a class.

(It uses statistics lists LD for the class midpoints (class midpoints) and LF for the computed frequencies.) For a simple list of numbers, the program will make only a histogram, not a frequency polygon. Optional extra: Tracing the histogram In first looking at that histogram, you might think there are four Ds, three Cs, two Bs, and one A. But when you check this by pressing TRACE you see that’s not correct. The highest class is the 15 to 16.5 class, since someone had a perfect score of 15.

(Remember: when a value is right on a class boundary, it is always assigned to the higher class.) So the top two classes in the histogram represent As (three students), the next lower is the three Bs, the next lower (shown at right) is four Cs, the next lower is two Ds, and the rest are Fs. Ungrouped Frequency Data. Children per family Freq. 0 9 1 6 2 10 3 2 4 2 5 1 You have recorded the numbers of children in 30 randomly selected families that used a community center in a given week, and you want to show a picture of the discrete distribution. There are only a few different values (numbers of children per family), so you choose an ungrouped frequency distribution.

Enter the data in two lists such as L3 and L4, run the MATH200A program, and select 1:Histograms etc. For Data Arrangement, select 2, ungrouped distribution. Enter your data list and frequency list as usual. When prompted, select whether you want a histogram, a frequency polygon, or both. I’ve selected a histogram, and the results are shown below.

The vertical line in the histogram is the y axis — you can remove it, if you want, by pressing 2nd ZOOM makes FORMAT and selecting AXES OFF. Also in the histogram, notice that the vertical rows of dots run through the center of each bar rather than along the edges.

This reminds you that you should label the bars of an ungrouped frequency distribution under the centers, not the edges as you would label a grouped histogram. If you want to trace the ungrouped frequency histogram, follow the same procedure as for. Relative Frequencies or Probabilities If you have a relative frequency distribution or probability distribution, you plot it in almost the same way as a frequency distribution. The main difference is that the relative frequencies or probabilities must add up to 1, and the program checks this for you. (There’s a special case: the binomial probability distribution.

For this, please see below.) Number of dice alike Probability 1 720/7776 2 5400/7776 3 1500/7776 4 150/7776 5 6/7776 Here’s an example of a general discrete probability distribution, drawn from the rainy-day game Yahtzee. In Yahtzee you roll five dice and try to make various combinations. Shown at right are the probabilities for number of dice alike when you roll five standard six-sided dice. (Thanks to Paul Sperry for help with the probabilities.) You can make a histogram of this probability distribution. Notice, by the way, that you’re more than twice as likely to roll three of a kind as to roll “none of a kind” or all five different: P(3) = 1500/7776, and P(1) = 720/7776.

And you’re over seven times as likely to roll two of a kind (either two the same and three all different, or two pairs with the fifth die different): P(2) = 5400/7776. Of course, Yahtzee isn’t just about the initial roll. You get two tries to improve your combinations by re-rolling some of your dice.

As before, put the x’s (1–5 this time) in one list and the p’s in another. Run the MATH200A program and select 1:Histograms etc. The data arrangement this time is a probability distribution, so specify 4 and then enter your data list and probability list. The finished probability histogram is shown at right. (For a probability distribution, the program automatically makes a histogram, with no option to make a polygon.) Notice that the bar for two of a kind is much higher than any of the others. If it wasn’t obvious from the numbers, you can see from the histogram that this distribution is skewed right. If you like, you can press the TRACE button and see the numerical value of the probability for each outcome.

For example, P(2) = 0.6944, meaning that when you roll five dice you have almost a 70% chance of getting two of a kind. What about two of a kind or better? This is a classic “at least” problem, and the complement is your friend. P(1) is 0.0926 and 1−P(1) = 0.9074. You have better than a 90% chance of getting at least two of a kind when rolling five dice.

Histogram of a Binomial Distribution A will show how to compute the binomial probability of a particular x or a range of x. But for an overview of a distribution, a histogram is most helpful. This part of the program creates one for you. Example: Suppose 65% of registered voters in Dryden are Republicans. Plot the binomial probability histogram for the random variable “number of Republicans in a random sample of ten registered voters“. Solution: Unlike the other cases, for the binomial probability histogram you don’t enter anything in statistics lists.

Instead, the program will prompt you for the necessary numbers. Run the MATH200A program and select 1:Histograms etc. When the program asks your data arrangement, select 5 for a binomial probability distribution. The program will then ask for number of trials (n) and probability of success (p).

You can select a histogram or polygon as usual; the histogram is shown at right. Interpreting the histogram: The bars are each one unit wide, running from 0 to n (0 to 10, for this example), with the dots marking each integer and the y axis marking the minimum possible value of successes, 0. You can see that the most likely result for a sample of ten registered voters is seven Republicans, with six being just slightly less likely. Eight and five are next most likely, then nine and four. A sample of ten is quite unlikely to contain ten or three Republicans, and fewer than three Republicans are extremely unlikely. To see the probabilities as numbers, or use to compute them.

Box-Whisker Plot Showing Outliers Summary: A modified box-whisker plot is a quick graphical representation of a data set. It plots the five-number summary (minimum, first quartile, median, third quartile, maximum) and also shows outliers if the data set contains any. The 2:Box-whisker part of the MATH200A program makes a modified box-whisker diagram for one data set or compares two or three data sets by stacking box-whisker plots.

One Sample, List of Numbers. 11 15 14 12 9 8 7 5 6 11 10 10.5 12 11 13 2 6 4 13.5 Here again is the set of quiz scores. You’ve already seen them graphed as a, but a box-whisker plot is another way to get a sense of the shape of the data.

Enter the data in any statistics list and run the MATH200A program. Select 2:Box-whisker and you see a prompt for the number of samples. You can make a box-whisker plot of a single data set, or you can. The quiz scores are a single sample, so you choose 1. Then the program asks whether you have a plain list of numbers or an ungrouped frequency distribution. Here you have a plain list of numbers, so you choose data arrangement 1. Finally, the program asks you for your data list.

Caution: Never make a boxplot of a grouped frequency distribution. Only a simple list of numbers or an ungrouped frequency distribution is suitable for a box-whisker plot. If you have a grouped frequency distribution, a box-whisker plot won’t be accurate and you should be using a. The box-whisker plot now appears (below left). You can see at a glance that it has no outliers, and that it’s slightly skewed left. You can also trace the box-whisker, to see the five-number summary and the values of any outliers.

Press the TRACE key and then use ◄ and ► to move left and right. In the illustration below right you can see that the median quiz score was 10.5. 11 15 14 12 9 8 7 5 6 11 10 10.5 12 11 13 2 6 4 13.5 22 What would an outlier look like on a box-whisker plot? Any outliers show up as isolated points separate from the main diagram. For example, take the same set of data, but append a twentieth data point, 22. Now make the plot and you’ll see the result at right. The TRACE key and arrow keys will display the values of any outliers as well as the five-number summary.

Stacked Boxplots to Compare Data Sets full citation at page 168, shows data for two groups of rats. One group was sent into space; the control group was treated the same except for the space flight. Their red blood cell mass was measured in milliliters.

Flight Control 8.59 6.87 7.00 6.39 7.43 9.79 9.30 8.64 7.89 8.80 7.54 7.21 6.85 8.03 8.65 7.62 7.33 7.14 8.40 8.55 9.88 6.99 7.44 8.58 9.14 9.66 8.70 9.94 Plotting the two data sets as two boxplots on the same screen is a good way to get a sense of whether there is a difference between the samples. (By the way, these two samples happen to be the same size, but you can just as well stack boxplots with samples of different sizes.) Put the flight group in one statistics list and the control group in another. Then run the MATH200A program, select 2:Box-whisker, and select 2:Compare 2 smpl. For each sample you’ll be asked your data arrangement. (That lets one sample be a plain list and the other an ungrouped frequency distribution, but in this example both are plain lists.) Enter the names of the two lists, as shown below right.

The results are shown at right. You can see that the flight group, as a group, had lower blood-cell mass than the control group, even though some individuals from the two groups had equal blood-cell mass. Look particularly at the medians: the median for the control group is about equal to the third quartile of the flight group, meaning that about three quarters of the flight rats had blood-cell mass lower than the median of the control group. Is that enough to say that space flight lowers blood-cell mass in rats in general? Later in the course, you’ll learn how to use a two-sample t test to tell whether space flight lowers blood-cell mass in rats — whether there is a difference between the populations of all space rats and all earthbound rats. Probability of a Binomial Experiment Summary: If you have a fixed number of trials n, and each trial has only two outcomes (called success and failure), and the probability of success p is the same on each trial, then you have a binomial probability distribution.

This part of the program computes the probability of a specific number of successes or the probability of a range of numbers of successes. TI-89 users: Please see.

See also: You can also make a. Binomial Probability for One x Value The program always asks you the number of trials n, the probability p of success on one trial, and the number of successes from and to. If you want a specific number of successes rather than a range, then from and to will be the same number.

Example 1: Larry’s batting average is.260. If he’s at bat four times, what is the probability that he gets exactly two hits? Solution: n = 4, p = 0.260, x = 2 Note: Some textbooks use r for number of successes, rather than x. Here you want the probability of exactly two successes, so FROM and TO will both be 2. In other words, you are computing the probability of 2 through 2 successes. Run the MATH200A program, select 3:Binomial prob, and specify n = 4, p =.26, from = 2, to = 2. The input and output screens are shown at right.

Answer: P(2) = 0.2221 Caution: Sometimes the probability is very small and the calculator reports it in scientific notation, such as 5.4189E-6. The exponent is not a decoration, and you must not report the probability as simply 5.1489.

Probabilities are never greater than 1! Conventionally, we round probabilities to four decimal places. If the probability is smaller than 0.0001 (smaller than 1E-4), you can show it to two significant figures, such as 1.3×10 -6, or report it as. Summary: By plotting data on your TI calculator, you can easily see how close they are to a normal distribution. The special quantile plot or normal probability plot asks what the distribution would look like if it were normal, and plots that against the actual distribution. The closer the points seem to be to a straight line, the more nearly normal the original distribution.

The most common application is in inferential statistics: with a small sample (less than about 30), you need to make sure that the population is normally distributed before you perform a Student’s t test. Example: Consider these vehicle weights (in pounds): 2500, 3250, 4000, 3500, 2900, 4500, 3800, 3000, 5000, 2200 Construct a plot to decide whether these vehicle weights seem to be normally distributed. Solution: Put the data in any statistics list, then press PRGM, scroll down to MATH200A, and press ENTER twice. Select 4:Normality chk.

The program will make the plot and display the correlation coefficient r and sample size n, as shown at right. If the points are clearly linear (close to a straight line), r is close to 1 and you know that the sample was nearly a normal distribution; if they are clearly not near a straight line, r is far from 1 and you know that the sample is non-normal. How far from r=1 is too far? The program also displays “critical r”, from. If r≥crit, it’s close enough to 1; if r. Region 2000 Census current sample Northeast 19.0% 274 Midwest 22.9% 303 South 35.6% 564 West 22.5% 359 “An urban economist wonders if the distribution of residents in the United States is different today than it was in 2000. In 2000,” the proportions were as shown in the table.

“The economist randomly selects 1500 households and obtains the frequency distribution shown. Conduct the appropriate test to determine if the distribution of residents in the United States is different today from the distribution in 2000, using the α = 0.05 level of significance.” Solution: H 0 is that the model is still good, and H 1 is that the distribution of US population has changed. The model is the percentages shown, and you are determining whether the current sample is enough different from the model for you to conclude that the model is no longer correct.

Put the percentages in L1; there’s no need to convert them to decimals but if you do then you must convert all of them. The current sample is the Observed numbers, and they go in L2. Caution: The Observed numbers must always be actual counts; never convert them to percentages. Caution: Some problems will give you total observations, but you never enter the totals in your calculator. Run the MATH200A program, select 6:GoF test, and press 9 to confirm that you have data in the correct lists. The results screens are shown at right. None of the expected values are below 5, so that requirement for the goodness-of-fit test is met.

There are three degrees of freedom (four categories minus one). The χ² test statistic is 8.25, and the p-value is 0.0410. Conclusion: Since p.

Sample of Birth Records Day of Week Freq- uency Sunday 57 Monday 78 Tuesday 74 Wednesday 76 Thursday 71 Friday 81 Saturday 63 “An obstetrician wants to know whether or not the proportion of children born each day of the week is the same. She randomly selects 500 birth records and obtains the data shown.” Clearly the frequencies in the sample vary from day to day, but do they vary enough that you can say babies in general are born with different frequencies on different days, at the 0.05 level of significance? Solution: H 0 is that babies are born with equal frequency on all days of the week, and H 1 is that they are not. You’re not given a numerical model because “equal frequencies” means that all model numbers are the same. Since there are seven categories, your model is seven 1’s in L2. The Observed numbers go in L2, as usual. The results are shown at right.

Here no E’s are α in a χ² test. Strictly speaking, the conclusion should be in neutral language as usual: you can’t determine from the data whether babies are born with equal frequency on all the days of the week or not. But traditionally, the conclusion is often stated in some words equivalent to “the data do not rule out the model”. This is the scientific method. If the experiment is repeated multiple times and H 0 is never rejected, we begin to have more and more confidence that H 0 is actually true.

We can’t accept H 0 from a single experiment, but the more times it’s not rejected, the more we believe that it may never be rejected. Χ² Test for 2-Way Tables.

Summary: The test of independence and the test of homogeneity have exactly the same calculations, and are often lumped together as a χ² test for 2-way tables. As with the one-way table in, the program computes and sums weighted differences between what the observations actually are, and what they would be if H 0 is true.

(H 0 is that two traits are independent or the populations all have the same proportion.) This section of the program includes the calculations of the TI-83’s χ²-Test and provides additional information. Example 2—side effects of drugs, adapted from full citation at pages 579–581: In clinical trials, patients with high cholesterol were randomly assigned to three groups. One group was given Zocor, one group was given a placebo, and one group was given cholestyramine (a generic drug).

Below are the numbers of those in each group who did and did not experience abdominal pain. At the 0.01 significance level, is a person equally likely to have abdominal pain on all three regimes?

Zocor Placebo Cholestyramine Abdominal pain 51 5 16 No abdominal pain 1532 152 163 Solution: This is a test of independence. H 0 is “abdominal pain is equally likely with all three treatments”, and H 1 is “abdominal pain is not equally likely with all three treatments”—not just in this sample of patients, but in cholesterol patients generally. Press 2nd x -1 makes MATRIX (on the classic TI-83, press MATRX), then ◄ ENTER to edit matrix A.

You have 2 rows and 3 columns. (In this problem, row and column totals are not shown, but even if they were you would ignore them.) Run the MATH200A program and select 7:Two-way table. Matrix A is the only input; the program will immediately start computing and in a few seconds will display the results. What is this screen telling you?

First, requirements for the χ² test are checked. As you know, all expected values must be at least 5. (Some authors give a looser requirement: no more than 20% of E’s below 5, and no E’s below 1. When there are expected values below 5, MATH200A also shows the number of expected values that are below 1.) In this case no E’s are below 5, so you’re fine. The expected values themselves can be seen in matrix B. There are two degrees of freedom ( df = rows−1 times columns−1). The test statistic is χ² = 14.71, quite large for df=2.

The p-value is therefore quite low, 6.4×10 -4 or 0.0006. Conclusion: Whether you take Zocor, nothing, or cholestyramine does make a difference to your likelihood of experiencing abdominal pain. The last two line tells you that the program has put additional information in matrix C, so let’s have a look there.

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Press 2nd x -1 makes MATRIX (on the classic TI-83, press MATRX), then ◄ 3 to view matrix C. Unfortunately you can see only three columns at a time on your calculator, but I’ve pasted two screens together here for convenience: The program dimensions matrix C two rows and two columns larger than input matrix A. This is what matrix C tells you:.

The upper left rectangle, of the same size as matrix A, is the contribution from each cell to the total χ² for the table. These χ² contributions are also known as the squares of standardized residuals, and as a rule of thumb a value above 4 is probably significant. Since matrix A was 2×3, look at the first 2 rows and 3 columns of C. All the χ² contributions are quite modest, except for row 1, column 3. Referring back to the original data table, you see that that’s the number of people in the cholestyramine group who experienced abdominal pain.

This suggests that abdominal pain with cholestyramine is either more likely or less likely than with either nothing or Zocor, not just in the sample, but in people in general. But be careful! Your significance level relates to the whole χ² test, and the p-value apples to the table as a whole, not to any one cell. The next row and column are the totals from the original matrix. Since the original matrix was 2×3, its totals are in row 3 and column 4 of matrix C. Look at the column totals (row 3 of this matrix). 1583 patients were in the Zocor group, 157 in the placebo group, and 179 in the cholestyramine group.

The row totals (column 4), show that in the entire sample, 72 patients had abdominal pain and 1847 did not. And whichever way you add, the total number of patients in the study was 1919.

Finally, the last row and last column of matrix C are the overall row and column percentages. (The percentages add to 100 in either direction, apart from rounding errors.) If H 0 is true, you would expect to see these same percentages in each row and each column, not just in the totals.

Your H 0 was that all three regimes are equally likely to lead to abdominal pain. In the sample overall, 3.752% (about 4%) had abdominal pain, and 96.248% (about 96%) did not. If H 0 is true and the likelihood of abdominal pain is the same for all three regimes, then you would expect 4% of each group to experience abdominal pain: 4% of the Zocor group, 4% of the placebo group, and 4% of the cholestyramine group. But referring back to the original data, you see that 51/1583≈3% of the Zocor group, 5/157≈3% of the placebo group, and 16/179≈9% of the cholestyramine group experienced abdominal pain. This is why is so large in this study.

Example 2—marital status and happiness, adapted from full citation at pages 575–576: In the 2006 General Social Survey, a random sample were asked their marital status and level of happiness, with the results below. Can you say, at the 0.05 significance level, that happiness level and marital status are dependent? Married Widowed Divorced/ Separated Never Married Totals Very happy 600 63 112 144 919 Fairly happy 720 142 355 459 1676 Not too happy 93 51 119 127 390 Totals 1413 256 586 730 2985 Solution: H 0 is “marital status and happiness level are independent”, and H 1 is that they are dependent (or associated, or not independent). Enter the numbers in 3×4 matrix A, run the MATH200A program, and select 7:Two-way table. Even if you are given row or column totals, you are concerned only with the detailed numbers. You have three happiness levels and four marital statuses, so this is a 3×4 matrix. Results: No expected values are below 5.

The χ² statistic of 224.12 is enormous, and the p-value of 1.4×10 -45 is correspondingly tiny, so you conclude that happiness level and marital status are indeed dependent. Matrix C looks like this (again, after pasting two screens together): Since matrix A is 3×4, the first 3 rows and 4 columns of matrix C are the χ² contributions, or squares of standardized residuals. Many of them are quite large, showing that “happiness is independent of marital status” fails badly as a model.

The row and column totals are next, but in this case you already know the totals because they were given in the problem. Finally, look at the row and column percentages.

About 31% of the sample described themselves as very happy, 56% as fairly happy, and 13% as not too happy. If H 0 is true, you would expect 31% of married people to be very happy, 56% of married people to be fairly happy, and 13% of married people to be not too happy; and you would expect those same percentages for widowed people, for divorced or separated people, and for never-married people.

In fact, the percentages are quite different, as you can see if you calculate a few of them. The column percentages can be interpreted in the same way. About 47% of the sample was married, so if H 0 is true you would expect 47% of very happy people, 47% of fairly happy people, and 47% of not-too-happy people to be married. (The same statement applies in the other columns, with their percentages.) Again, this is obviously not the case if you look at the original observed data. What’s New 30 Mar 2016: Add the for critical value in the normality check, for users of the “TI-83 classic”. 27 Dec 2015:.

Program version 7.2: sense the calculator’s screen resolution, using a technique I learned in the forums at, and adjust the and layouts accordingly. Numerical results from the normality check are now shown on the graph screen again. The vamps can we dance ep rar.

Simplify instructions for. Show input screen for the. Shrink most screen shots by 25%.

13 Dec 2015: Program version 7.1: update Web address, show compatibility on splash screen, and shorten menu header. 17 Oct 2015: Add TI Connect CE in. 16 May 2015: Mention binomial probability distribution in the that the program can create. (intervening changes suppressed) Dec 2008: Program version 1.