(Statistics) Using the TI-83 to find z-scores?
918506
Here is your answer - the normalcdf function is used:\nhttp://webserver3.lanecc.edu/math/staff/ShellabD/calculatorCentralFolder/TI-83NormCDF&Invnorm.html
986574
if you are using a TI-83, use the Z-Interval test funcition.\n\nManual working:\n\nSampling error = 2.06(SD/sqrt n)\n\n\nlower boundary =  20 - 2.06(3/sqrt 25) = 18.764\nupper boundary =  20 + 2.06(3/sqrt 25) = 21.236\n\n96% confidence limits is\n18.764 < mean < 21.236
278272
You'll use Z-scores and the Z-table. Z-scores convert deviations to standard deviation units. \n\nIn this case, you're looking for the range of scores that makes up the middle 88% of the distribution - 44% on either side of the mean. \n\nThe z-score is 1.56 (look in the table for the z-score that cuts off 44% of the distribution.) This tells you that 44% of the scores fall between the mean and 1.56 standard deviation units above the mean - and 44% of all scores fall between the mean and 1.56 standard deviation units below the mean. \n\nNow, since each standard deviation unit in this particular distribution is 4ml (as per your question), 1.56 standard deviation units is equal to 1.56 x 4, which is equal to 6.24ml.\n\nBack to the problem - your range is 6.24 ml above and below the mean - which is 250ml - therefore the range of volumes is from 6.24ml less than 250 to 6.24ml more than 250 - or in other words - 243.76ml to 256.24 ml.\n\nHope that helped!
875098
Depends on the model...\n\nOn the Texas Instruments (TI) models, you would use the GCD function.\n\nGCD is found under MATH → NUM, arrow down to gcd( \nFrom the home screen, choose gcd(  followed by two numbers separated by a comma.\n \nNOTE:  gcd( accepts non-negative integers only and works with only two entries at a time.  If you need to figure the GCF or more than two numbers, do them a pair at a time.
211283
This link has statistical calculators:\n\nhttp://calculators.stat.ucla.edu/\n\nA webpage that will lead you to more statistic links.\n\nhttp://math.about.com/od/statistics/\n\nI hope this helps.
1178089
I think what you should do is zoom standard, I had the same problem on Ti 83 and it worked well after i resetted the zoom. So I think it should work on urs too. Good luck.
509283
In statistics we talked about residuals. a Residual is anestimate on how well your regression line fits your data. The residual is often notes as r^2. I'm not sure on exactly how to find it, but on the TI 83 plus, the residual is a flat line and the data is scattered around it. The data should be spread equally on both sides of the line. the more equal the data is spread, the better your residual line. When you find the "VarsStat1" on the TI 83, the residual value will be there noted as r^2.
308472
Find yourself a z-statistic table and look up critical value in the table.  Check your stats book on how to interpret the table.\nhttp://www.math2.org/math/stat/distributions/z-dist.htm
808315
I havn't done it mysqlf but a good place to start is with the TI-83 SDK and the documentation. The headers and required includes usually have a lot of information on how they can be used and often have a brief tutorial/starter project to build off of.
838358
i can't give you the answer but i used my ti-83 and i can tell you the points of the graph...\n(1,0) (2,4) (3,8) (4,0)\n\nHope this helps
550037
Press 2nd Math  to view the test menu. \n\nPress the number associated with a symbol or use the cursor keys to select one and press enter. \nThe relationship between any two numbers, expressions, lists, or matrices can be compared by selecting one of the options. A 1 will be returned if the test is true and a 0 if the test is false. For example, the test, 4 > 7, returns a value of 0 since it is a false statement. Tests can also be made on statements such as\nx = 9 or x < 2.5\n\nThe result of such a test depends upon the value of the variable x. This type of comparison is often used to control the domain over which a graph is drawn. \nWhen comparing lists and matrices, each element in the first list or matrix is compared to its corresponding element in the second list or matrix. For example, note the results of the following comparison between the two 3 element lists. \nThe {1 1 0} indicates the first two comparisons are true, while the third is false.
965376
z=w+whx\nz=w(1+hx)\nz/(1+hx) = w &#92;
309011
Using the z-scores corresponding to the given areas, the cutoff scores would be:\nA: 93% (93.2233%)\nB: 80% (79.7798%)\nC: 68% (68.2202%)\nD: 58% (58.4535%)\nF: <58%
1078373
2z+7<z+10\nyou need to get like terms together so first...\n2z+7+-7<z+10-7\n2z<z+3\nnow get z over to the  2z\n2z-z<z-z+3\nz<3\n\ncheck ya later ♥
863034
This is from memory so:\na z-score is the number of std deviations away from the mean. This allows you to compare data points from different populations as how many std-dev they are away from the mean.\n\nTherefore the z-score at the mean = 0.\nz-score one std deviation away = 1.\n\nFrom wiki:\n"In statistics, a standard score (also called z-score or normal score) is a dimensionless quantity derived by subtracting the population mean from an individual (raw) score and then dividing the difference by the population standard deviation.\n\nThe z score reveals how many units a case is above or below the mean. The z score allows us to compare the results of different normal distributions, something done with frequent research."
736646
The 100(1 - α)% prediction interval (PI) is calculated as\nYfit ± t(α,n-1)*S*sqrt(1 + 1/n + (X_0 – X_bar) / ((n-1)S^2) )\n\nwhere Yfit is the fitted linear regression model; t(α,n-1) is the 100(1 - α /2)th percentile of the Student’s t distribution with n – 1 degrees of freedom; S is the standard deviation and X_bar is the mean.\n\nSince we are given the following facts:\n(a) Yfit = 140.53 (value for the 94-day linear regression model)\n(b) X_0 = 140.91 (current daily value)\n(c) X_bar = 132.55 (sample mean/average for the 94 days)\n(d) S = 4.72 (sample standard deviation for the 94 days)\n(e) α = 0.05 (level of significance)\n(f) n = 3483 observations\n\nThen, t(α,n-1) is approximated by the Gaussian (normal) distribution; that is,\nt(α,n-1)  is approximated by Zα = 1.960645. So, by plugging all these informations, we get the upper and lower prediction limits (UPL and LPL, respectively) as follows:\n\nUPL = 153.2923114 and\nLPL = 127.7676886.\n\nThat is, we are 95% confident that the on value of the LRT on the 95th day will be between 127.77 to 153.29.\n\nNote that PI’s are always gives wider interval than that of the confidence intervals (CI’s). This happens because the standard error of Yfit in the PI is expressed as sqrt(1 + 1/n + (X_0 – X_bar) / ((n-1)S^2) ) is greater than the standard error of Yfit in the CI is expressed as sqrt(1/n + (X_0 – X_bar) / ((n-1)S^2) ).\n\nIn the second question, it is indeed that a rough estimate of the 95% PI is 2 STANDARD ERRORS above and below the LRT. Note that there is a result from statistical theory that given a large sample size (as what happened to your data), the t-distribution is being approximated by the Gaussian distribution. That is with large sample size (n),  T_0.95 is the 97.5th percentile of the Student's t distribution and is about 1.959963985. This approximately 2. On the other hand, the 97.5th percentile of the Gaussian distributio is 1.959963985 which is approximately 2.
740147
If you have a TI-83 you can graph it and find it easily...\n\ny= button (plug in equation) ---> hit graph\n\nthen hit the second button(yellow)--->calc(same as trace button)\n\nbecause the graph was concave up you want to find a minimum...so click that...and simply enter once on the left side of the graph, once on the right, and then guess in the middle.  It will then give you the exact answer.\n\n(1,6) should be the correct coordinate.