How do i perform a 2-way anova using only SPSS software?
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SPSS\nWith your data entered, go to ANALYZE > GENERAL LINEAR MODEL > UNIVARIATE \nEnter dependent variable in Dependent List \nEnter independent variables in Factor \nContinue – OK \nOutput:\n(A) Descriptives (N, Mean, Std. Dev., etc) \nThis will give you a table. On the right hand side, the significances will be listed, along with the F values. On the far left side, under the column “source,” your variables will be listed. The corresponding F values and significances are for main effects and interaction effects. If the significance reading is less than .05, you are able to reject the null hypothesis.
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I've just answered this in your previous question: "Statistics: Variability & SRS Samples of Population?"\n \nIf you look there, you'll find the formula for the s.d. of p_hat. So that increasind n by multiplying it by "m", reduces (devides) sd (or se) by sqrt(m).\n \nFor example, lets say the sd is 0.2 and the sample size is 100. If you increase the sample to 400 (multiply it by 4), the sd is devided by sqrt(4) = 2, i.e., sd = 0.1.
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What are the main topics of Biostatistics? ANOVA?
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If you are talking about z scores for one particular variable, then you should use a histogram.  Excel does not do histograms, so you should probably use MINITAB or some other statistics program.\n\nThe previous answer only applies if there are two variables involved.
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Check out this website it might help a bit possibly!?  I use SPSS 14.0 so I'm not sure about the differences but this site is for 13.0 ...http://www.ats.ucla.edu/STAT/SPSS/output/reg_spss.htm\n\nHowever, I'm not sure how familiar you are but the stepwise multiple regression is selected in SPSS by using the Method text box on the Linear Regression box.  You need no particular order but the variables are entered into the analysis such that those that produce the largest increase in the multiple R are entered first.  For instance, if you did not have a particular theory but you wanted to see which of the variables best predicted the outcome variable, SPSS will do this for you based on the extent to which they increased the mutiple R. \n\nHope this helps! Good luck!!
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a 2x3 ANOVA has two factors one with 2 levels and one has 3 levels a 1x6 would mean that one factor only has one level which doesn't make sense because then it isnt a variable
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I find it hard to believe that the pairs are treated independently. There's no point to that, it doesn't give you any useful statistical information.\n\nHowever, if you're trying to compare all the values on the left side as a group with their pairs on the right side as a group, then you could use a modified version of the Chi^2 test. Basically multiply everything by 100, and then do a normal Chi^2 test.\n\nYour degrees of freedom will be (6-1)(2-1) = 5
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I was answering another similar question, and I came across a website that has a downloadable sample spreadsheet for doing statistical distributions.  Play with this and see if it helps you out.
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Well, your difference in results could be a number of a things. As far as Excel is concerned, pay close attention to how many decimal places you allow in each cell. If your cell is too small, the number will be rounded up and it's possible it could be radically different from the number SPSS & MINITAB give you.\n\nI like SPSS because the syntax editing is pretty simple. At work, I use SPSS, Excel and STATA, and honestly, I prefer STATA. It's very flexible, even if it is trickier to use.\n\nAnyway, It's hard to answer your questio without more details. Your values can also vary if you haven't done the right command. And finally, as one person mentioned above, each program uses a different algorithm to compute the values your asking for, so your answers will probably vary slightly from time to time. However, if there are big differences, then you should check your method or your data, because something in the command you executed or missing data could be causing you problems.
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You need a variable in any experiment to compare results. Perhaps you could use cold (ice) water in one container, warm in another and hot or boiling in another. You can then make your prediction on which will dissolve first and have an experiment that will verify whether you have made an accurate prediction.
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The first sample proportion is p1 = 30/100. Its variance is (30/100)*(1-30/100)/100.\n\nThe second sample proportion is p2=60/80. Its variance is (60/80)*(1-60/80)/80.\n\nWhen you look at the difference, p1-p2, and assume that the two proportions are independent then the standard deviation is\nsqrt(var1+var2)=\nsqrt(  (30/100)(1-30/100)/100 + \n          (60/80)(1-60/80)/80)\nwhich is approximately 0.06666146.\n\nHere you can use the normal approximation for both proportions, and the difference is \n30/100-60/80 = -.45.\nIf you use a two sided test then the z-score is -.45/0.06666146 = -6.75, and the p-value is 1.473087e-11. It is highly likely that the difference is zero! Please note that this last part is not quite right. If you want to do a formal hypothesis test then you need to do a pooled estimate.
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If the two numbers across a row refer to the same plant then you have paired data. You can think of it as "before and after." So if you have \n\nx_1  | y_1\nx_2  | y_2\n...\nx_N  | y_N\n\nthis is paired data and you can replace it with a single variable which is the difference:\n\nw_1 = x_1 - y_1\nw_2 = x_2 - y_2\n...\nw_N = x_N - y_N\n\nNow you can do a standard statistical test on w. For example\n\nh0 : mean of w = 0\nha: mean of w is not 0\n\n(If mean w < 0 then mean x < mean y)\n\nOn the other hand, if the two sets of numbers are for two different sets of plants then you do not have paired data. To make matters confusing, in this case you have to used a paired t-test! (The worst thing about statistics is the language.)\n\nFor example, if you have a set of plants, x, that you raise in one condition, and another set of plants, y, that you raise in a different set of conditions then you do not have paired data and have to look at the difference in the means,\n\nLet w = x-y\nthen the mean of the sample mean of w = mean x - mean y\nstd dev of the sample mean of w = sqrt( var(x)/N_x + var(y)/N_y)\n\nNow you do your statistical test on w,
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http://www.utexas.edu/its/rc/tutorials/stat/spss/spss1/
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http://www.statsoft.com/textbook/stathome.html
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Does your school give you access to SPSS?\n\nCheck with one of the librarians at your campus library....they can really help you out with this type of stuff.\n\nGood Luck!
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In a one-way ANOVA, you are testing.\n\nH0: mu1 = mu2 = ... = mk\nHA: At least one mean is different,\n\nwhere k is the number of groups.\n\nThe F statistic, sB^2/sW^2, has an F distribution with k-1 and n-k degrees of freedom under the null hypothesis, where n is the total number sampled.  You will reject the null if the F statistics is too large, since sB^2 would become larger than sW^2 if the null is false.  You would pick an F value such that your significance level (typically called alpha) is in the tail of the F distribution.
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yes but it doesnt do a real good job of it. you need the data analysis tool pack and then just run a regression and see if it works out for you. i use eviews for mine but there are a ton of other regression analysis tools to use. the only difference is the interface and the amount of the commands that you have to type in.