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You have more than 1 000 000 rows of data?

I find it hard to share your indignation. but perhaps it is justified if you know the context and think that the author is somehow using the graph to mislead the reader. Perhaps 0, 1, 2, 3 would have been better put as 0%, 50% 67%. 75% but that might not have got the authors point across as...

You can look at it as being linear in P for a given amount of C. I'm not familiar with the context so I cannot say more, but how would you improve the graph so that it illustrates what you think is really the connection between Risk and Diet.

Can you tell us what the graph seems to be saying at first glance, and what it actually is saying.

OK, you're right. They still have to justify all the steps, and it seems unlikely to me that Markov's inequality will help here.

In Excel type =1-GAMMADIST(6,3,1,1) ( I think... You're unlikely to get more efficient than that if it's true.)

Exponential with rate 1, perhaps? So something to do with gamma. It's been so long....

Have you considered a Monte Carlo approach where you use your accumulated data to generate a population? I feel that you current approach is asking for trouble in the future. I understand that you intend to publish your findings, at which point your work will be scrutinized by hostile critics...

Most designs have groups of data points recording the response for several subjects. These data points are assumed to be normal so that the anova theory works. As I understand it, it is not essential for these groups to be normal - it is the sampling distribution of the sample means of these...

Independence is difficult to check. The best way is a well designed experiment with randomization of subjects, but even so you need to avoid, for example, twins as subjects (unless they are part of the design.) For the rest, most researchers will make the best design possible, do the...

A simple view is a combination of regression (continuous predictors) and anova (categorical predictors) although the internal workings use just what you have said - regression with dummy predictors for the categories.

I'm not sure exactly what you want but d(logX) can be written dX/X

Now you know they're significant, try doing them separately.

You're perfectly right. Sorry. I think I probably got mixed up with the word "observed" which usually means the number seen which in this case is "count". In any event, with chi square, it only works with the raw count data, never averages or percentages.

So I guess the question boils down to "how many can you take and have the chance of all negatives = 0.01" The chance of any one choice negative is 1 - 0.06 = 0.94 so the chance of getting k negatives in a row is 0.94^k = 0.01 Does that sound right? Over to you ...

It looks to me like - get obs and count for each day - get total obs and total count - Then for each day, expected = day count/total count*total obs. Observed is always the actual count.

This is a rather oddly framed question. Do they mean "exactly 300/5000 bottles are infected" or "any particular bottle has a probability of 300/5000 or 0.06 of being infected"?

582 I assume you have Excel. Google how to get the developer tab up and create a macro, type or copy, then run it. Sub count() sol = 0 For a = 1 To 10: For b = 1 To 10: For c = 1 To 10: For d = 1 To 10: For e = 1 To 10 If 10 * a + 9 * b + 7 * c + 3 * d + 5 * e = 229 Then sol = sol + 1 Next...

This is almost certainly done most quickly by writing a brute force program with 5 loops. 3 min for program, 2 sec for solution.

Perhaps the 33/81 is for suspension helmets alone.