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The given data below shows the time in intervals in seconds between successive white cars in flowing traffic in an open road.Can these be modeled by an exponential distribution.

Time        0-     20-      40-     60-      90-     120-180
Frequency   41     19       16      13        9        2

My question is when I calculate Expected frequencies though the total should add up to 100 it does not do so because of rounding off errors right?
I get a value of 98.89.
Here should I need to add another category myself as the interval being 180-infinity and get its expected value as 100-98.89=1.11. Is it necessary to add this category as this is an exponential distribution which goes for infinity?
Of course as the Expected value of the created category is less than 5, in this case I have to add this up with the upper category.
But if I do not consider this new category then the degrees of freedom changes.Is it necessary to add this new category as 180-infinity

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    $\begingroup$ Let's back-up: do you have the raw data, because if so, we wouldn't start from here? $\endgroup$ – Nick Cox Dec 6 '13 at 14:52
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    $\begingroup$ The raw data would be the individual times. This is a coarsely binned distribution, which is less satisfactory. These "Frequencies" look like percents, not counts, as they add to 100. You need the total number of measurements to apply a goodness of fit test. Where does 180 come from? You need at least one bin 120 up. Also, we can't comment on your calculations if you don't show us what they are. $\endgroup$ – Nick Cox Dec 6 '13 at 15:02
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    $\begingroup$ Unless you are using a broken abacus, the sum of expected frequencies should be indistinguishable from $1$. But for us to help you, you will need to describe how you are computing the expected frequencies. As far as the degrees of freedom go, the correct value depends on how the bin endpoints ($0,20,40,60,80,90,120$) were chosen: if they were fixed before observing the data, you're fine; but if they were determined based on the data (as it would seem from their irregular spacing), the situation is complicated. And please heed @Nick Cox's caution. $\endgroup$ – whuber Dec 6 '13 at 15:05
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    $\begingroup$ Is this seemingly textbook-style question for some subject? Is it for your own study? $\endgroup$ – Glen_b Dec 6 '13 at 15:47
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    $\begingroup$ This looks like a homework problem, which would explain why raw data aren't available and why the problem seems narrowly framed. If this were a real data analysis, I would (1) probably include the additional ">180" empty bin; (2) figure that it was unlikely to make much difference because it's a small fraction of the data; (3) worry about computing the mean of the exponential from the midpoints of the bins. If it were a homework problem, I would ask the instructor for clarification. $\endgroup$ – Ben Bolker Dec 6 '13 at 15:48
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Firstly, Rearrange the table so it makes better sense: and calculate mean via calculator.calculator. In order to do this take the midpoint of each interval for time and enter this on the display alongside the frequencies. you should get 40. this is the MEAN however, so to get lambda, use the following formula lambda = 1/mean this should give you a lambda value of 0.025. use this and the cumulative pdf (1-e-^lambda x time interval) function to calculate p(X=x) for each and x by total to get expected for each. Because we are dealing with a continuous distribution, the first and last intervals should be calculated as less than 20 and more than 120.i.e first calculation will be 1-e-^0.025x20= 0.3935 (keep to at least 4dp).. second will be (1-e-^0.025x40) - (1-e-^0.025x20) = 0.2387.. carry on until ll intervals have been done as shown above. times each p(X=x) by 100. keep expected to 2dp.

Time           0-20   20-40    40-60    60-90   90-120       120-180     total
Frequency        41      19       16       13        9       (11)  2       100
p(X=x)       0.3935  0.2387   0.1447   0.1177   0.0556        0.0497         1 (about!)
Expected      39.35   23.87    14.47    11.77     5.56  (10.53) 4.97   (combine last two)
(O-E)^2/E   0.06919 0.99359  0.16178  0.12854  0.02098    1.37 (3sf)

H0: Data follows an Exponential Distribution
H1: Data does not follow an Exponential Distribution (i.e Use 5% SL)

DF = 5-1-1 = 3 (1 for totals and 1 for estimating lambda from population)
X (c.v)= 7.815

1.37 < 7.815 so accept H0. Conclude time in intervals in seconds between successive white cars in flowing traffic in an open road can be modeled by an exponential distribution.

I did A2 Statistics, a reliable source in itself.

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  • $\begingroup$ (+1) This is basically a good summary--although it would be nice to see some abbreviations explained--but there is one small technical fault: for the chi-squared distribution to be valid, the exponential parameter has to be fit using maximum likelihood rather than this approximate method of moments. It turns out also that your result is fairly sensitive to the post hoc decision to combine the last two categories. These increase the $\chi^2$ statistic to $5.402$ (based on an estimated exponential rate of $\hat\lambda=0.02556$), corresponding to $p=0.37$--still not significant by any standard. $\endgroup$ – whuber Jun 27 '14 at 16:57
  • $\begingroup$ (+1) How exactly are you ending up with a mean of 40? I don't fully understand your arguments for that. $\endgroup$ – k.dkhk Jun 11 '17 at 17:11

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