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# X^2 (kye squared) tests watch

1. Hi guys i have a quick question about q11 part c. If the probability of success changes by calculating it from the data, then the expected values will change.

So the test statistic will change as well as the critical value as the degree of freedom changes.

So do we assume that the change in expected values and thus the test statistic are negligible so that we can reach the answer in part c in solution bank?

Thanks
2. (Original post by Shaanv)
Hi guys i have a quick question about q11 part c. If the probability of success changes by calculating it from the data, then the expected values will change.

So the test statistic will change as well as the critical value as the degree of freedom changes.

So do we assume that the change in expected values and thus the test statistic are negligible so that we can reach the answer in part c in solution bank?

Thanks
Blows my mind for sure.

The only thing I can think of is that you would expect the estimate to be fairly close to 0.5 (analogous to throwing say 100 times a fair coin, you might get 47 heads and hence estimate a probability of 0.47, not far off 0.5), hence expected values similar hence ...

Seems like a stinker of a question.
3. (Original post by Shaanv)
Hi guys i have a quick question about q11 part c. If the probability of success changes by calculating it from the data, then the expected values will change.

So the test statistic will change as well as the critical value as the degree of freedom changes.

So do we assume that the change in expected values and thus the test statistic are negligible so that we can reach the answer in part c in solution bank?

Thanks
You are quite right. If you plow through the calculations, you get a value for the binomial parameter of 0.486875, and a value of of 4.4778, and therefore the conclusion of the question stands.

There is a sneaky way to answer the question fully, though, without re-doing the calculations this way. You know that the value of must be less for the estimated binomial parameter (can you see why?) and therefore the conclusion stands if the previous version of is less than the new critical value.

4. (Original post by Gregorius)
You are quite right. If you plow through the calculations, you get a value for the binomial parameter of 0.486875, and a value of of 4.4778, and therefore the conclusion of the question stands.

There is a sneaky way to answer the question fully, though, without re-doing the calculations this way. You know that the value of must be less for the estimated binomial parameter (can you see why?) and therefore the conclusion stands if the previous version of is less than the new critical value.

Because binomial parameter is calculated from the data the expected values are closer to the observed so the \chi^2 value decreases and must be less than the old \chi^2 value.

As old \chi^2 value is less than new critical value, then new \chi^2 value must also be less than new critical value.

Is that correct?

Thanks for the help.
5. (Original post by Shaanv)
Because binomial parameter is calculated from the data the expected values are closer to the observed so the \chi^2 value decreases and must be less than the old \chi^2 value.
Yes, that's it, though I realized after posting that this is trickier to justify than I thought!

As old \chi^2 value is less than new critical value, then new \chi^2 value must also be less than new critical value.

Is that correct?
Yes.

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