Q&A Bank4 Q4.28

Hello I have a few questions here on the solution given,


Chi square test: "the mean of the sampling distribution is 7....". May I know where does the sampling distribution comes from and how do you derive the 7?

Individual standardised deviations: The solution uses the formula of standardised deviation which should be used only when Poisson model is used in the graduation. However, we are given 'Exposed to Risk' (Ex) and graduated initial mortality. Shouldn't the underlying model be a Binomial model? Am I right to say that both can be accepted since we learnt in CT3 that Poisson distribution can be used to approximate a Binomial distribution?

Also, the solution gives a table of values under the formula. Can someone explain to me how the figures are derived? I couldn't seem to get the figures. Probably someone can just show me how one or two of the figures are derived.

Cumulative deviations: I do not understand this part. Seems like the answer is using a different formula from the reading material. Can anyone explain to me how the observed value (-24), standard deviation (103.6), observed test statistic (-0.23) are derived? Or why they are derived this way?

What does it mean when it says "the observed test statistics is -0.23, which has a unit normal sampling distribution". Unit normal distribution refers to a standard normal distribution. Why do we have a unit normal sampling dist?

Hope these questions are not too much

 

One more question

Why does the solution skip the serial correlation test?

 

Hi - here we go!

Note this is Q4.29 in my 2008 version!

Chi-square - It's basically just referring to the fact that the expected value of a variable that has a chi-squared distribution is equal to the number of degrees of freedom - refer back to your CT3 notes or any stats text book for this.

Std devs - It is using the binomial - but is using an approximation to the variance in the denominator - should be Eqp but assumes that p is very close to 1 so ignore it.

Table of values - expected values are the number of ages multiplied by the expected (Unit Normal proportions) - this is explained in the notes Ch 12 Section 7.2 Step 3 - and for the observed just count them up - eg (-2,-1) group - the 2nd, 4th and last values in the list of calculated std devs are in this range, ie 3 in all.

Cum devs - the observed value is simply the total actual deaths minus the total expected deaths = 10710 - 10734 = -24. Divide by the standard deviation of this (ie sqaure root of 10734 = 103.6), giving a standardised value (test statistic) of -24/103.6 = -0.23. This is within the acceptable 2-tailed range of -1.96 to +1.96. To be honest, I don't see how this differs from the way it is done in the course. As to why this in unit-normal - the approach I've just described is shown in Section 7.4 of Chap 12. Whenever you take a Normal variable (-24 here), take off its mean (zero) and divide by its standard deviation (103.6 here), then a standard normal variable is the result. Go back to CT3 again if you're not sure!

As for the serial correlations test - I have no idea why this isn't in, to be honest. If this was an exam question then it would be a perfectly viable test to choose. One thing in the exam though, if a question just says "test adequately", or similar, then you don't have to do every test. Eg cum devs and signs tests are broadly equivalent, and serial correlations and grouping of signs are broadly equivalent. (However, this argument doesn't apply to this Q because the solution includes both the signs and cum dev tests but only the grouping of signs test!) Comments from others very welcome here!