standardised deviations test: normal dist percentages

[Molly]

Hi all,

Have just had a thought - when calculating the standardized deviations test, can the percentages used in the expectation calculations change?

Ie for the interval (0inf, -3) -> (3, inf) we use 0, 0.02m, 0.14m, ... in our calculations of the expectation - would this change if the interval changes? Does anyone know where i could find the new percentages?

Thanks
Molly

 

[Andrew Martin]

Hi Molly

The percentages depend on the intervals. If we have Z ~ N(0,1), then:

P(- inf < Z < 3) = 1 - 0.99865 = 0.135%, which is rounded to 0% in your reference above.

The 0.99865 comes from the tables for looking up P(Z <= 3), we could also use R / Excel to calculate this.

Also:

P(2 < Z < 3) = P(Z <= 3) - P(Z <= 2) = 0.99865 - 0.97725 = 2.140%, which is rounded to 2% in your reference above.

Again, the 0.99865 comes from the tables for looking up P(Z <= 3) and the 0.97725 comes from P(Z <= 2).

If we were doing this in a paper B, we could use R to calculate all the relevant probabilities as follows:

> intervals <- c(-Inf, -3, -2, -1, 0, 1, 2, 3, Inf)

> (CDF.probs <- pnorm(intervals))
[1] 0.000000000 0.001349898 0.022750132 0.158655254
[5] 0.500000000 0.841344746 0.977249868 0.998650102
[9] 1.000000000

> diff(CDF.probs)
[1] 0.001349898 0.021400234 0.135905122 0.341344746
[5] 0.341344746 0.135905122 0.021400234 0.001349898

We could then use these more accurate probabilities rather than the rounded 0%, 2%, 14% etc

Hope this helps!

Andy

 

[Molly]

Hi Molly

The percentages depend on the intervals. If we have Z ~ N(0,1), then:

P(- inf < Z < 3) = 1 - 0.99865 = 0.135%, which is rounded to 0% in your reference above.

The 0.99865 comes from the tables for looking up P(Z <= 3), we could also use R / Excel to calculate this.

Also:

P(2 < Z < 3) = P(Z <= 3) - P(Z <= 2) = 0.99865 - 0.97725 = 2.140%, which is rounded to 2% in your reference above.

Again, the 0.99865 comes from the tables for looking up P(Z <= 3) and the 0.97725 comes from P(Z <= 2).

If we were doing this in a paper B, we could use R to calculate all the relevant probabilities as follows:

> intervals <- c(-Inf, -3, -2, -1, 0, 1, 2, 3, Inf)

> (CDF.probs <- pnorm(intervals))
[1] 0.000000000 0.001349898 0.022750132 0.158655254
[5] 0.500000000 0.841344746 0.977249868 0.998650102
[9] 1.000000000

> diff(CDF.probs)
[1] 0.001349898 0.021400234 0.135905122 0.341344746
[5] 0.341344746 0.135905122 0.021400234 0.001349898

We could then use these more accurate probabilities rather than the rounded 0%, 2%, 14% etc

Hope this helps!

Andy

Hi Andy,

That is really really helpful, thank you so much. How do you normally decide on which interval to use?

 

[Andrew Martin]

Hi Molly

The intervals used in my post are the example intervals given in the Core Reading (see Section 7.2 in Chapter 10). I'd use these as a starting point but then see whether we have any issues with any expected values being too small, which causes issues with the robustness of the test (as per any test of this nature). If so, we would need to combine some of the intervals, eg we may need to try -inf to -2 rather than -inf to -3. Again this is mentioned in Section 7.2.

Hope this helps!

Andy

 

[Molly]

Hi Molly

The intervals used in my post are the example intervals given in the Core Reading (see Section 7.2 in Chapter 10). I'd use these as a starting point but then see whether we have any issues with any expected values being too small, which causes issues with the robustness of the test (as per any test of this nature). If so, we would need to combine some of the intervals, eg we may need to try -inf to -2 rather than -inf to -3. Again this is mentioned in Section 7.2.

Hope this helps!

Andy

Amazing, thank you so much