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# S2 - Normal Distribution Hypothesis Testing watch

1. I'm a bit confused about one thing in normal distributions for S2 (for OCR MEI), so was wondering if someone could help.

When you perform a hypothesis test on a sample distribution against a population distribution (to see if the population mean is wrong using a sample from the distribution), I understand how to find the test statistic but I am having problems understanding how to find the critical value.

My current method is to find the inverse normal of (1-SL%) for one tailed tests or (1-0.5*SL%) for two tailed tests. However, this only gives positive values using the inverse normal tables and sometimes the answer requires a negative value. I think it has something to do with the tail of the distribution, but don't really understand when my critical value should be negative.

Could anyone help explain this to me (using an example would be awesome)?

Thanks.
2. Okay, example.

You have a distribution you believe to have mean 0, s.d. 10. You want to test if the mean is greater than or less than 0, with a significance level of 10%. Now, the value on a normal distribution of mean 0, s.d. 1 for which the probability of an observation being less than that is 0.05, can be found with a simple sketch of the bell-curve. As it happens, that value is -1.645 (to 3 s.f.). Now, we can just convert that to the value we want for our normal distribution, by doing: (a - 0)/10 = -1.645; where a it our critical value. So a would be -16.45. How could you be getting a positive value?

I have not explained that very well, sorry. What are these inverse normal tables, by the way?
3. (Original post by StrangeBanana)
Could you give us an example question (sorry for being no help, but I'm having a hard time understanding your problem)?

Sure:

I started by defining both situations (X-N(7000,10000)) and (Xbar-N(6972,400)) and defining my null and alternative hypothesis (Ho: miu=7000, H1: miu does not equal 7000 - where miu = population mean lifetime of low energy bulbs).

I then found the test statistic z using z=(6972-7000)/20 = -1.4

Then I tried to find the critical value, and this is where I have a problem. My method is that for one tail test I find the inverse normal (from tables) of (1.00-SL) and for two tail tests I find the inverse normal of (1.00-0.5xSL). So for this case the SL = 10% and it is two tail, so I found the inverse normal of 1-0.05 = 0.95. This gives me a value of 1.645. However, it should be -1.645, my issue is why?

How do I know when it's negative or not? Because, as you probably know, the inverse normal tables only give positive values therefore using the table alone you can't get negative values.

If you'd like me to clarify anything, just ask.

Thanks again
4. (Original post by StrangeBanana)
Okay, example.

You have a distribution you believe to have mean 0, s.d. 10. You want to test if the mean is greater than or less than 0, with a significance level of 10%. Now, the value on a normal distribution of mean 0, s.d. 1 for which the probability of an observation being less than that is 0.05, can be found with a simple sketch of the bell-curve. As it happens, that value is -1.645 (to 3 s.f.). Now, we can just convert that to the value we want for our normal distribution, by doing: (a - 0)/10 = -1.645; where a it our critical value. So a would be -16.45. How could you be getting a positive value?
This appears to be a completely different method but I'll see what happens!

Why is s.d. 10?

Hmm using your method this is what I can follow:

P(Z<k)=0.05
Inverse normal(0.05) = -1.645

Once I have this value, I can see that [-1.4] < [-1.645] (where [ denotes a modulus) therefore the test statistic < critical value therefore result is not significant.

And this is exactly what the mark scheme states...

I don't understand where your value of -16.45 comes into this?

I have not explained that very well, sorry. What are these inverse normal tables, by the way?
I think you did a good job, I've just never seen your method before!

In the formula booklet it is labelled "The Inverse Normal function: values of Inverse(P)=z"

Basically, if P(Z<a)=b
The inverse normal of b = a *

* I think so, anyway!

Thanks
5. (Original post by Law-Hopeful)
This appears to be a completely different method but I'll see what happens!

Why is s.d. 10?

Hmm using your method this is what I can follow:

P(Z<k)=0.05
Inverse normal(0.05) = -1.645

Once I have this value, I can see that [-1.4] < [-1.645] (where [ denotes a modulus) therefore the test statistic < critical value therefore result is not significant.

And this is exactly what the mark scheme states...

I don't understand where your value of -16.45 comes into this?

I think you did a good job, I've just never seen your method before!

In the formula booklet it is labelled "The Inverse Normal function: values of Inverse(P)=z"

Basically, if P(Z<a)=b
The inverse normal of b = a *

* I think so, anyway!

Thanks
I've been studying using Edexcel, not OCR MEI, that may be the reason behind the differences.

I was just using that as an example. You're rarely going to get distributions with a mean of 0 and s.d. of 1, so I tried to make it a bit more realistic.

Well if it's what the mark-scheme says, then hey-ho!

-16.45 would have been the critical value for the distribution with s.d. 10. -1.645 would be it for a s.d of 1.

:O I've never seen tables like those before.

Anyway, if this has helped, then hooray; I'm slightly confused myself, to be honest. I shall have to do some work through the OCR MEI version of S2!
6. (Original post by StrangeBanana)
I've been studying using Edexcel, not OCR MEI, that may be the reason behind the differences.

I was just using that as an example. You're rarely going to get distributions with a mean of 0 and s.d. of 1, so I tried to make it a bit more realistic.

Well if it's what the mark-scheme says, then hey-ho!

-16.45 would have been the critical value for the distribution with s.d. 10. -1.645 would be it for a s.d of 1.

:O I've never seen tables like those before.

Anyway, if this has helped, then hooray
Oh ok, if it's not too much trouble, could you adapt your example and go through it for the example I posted? (If it's too much trouble don't bother!)

Oh I see.

They are a central part of S2 in MEI!

It has helped a bit, thanks!

I didn't really understand this bit though: "Now, the value on a normal distribution of mean 0, s.d. 1 for which the probability of an observation being less than that is 0.05, can be found with a simple sketch of the bell-curve." - why is the value we want less than 0.05 when the SL is 10%, and it is a two tailed test?

Also, from P(X<0.05) how did you get -1.645 from the bell curve?

(Original post by StrangeBanana)
I'm slightly confused myself, to be honest. I shall have to do some work through the OCR MEI version of S2!

If you look through the mark scheme for that which you can find here (for Q3iv) perhaps you can follow what they did as I probably didn't explain it very well

Thank you for your help though, much appreciated!
7. (Original post by Law-Hopeful)
Oh ok, if it's not too much trouble, could you adapt your example and go through it for the example I posted? (If it's too much trouble don't bother!)
I'll have a crack in the morning, right now I just want to sleep. xD

(Original post by Law-Hopeful)
I didn't really understand this bit though: "Now, the value on a normal distribution of mean 0, s.d. 1 for which the probability of an observation being less than that is 0.05, can be found with a simple sketch of the bell-curve." - why is the value we want less than 0.05 when the SL is 10%, and it is a two tailed test?
Well if it's two-tailed, we want 5% on each side (for a 10% significance level).

(Original post by Law-Hopeful)
Also, from P(X<0.05) how did you get -1.645 from the bell curve?
We're not looking for P(X < 0.05), we're looking for P(X < a) = 0.05 (we want a).
8. (Original post by StrangeBanana)
I'll have a crack in the morning, right now I just want to sleep. xD

Well if it's two-tailed, we want 5% on each side (for a 10% significance level).

We're not looking for P(X < 0.05), we're looking for P(X < a) = 0.05 (we want a).
Yeah no problem, thanks!

That makes sense.

Ah yes, and we can use the inverse normal tables to find a by finding the inverse of 0.05 (which gives -1.645!). It's starting to come together slowly I think...

I'll mull everything over, thanks again for your help
9. (Original post by Law-Hopeful)
Yeah no problem, thanks!

That makes sense.

Ah yes, and we can use the inverse normal tables to find a by finding the inverse of 0.05 (which gives -1.645!). It's starting to come together slowly I think...

I'll mull everything over, thanks again for your help
Sometimes it can take significant hardship to unify two syllabi.

No problem!

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