Need pro statistician help to answer question

Okay I have read through, it is really helpful.

I have decide that I have two teeth (not three) with the two risks I described in the second part of Problem two, post #38. Here is my calculation:

No tooth requiring treatment = 0.94962 * 0.94962 = 90.18%

1 tooth = 0.94962 * 0.05038 * 2 (possibilities) = 9.56%

[the two possibilities are tooth A but not B or tooth B but not A]

2 teeth = 0.05038 * 0.05038 = 0.25%

It adds up to 100 which is reassuring, have I understood that correctly? And thank you.

jpod said:

Okay I have read through, it is really helpful.

I have decide that I have two teeth (not three) with the two risks I described in the second part of Problem two, post #38. Here is my calculation:

No tooth requiring treatment = 0.94962 * 0.94962 = 90.18%

1 tooth = 0.94962 * 0.05038 * 2 (possibilities) = 9.56%

[the two possibilities are tooth A but not B or tooth B but not A]

2 teeth = 0.05038 * 0.05038 = 0.25%

It adds up to 100 which is reassuring, have I understood that correctly? And thank you.

Click to expand...

Looks good to me

Top draw, what's the chances of that! Thanks fellas. pod

no worries - you're not alone btw.. it is a bit of an issue in healthcare in general. Most doctors struggle with basic probability, which is slightly worrying if it means they're not able to then explain basic things to patients like the implications of test results, risk of cancer etc..

http://www.bbc.co.uk/news/magazine-28166019

In 2006 and 2007 Gigerenzer gave a series of statistics workshops to more than 1,000 practising gynaecologists, and kicked off every session with the same question:

A 50-year-old woman, no symptoms, participates in routine mammography screening. She tests positive, is alarmed, and wants to know from you whether she has breast cancer for certain or what the chances are. Apart from the screening results, you know nothing else about this woman. How many women who test positive actually have breast cancer? What is the best answer?

nine in 10
eight in 10
one in 10
one in 100


Gigerenzer then supplied the assembled doctors with some data about Western women of this age to help them answer his question. (His figures were based on US studies from the 1990s, rounded up or down for simplicity - current stats from Britain's National Health Service are slightly different).
The probability that a woman has breast cancer is 1% ("prevalence")
If a woman has breast cancer, the probability that she tests positive is 90% ("sensitivity")
If a woman does not have breast cancer, the probability that she nevertheless tests positive is 9% ("false alarm rate")

In one session, almost half the group of 160 gynaecologists responded that the woman's chance of having cancer was nine in 10. Only 21% said that the figure was one in 10 - which is the correct answer. That's a worse result than if the doctors had been answering at random.

The fact that 90% of women with breast cancer get a positive result from a mammogram doesn't mean that 90% of women with positive results have breast cancer. The high false alarm rate, combined with the disease's prevalence of 1%, means that roughly nine out of 10 women with a worrying mammogram don't actually have breast cancer.
It's a maths puzzle many of us would struggle with. That's because, Gigerenzer says, setting probabilities out as percentages, although standard practice, is confusing. He campaigns for risks to be expressed using numbers of people instead, and if possible diagrams.

Click to expand...

You'd have thought that medical schools and dentistry schools would try and address this with some basic training in statistics.

You know exactly where you are coming from. I was just thinking today that these statistics skills would make a good 2 or 3-hour study sessions for dentists.

I am writing detailed prognoses for personal injury cases, this involves predicting forward what may happen with and without treatment. It is interesting and I have never understood the maths aspect like this before. Makes me better at it, it is all thanks to you and your maths rival there

Clinically I am not really giving much information in statistic form. Some people ask for percentages, the 'phrase more likely than not' is useful in law and to explain things for patients in terms they understand.

Your articles refers to dental radiographs. The information on risk of dental radiographs is here:
http://www.fgdp.org.uk/publications...ation-doses-and-risks-in-dental-practice.ashx

There are exquisitely clear guidelines for when dentists should take radiographs and about 15 years ago there were two UK laws which apply to dentists taking dental x-ray pictures in clinics.

It is important to limit exposure to infants and children. Old-timers have nothing to worry about.

We are taught the skill of critically appraising scientific studies but the statistics is hard. I am a bit pants at maths and the written word, I have to work very hard at it.

Children's dentistry in actual fact has some of the best evidence base, but often dental studies are too 'low powered', study groups n= 20 or 100 and the studies are meaningless or worse, poorly designed and so mismatched that few can be used for meta-analysis.

Another interesting things Dowie is 'changing clinicians perspective', you give older clinicians (dentists and doctors) statistics and guidelines and their prescribing moves a little bit over, it is shifts toward 'the evidence'. You give the same to a younger clinician, or someone who is has an evidence-based practice and they adopt it immediately.

Making the material easy for the clinicians to use, appreciate and buy into can only help.

Thanks

Hello again, ooh my head hurts.................maths.........Please could a maths Pro check my calculations

3 teeth, each tooth has a 12.5% chance of a complication (0.125) and therefore 0.875 chance of no complication.

Therefore the % chances are:

0 tooth needing treatment 67%
1 teeth needing treatment 29%
2 teeth needing treatment 4%
All 3 teeth needing treatment 0.02%

How did I do?

Many thanks, pod

assuming all independent events then looks good... though you've made a typo with the last one and whacked a zero behind the decimal point where it isn't needed

should be approx 0.2% not 0.02%

Dowie you are a star, thank you so much.

I really appreciate you checking that over and coaching my probabilities. I am understanding it a bit better.

Thanks

42

no worries

I am stuck again, please help.

Problem is 4 teeth
3 complications all independent risks and independent of each tooth 5%, 5%, 7.5%

I need an explanation like i am a 10 year old

Chance of 0 complications
Chance of 1 tooth
2 teeth
3 teeth
all 4

I made this but it doesn't add up to 100! Sniff

Any maths Pros? Help needed, sniff............

Think I have figured it out

Chance of 0 teeth complication 48.5%
Chance of 1 tooth 38.5%
Chance of 2 11.5%
Chance of 3 1.5%
Chance of 4 approaching 0%


There were 16 possibilities

Where is Dowie and the maths Pros?

Yes. Look up combinations and probability. You know that there will be 6 combinations for 2 complications from 4 teeth because 4 choose 2 = 6

https://en.wikipedia.org/wiki/Binomial_coefficient
http://www.calculatorsoup.com/calculators/discretemathematics/combinations.php

Hi Amleto
Hmm thanks for this I will have a read. There are 3 possible complications and 4 teeth...have a look at the table I made......did I get the write answers?

you answers are close to correct, however you have got some rounding errors (assuming you haven't intentionally rounded to the nearest .5%, which would not be good form anyway).

Also, you should only really state your final values to the same precision as your input values (2 significant figures). Strictly speaking it should only be to a single figure as 5% is given as the input value rather than 5.0%, but I'd stick to 2sf personally.

0: 49%
1: 38%
2: 11%
3: 1.5%
4: 0%

Hi Bill thank you so much, that is very reassuring indeed.

I actually can't remember how to round up - I will have to research that once again.

When the probability of something is 0.0007 is it correct to say the chance of it occurring is approaching zero.

Because although the chance of 4 complications is O in this instance, it is a possibility..we can't say it would never happen.

Thanks again pod

To save confusion, I would avoid the phrase 'approaching zero', as it is used in mathematics to mean something other than what you are trying to describe. (i.e. if you take a number and repeatedly half it, then it approaches zero). In the situation you are describing, the probability of 4 complications is a fixed value, rather than one that changes.

You could say that it is 'close to zero', or explicitly state the degree of accuracy (i.e. that it is 0.0% to 2 significant figures, or that it is 0.0% to 1 decimal place).

Bill that is very helpful indeed. Thank you for taking the time to provide such clear explanations. I learnt something here and it is appreciated because it helps me with my job.