Question

In Desperate Need Of Help!!!
when a fair coin is flipped, we all know that the probability the coin lands on heads is 0.50. however, what if a coin is spun? according to an article, 2 polish math professors and their students spun a Belgian euro coin 250 times. it landed heads 140 times. one of the professors concluded that the coin was minted asymmetrically. a representative from the Belgian mint said the result was just chance.
(a) state the hypotheses we are interested in testing

Answer 1

There is a higher chance on landing on heads than tails when a coin is spun.

Answer 2

Or something like that... You need to state: "it is hypothesized that...".

Answer 3

Whilst I can't do the maths right now - 140/250 as opposed to 125/250 does not seem to be a statistically unlikely outcome. Maybe you could evaluate the probability of 140/250 heads? I'd guess it's fairly high.

Answer 4

(b) show that the conditions for performing a one sample z-test for a proportion are satisfied

Answer 5

Here is a detailed analysis of exactly the question:

http://www.cs.toronto.edu/~mackay/euro.pdf

The stats are beyond my scope - but the conclusion seems to be that 140/250 is not a significant variation from the 'norm' i.e. does not indicate a bias to heads

Answer 6

I don't understand what that is. I've never seen that equation before

Answer 7

That paper is doing some kind of Bayesian inference. I don't think you need to approach the problem in that way though.

Answer 8

Usually the conditions for performing a one sample z-test for a proportion are:
- a random sample is selected from binomial distributed data
- the sample size is large (usually this is checking that \(np\ge15\) and \(n(1-p)\ge 15\), where the \(p\) here is the one from your null hypothesis

Answer 9

The 7% figure in that paper should be the classical p-value you would obtain (in a "frequentist" framework vs. a Bayesian one)

Answer 10

what you said is gibberish to me. I haven't understood this class from day 1.

Answer 11

Do you know what a binomial distribution is?

Answer 12

no

Answer 13

An experiment can be said to have a binomial distribution if:
-You have successive trials that are independent of each other (that means the previous event doesn't affect the outcome of the next event)
-The trials are binary in nature. That means that each trial can have 2 outcomes. Usually a "Yes/No" outcome, or "Alive/Dead", or "It rained/It didn't rain", etc. In math talk, we abstractly refer to these 2 types of events as "Success/Failure" to denote any 2 outcomes in a general sense.
-Each event has a probability of occurrence. The probability of a success might be "p" and the probability of failure as "1-p". Also, this probability of success (and of failure) do not change in each trial.

Let me illustrate the above with a simple example:

We toss a die 10 times, and we want to know the probability that it will show up "Heads" 4 times.
-We have 10 trials here.
-Each trial has 2 outcomes: Heads or Tails. We denote the outcome of interest (the number of heads) as the success. So, getting a heads = a "success". Getting a Tails = a "failure." So, the goal is to find the probability of getting 4 successes.
-The probability of success (getting a "heads) is p=1/2 , and the probability of getting a failure (getting "a tails") is 1-p= 1/2.
-Each trial is independent of each other. So, let's say on trial 1, you flip the coin and get "Heads". Does this mean that you are more likely to get tails on the 2nd trial? NO!! Because the coin has no memory... it doesn't know if you have heads or tails before, so the outcomes are independent of each other.
-Also, note that obviously the probability within in trial itself doesn't change (the probability of getting a heads is always 1/2 on a given trial).

Now, there is a formula that tells you the probability of getting a certain number of successes. If you denote the number of successes by \(X\), then the probability of getting a certain number of successes has the formula:
\[P(X=x)={n \choose x}p^x(1-p)^{n-x}\]
where \(n\) is the number of trials, \(p\) is the probability of a success.

So, if you're interested in knowing the probability of getting 4 heads in 10 tosses, then you would plug in:
\[P(X=4)={10 \choose 4}\left(\frac{1}{2}\right)^4\left(1-\frac{1}{2}\right)^{10-6}\]

Answer 14

Sorry I wrote "We toss a die 10 times" when I really mean "We toss a coin 10 times"