# In 200 tosses of a coin,

• December 29th 2009, 08:39 PM
zorro
In 200 tosses of a coin,
Question : In 200 tosses of a coin, 115 heads and 85 tails were observed. Test the hypothesis that the coin is fair , using significance level 0.05
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Should i check for mean or should i check for variance
• December 29th 2009, 11:48 PM
mr fantastic
Quote:

Originally Posted by zorro
Question : In 200 tosses of a coin, 115 heads and 85 tails were observed. Test the hypothesis that the coin is fair , using significance level 0.05
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Should i check for mean or should i check for variance

You should calculate $Pr(X \leq 85)$ under the null hypothesis X ~ Binomial(n = 200, p = 1/2) and draw a conclusion.
• December 31st 2009, 02:03 AM
zorro
I thinks i might be wrong so i am posting the
Quote:

Originally Posted by mr fantastic
You should calculate $Pr(X \leq 85)$ under the null hypothesis X ~ Binomial(n = 200, p = 1/2) and draw a conclusion.

Mr fantastic I think i might be wrong thats why i am posting the starting bit please let em know if i am correct or no???

$H_0 : X \le 85$
$H_a : X > 85$

$z_{0.05}$ : 1.645

$z = \frac{\bar x - \mu}{\sigma/ \sqrt{n}}$

I am pretty sure that this is not correct.....need ur guidance
• December 31st 2009, 04:51 AM
Ond
I would have thought that you needed a test for the proportion. That is:

$H_0 : \pi=0,5$
$H_A : \pi<0,5$

$Z=\frac{p-\pi}{\sqrt{\pi (1-\pi )/n}}$

where $p=85/200, \pi=0,5$ and $n=200$
• December 31st 2009, 07:50 AM
ANDS!
Quote:

Originally Posted by zorro
Mr fantastic I think i might be wrong thats why i am posting the starting bit please let em know if i am correct or no???

$H_0 : X \le 85$
$H_a : X > 85$

$z_{0.05}$ : 1.645

$z = \frac{\bar x - \mu}{\sigma/ \sqrt{n}}$

I am pretty sure that this is not correct.....need ur guidance

If you are going to be testing the mean, then you need to test with a null hypothesis of X equal to 100, and an alternate hypothesis of X not equal to 100.

The hypothesis you have set up now, is saying that this "fair coin" observance comes from a population with a mean of 85 tails. . .that doesn't make any sense. You are testing under the assumption that coin IS fair, and that this collection of 200 tosses you have sampled comes from a population of "coin tosses" with a mean of 100 tails and 100 heads.

When you are running your hypothesis test, you are essentially trying to figure out what are the chances that you grabbed one of the samples of 200 that have such an unbalanced distribution of heads and tails (assuming the coin is fair). If the probability is low (your p-value), or if its just lower than your significance level, then there might be good evidence that the sample you pulled from isn't one with an even distribution of heads and tails.
• December 31st 2009, 01:58 PM
zorro
Quote:

Originally Posted by Ond
I would have thought that you needed a test for the proportion. That is:

$H_0 : \pi=0,5$
$H_A : \pi<0,5$

$Z=\frac{p-\pi}{\sqrt{\pi (1-\pi )/n}}$

where $p=85/200, \pi=0,5$ and $n=200$

i have a different formula for test of proportion .

$z = \frac{x-n \theta}{\sqrt{n \theta (1 - \theta)}}$

Is this the same one which u have mentioned in ur post and could i use this one instead of the one which u posted or are u using a totally different formula
• December 31st 2009, 07:47 PM
mr fantastic
Quote:

Originally Posted by Ond
I would have thought that you needed a test for the proportion. That is:

$H_0 : \pi=0,5$
$H_A : \pi<0,5$

$Z=\frac{p-\pi}{\sqrt{\pi (1-\pi )/n}}$

where $p=85/200, \pi=0,5$ and $n=200$

If the hypothesis that the coin is fair is to be tested, then
$H_A : \pi {\color{red}\neq} 0,5$. The test will be two-tailed.

All the ideas necessary to answer the question are now in this thread.

The trouble is not that help is needed with this question (which is similar to some otheres that have previously been posted). The trouble is that all this material has to be presented and taught from scratch. MHF cannot do this. That is the job of the instructor, classnotes and textbook.
• December 31st 2009, 07:51 PM
mr fantastic
Quote:

Originally Posted by zorro
i have a different formula for test of proportion .

$z = \frac{x-n \theta}{\sqrt{n \theta (1 - \theta)}}$

Is this the same one which u have mentioned in ur post and could i use this one instead of the one which u posted or are u using a totally different formula

You should realise that the two formulae are equivalent. There are just notational differences. If you multiply the numerator and denominator of the formula quoted by Ond you get the formula quoted by yourself, aside from the notational differences.

You should also realise that:

1. $p$ and $\theta$ are the same thing (just different notation).

2. The estimator of $\theta$ is $\hat{\theta} = \frac{X}{n}$.

3. $\hat{\theta}$ ~ Normal $\left(\mu = \theta, \, \sigma^2 = \frac{\theta (1 - \theta)}{n} \right)$.

Your formula follows directly from this.

I have said this before: It is important to understand where a formula comes from and exactly what the symbols in it mean.
• January 1st 2010, 02:49 AM
zorro
what is $\pi$ then????
• January 1st 2010, 03:05 AM
mr fantastic
Quote:

Originally Posted by zorro
what is $\pi$ then????

Quote:

Originally Posted by Mr Fantastic
[snip]
multiply the numerator and denominator of the formula quoted by Ond and you get the formula quoted by yourself, aside from the notational differences

Compare the two formulas. What do you think?
• January 1st 2010, 03:43 AM
zorro
from the look of it i think $\pi$ is $\theta$