Tjyylli

How to determine omitted units in a publication

Did the new image of black hole confirm the general theory of relativity?

What was the last x86 CPU that did not have the x87 floating-point unit built in?

What force causes entropy to increase?

How to read αἱμύλιος or when to aspirate

Circular reasoning in L'Hopital's rule

How to make Illustrator type tool selection automatically adapt with text length

Did the UK government pay "millions and millions of dollars" to try to snag Julian Assange?

Can the Right Ascension and Argument of Perigee of a spacecraft's orbit keep varying by themselves with time?

Can we generate random numbers using irrational numbers like π and e?

Is there a way to generate uniformly distributed points on a sphere from a fixed amount of random real numbers per point?

Do ℕ, mathbb{N}, BbbN, symbb{N} effectively differ, and is there a "canonical" specification of the naturals?

How do you keep chess fun when your opponent constantly beats you?

Can a flute soloist sit?

Can each chord in a progression create its own key?

Can I visit the Trinity College (Cambridge) library and see some of their rare books

Using dividends to reduce short term capital gains?

Word for: a synonym with a positive connotation?

What's the point in a preamp?

Make it rain characters

Loose spokes after only a few rides

Why can't devices on different VLANs, but on the same subnet, communicate?

Didn't get enough time to take a Coding Test - what to do now?

Do I have Disadvantage attacking with an off-hand weapon?

two types of coins, decide which type it is based on 100 flips

.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty{ margin-bottom:0;
}

2

$begingroup$

I have two types of coins, one is fair(0.5 chance to fall on H), and the other is biased(0.7 chance for H). I have a coin, and I need to determine, based on a 100 flips - which type is my coin.
the hypothesis is that the coin is fair only if it fell on heads no more than 62 times out of 100.
what are the chances that I got to the wrong conclusion and the coin is fair?
what are the chances I got to the wrong conclusion and the coin is unfair?

I tried to calculate the chances but I got some prety ugly numbers...

hypothesis-testing

share|cite|improve this question

asked yesterday

Benaya TrabelsiBenaya Trabelsi

111

New contributor

Benaya Trabelsi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Can you lay out your calculations?
  $endgroup$
  – bi_scholar
  yesterday
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I am trying but still can't get how to write math equations here...sorry. in words: sum K from 0 to 62 of: 100 choose k * 0.5^k * 0.5^100-k.(for the first part)
  $endgroup$
  – Benaya Trabelsi
  yesterday
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  "the hypothesis is that the coin is fair only if it fell on heads no more than 62 times out of 100. " This is a strange and unclear hypothesis. What do you mean with this sentence? Maybe you meant something like "the decision rule is to accept the hypothesis that the coin is fair when it fell on heads no more than 62 times out of 100"
  $endgroup$
  – Martijn Weterings
  yesterday
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  "what are the chances that I got to the wrong conclusion and the coin is fair?" You can not compute this with the given information. Did you maybe mean "what are the chances that I got to the wrong conclusion if the coin is fair?"
  $endgroup$
  – Martijn Weterings
  yesterday
  

  
  
  
  

add a comment | 

2

$begingroup$

I have two types of coins, one is fair(0.5 chance to fall on H), and the other is biased(0.7 chance for H). I have a coin, and I need to determine, based on a 100 flips - which type is my coin.
the hypothesis is that the coin is fair only if it fell on heads no more than 62 times out of 100.
what are the chances that I got to the wrong conclusion and the coin is fair?
what are the chances I got to the wrong conclusion and the coin is unfair?

I tried to calculate the chances but I got some prety ugly numbers...

hypothesis-testing

share|cite|improve this question

asked yesterday

Benaya TrabelsiBenaya Trabelsi

111

New contributor

Benaya Trabelsi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Can you lay out your calculations?
  $endgroup$
  – bi_scholar
  yesterday
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I am trying but still can't get how to write math equations here...sorry. in words: sum K from 0 to 62 of: 100 choose k * 0.5^k * 0.5^100-k.(for the first part)
  $endgroup$
  – Benaya Trabelsi
  yesterday
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  "the hypothesis is that the coin is fair only if it fell on heads no more than 62 times out of 100. " This is a strange and unclear hypothesis. What do you mean with this sentence? Maybe you meant something like "the decision rule is to accept the hypothesis that the coin is fair when it fell on heads no more than 62 times out of 100"
  $endgroup$
  – Martijn Weterings
  yesterday
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  "what are the chances that I got to the wrong conclusion and the coin is fair?" You can not compute this with the given information. Did you maybe mean "what are the chances that I got to the wrong conclusion if the coin is fair?"
  $endgroup$
  – Martijn Weterings
  yesterday
  

  
  
  
  

add a comment | 

$begingroup$

I have two types of coins, one is fair(0.5 chance to fall on H), and the other is biased(0.7 chance for H). I have a coin, and I need to determine, based on a 100 flips - which type is my coin.
the hypothesis is that the coin is fair only if it fell on heads no more than 62 times out of 100.
what are the chances that I got to the wrong conclusion and the coin is fair?
what are the chances I got to the wrong conclusion and the coin is unfair?

I tried to calculate the chances but I got some prety ugly numbers...

hypothesis-testing

share|cite|improve this question

asked yesterday

Benaya TrabelsiBenaya Trabelsi

111

New contributor

Benaya Trabelsi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

share|cite|improve this question

asked yesterday

Benaya TrabelsiBenaya Trabelsi

111

New contributor

Benaya Trabelsi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

share|cite|improve this question

asked yesterday

Benaya TrabelsiBenaya Trabelsi

111

New contributor

Benaya Trabelsi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked yesterday

Benaya TrabelsiBenaya Trabelsi

111

New contributor

Benaya Trabelsi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked yesterday

Benaya TrabelsiBenaya Trabelsi

111

3 Answers
3

active

oldest

votes

2

$begingroup$

Essentially, the upper tail (K > 62) of the head-distribution of the fair coin (P(head) = 0.5) gives you the probability that a fair coin is falsely labeled as biased while the lower tail (K <= 62) of the head-distribution of the biased coin gives you the probability that a biased coin will be falsely labeled as fair.

share|cite|improve this answer

answered yesterday

bi_scholarbi_scholar

41813

$endgroup$

add a comment | 

1

$begingroup$

Based on your comment, you're nearly correct for the first one, you just need to subtract the value from $1$ since for having wrong decision if the coin is fair
you need $>62$ heads (let $W$ be having the wrong decision, $F$ be the fact that coin is fair, and $X$ be the number of heads appeared):
$$P(W|F)=1-P(Xleq62|F)=1-sum_{k=0}^{62}{100 choose k}0.5^k0.5^{100-k}$$

Replicating the analysis for the unfair case:
$$P(W|F')=1-P(X>62|F')=1-sum_{k=63}^{100}{100 choose k}0.7^k0.3^{100-k}$$

Note: Your seem to ask for "having wrong decision and the coin is fair" for the first one (same situation for your second question), which means $P(Wcap F)$, not $P(W|F)$ as found above. So, if you need $P(Wcap F)=P(W|F)P(F)$, you need to use $P(F)$, which you didn't provide.

share|cite|improve this answer

answered yesterday

gunesgunes

7,3761316

$endgroup$

add a comment | 

0

$begingroup$

This is the binomial distribution. Doing a web search for "binomial calculator" should get an online calculator. You can then enter .5 for probability of success on a single trial for the fair coin, 100 for number of trials, and 62 as number of successes. You can also enter .7 as the probability of success on a single trial for the unfair coin.

share|cite|improve this answer

answered yesterday

AcccumulationAcccumulation

1,68626

$endgroup$

add a comment | 

Your Answer

StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "65"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});

function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: false,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: null,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});


}
});

Benaya Trabelsi is a new contributor. Be nice, and check out our Code of Conduct.

draft saved

draft discarded

Sign up or log in

StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});

Sign up using Google

Sign up using Facebook

Sign up using Email and Password

Post as a guest

Name

Email

Required, but never shown

StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstats.stackexchange.com%2fquestions%2f402413%2ftwo-types-of-coins-decide-which-type-it-is-based-on-100-flips%23new-answer', 'question_page');
}
);

Post as a guest

Name

Email

Required, but never shown

2

$begingroup$

Essentially, the upper tail (K > 62) of the head-distribution of the fair coin (P(head) = 0.5) gives you the probability that a fair coin is falsely labeled as biased while the lower tail (K <= 62) of the head-distribution of the biased coin gives you the probability that a biased coin will be falsely labeled as fair.

share|cite|improve this answer

answered yesterday

bi_scholarbi_scholar

41813

$endgroup$

add a comment | 

2

$begingroup$

Essentially, the upper tail (K > 62) of the head-distribution of the fair coin (P(head) = 0.5) gives you the probability that a fair coin is falsely labeled as biased while the lower tail (K <= 62) of the head-distribution of the biased coin gives you the probability that a biased coin will be falsely labeled as fair.

share|cite|improve this answer

answered yesterday

bi_scholarbi_scholar

41813

$endgroup$

add a comment | 

$begingroup$

Essentially, the upper tail (K > 62) of the head-distribution of the fair coin (P(head) = 0.5) gives you the probability that a fair coin is falsely labeled as biased while the lower tail (K <= 62) of the head-distribution of the biased coin gives you the probability that a biased coin will be falsely labeled as fair.

share|cite|improve this answer

answered yesterday

bi_scholarbi_scholar

41813

$endgroup$

share|cite|improve this answer

answered yesterday

bi_scholarbi_scholar

41813

answered yesterday

bi_scholarbi_scholar

41813

answered yesterday

bi_scholarbi_scholar

41813

1

$begingroup$

Based on your comment, you're nearly correct for the first one, you just need to subtract the value from $1$ since for having wrong decision if the coin is fair
you need $>62$ heads (let $W$ be having the wrong decision, $F$ be the fact that coin is fair, and $X$ be the number of heads appeared):
$$P(W|F)=1-P(Xleq62|F)=1-sum_{k=0}^{62}{100 choose k}0.5^k0.5^{100-k}$$

Replicating the analysis for the unfair case:
$$P(W|F')=1-P(X>62|F')=1-sum_{k=63}^{100}{100 choose k}0.7^k0.3^{100-k}$$

Note: Your seem to ask for "having wrong decision and the coin is fair" for the first one (same situation for your second question), which means $P(Wcap F)$, not $P(W|F)$ as found above. So, if you need $P(Wcap F)=P(W|F)P(F)$, you need to use $P(F)$, which you didn't provide.

share|cite|improve this answer

answered yesterday

gunesgunes

7,3761316

$endgroup$

add a comment | 

1

$begingroup$

Based on your comment, you're nearly correct for the first one, you just need to subtract the value from $1$ since for having wrong decision if the coin is fair
you need $>62$ heads (let $W$ be having the wrong decision, $F$ be the fact that coin is fair, and $X$ be the number of heads appeared):
$$P(W|F)=1-P(Xleq62|F)=1-sum_{k=0}^{62}{100 choose k}0.5^k0.5^{100-k}$$

Replicating the analysis for the unfair case:
$$P(W|F')=1-P(X>62|F')=1-sum_{k=63}^{100}{100 choose k}0.7^k0.3^{100-k}$$

Note: Your seem to ask for "having wrong decision and the coin is fair" for the first one (same situation for your second question), which means $P(Wcap F)$, not $P(W|F)$ as found above. So, if you need $P(Wcap F)=P(W|F)P(F)$, you need to use $P(F)$, which you didn't provide.

share|cite|improve this answer

answered yesterday

gunesgunes

7,3761316

$endgroup$

add a comment | 

$begingroup$

Based on your comment, you're nearly correct for the first one, you just need to subtract the value from $1$ since for having wrong decision if the coin is fair
you need $>62$ heads (let $W$ be having the wrong decision, $F$ be the fact that coin is fair, and $X$ be the number of heads appeared):
$$P(W|F)=1-P(Xleq62|F)=1-sum_{k=0}^{62}{100 choose k}0.5^k0.5^{100-k}$$

Replicating the analysis for the unfair case:
$$P(W|F')=1-P(X>62|F')=1-sum_{k=63}^{100}{100 choose k}0.7^k0.3^{100-k}$$

Note: Your seem to ask for "having wrong decision and the coin is fair" for the first one (same situation for your second question), which means $P(Wcap F)$, not $P(W|F)$ as found above. So, if you need $P(Wcap F)=P(W|F)P(F)$, you need to use $P(F)$, which you didn't provide.

share|cite|improve this answer

answered yesterday

gunesgunes

7,3761316

$endgroup$

share|cite|improve this answer

answered yesterday

gunesgunes

7,3761316

answered yesterday

gunesgunes

7,3761316

answered yesterday

gunesgunes

7,3761316

0

$begingroup$

This is the binomial distribution. Doing a web search for "binomial calculator" should get an online calculator. You can then enter .5 for probability of success on a single trial for the fair coin, 100 for number of trials, and 62 as number of successes. You can also enter .7 as the probability of success on a single trial for the unfair coin.

share|cite|improve this answer

answered yesterday

AcccumulationAcccumulation

1,68626

$endgroup$

add a comment | 

0

$begingroup$

This is the binomial distribution. Doing a web search for "binomial calculator" should get an online calculator. You can then enter .5 for probability of success on a single trial for the fair coin, 100 for number of trials, and 62 as number of successes. You can also enter .7 as the probability of success on a single trial for the unfair coin.

share|cite|improve this answer

answered yesterday

AcccumulationAcccumulation

1,68626

$endgroup$

add a comment | 

$begingroup$

This is the binomial distribution. Doing a web search for "binomial calculator" should get an online calculator. You can then enter .5 for probability of success on a single trial for the fair coin, 100 for number of trials, and 62 as number of successes. You can also enter .7 as the probability of success on a single trial for the unfair coin.

share|cite|improve this answer

answered yesterday

AcccumulationAcccumulation

1,68626

$endgroup$

share|cite|improve this answer

answered yesterday

AcccumulationAcccumulation

1,68626

answered yesterday

AcccumulationAcccumulation

1,68626

answered yesterday

AcccumulationAcccumulation

1,68626