two types of coins, decide which type it is based on 100 flips The 2019 Stack Overflow Developer Survey Results Are In Unicorn Meta Zoo #1: Why another podcast? Announcing the arrival of Valued Associate #679: Cesar Manarahow do I interpret the following hypothesis test?Determine if many coins are fairYou observe k heads out of n tosses. Is the coin fair?Binary variance? Comparing two sacks of uneven coins or two heterogenous groups of peopleCheck whether a coin is fairHow to test if the proportion of subjects being significant - is significantly different than expected by chance?How can I compute the probability of type 1 error and power?Hypothesis testing: rejection region depends on significance level or vice versa?Bias caused by optional stoppingTest for average of independent, non-Identical binomial distributions
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two types of coins, decide which type it is based on 100 flips
The 2019 Stack Overflow Developer Survey Results Are In
Unicorn Meta Zoo #1: Why another podcast?
Announcing the arrival of Valued Associate #679: Cesar Manarahow do I interpret the following hypothesis test?Determine if many coins are fairYou observe k heads out of n tosses. Is the coin fair?Binary variance? Comparing two sacks of uneven coins or two heterogenous groups of peopleCheck whether a coin is fairHow to test if the proportion of subjects being significant - is significantly different than expected by chance?How can I compute the probability of type 1 error and power?Hypothesis testing: rejection region depends on significance level or vice versa?Bias caused by optional stoppingTest for average of independent, non-Identical binomial distributions
.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;
$begingroup$
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
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.
$endgroup$
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
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
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.
$endgroup$
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
hypothesis-testing
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
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
asked yesterday
Benaya TrabelsiBenaya Trabelsi
111
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.
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.
Benaya Trabelsi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$begingroup$
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 |
$begingroup$
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
$begingroup$
Can you lay out your calculations?
$endgroup$
– bi_scholar
yesterday
$begingroup$
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
$begingroup$
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
$begingroup$
"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
$begingroup$
"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 |
3 Answers
3
active
oldest
votes
$begingroup$
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.
answered yesterday
bi_scholarbi_scholar
41813
$endgroup$
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):
$$P(W|F)=1-P(Xleq62|F)=1-sum_k=0^62100 choose k0.5^k0.5^100-k$$
Replicating the analysis for the unfair case:
$$P(W|F')=1-P(X>62|F')=1-sum_k=63^100100 choose k0.7^k0.3^100-k$$
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.
answered yesterday
gunesgunes
7,3761316
$endgroup$
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.
answered yesterday
AcccumulationAcccumulation
1,68626
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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.
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.
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.
answered yesterday
bi_scholarbi_scholar
41813
$endgroup$
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.
answered yesterday
bi_scholarbi_scholar
41813
answered yesterday
bi_scholarbi_scholar
41813
answered yesterday
bi_scholarbi_scholar
41813
answered yesterday
bi_scholarbi_scholar
41813
41813
add a comment |
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):
$$P(W|F)=1-P(Xleq62|F)=1-sum_k=0^62100 choose k0.5^k0.5^100-k$$
Replicating the analysis for the unfair case:
$$P(W|F')=1-P(X>62|F')=1-sum_k=63^100100 choose k0.7^k0.3^100-k$$
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.
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):
$$P(W|F)=1-P(Xleq62|F)=1-sum_k=0^62100 choose k0.5^k0.5^100-k$$
Replicating the analysis for the unfair case:
$$P(W|F')=1-P(X>62|F')=1-sum_k=63^100100 choose k0.7^k0.3^100-k$$
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.
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):
$$P(W|F)=1-P(Xleq62|F)=1-sum_k=0^62100 choose k0.5^k0.5^100-k$$
Replicating the analysis for the unfair case:
$$P(W|F')=1-P(X>62|F')=1-sum_k=63^100100 choose k0.7^k0.3^100-k$$
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.
answered yesterday
gunesgunes
7,3761316
$endgroup$
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):
$$P(W|F)=1-P(Xleq62|F)=1-sum_k=0^62100 choose k0.5^k0.5^100-k$$
Replicating the analysis for the unfair case:
$$P(W|F')=1-P(X>62|F')=1-sum_k=63^100100 choose k0.7^k0.3^100-k$$
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.
answered yesterday
gunesgunes
7,3761316
answered yesterday
gunesgunes
7,3761316
answered yesterday
gunesgunes
7,3761316
answered yesterday
gunesgunes
7,3761316
7,3761316
add a comment |
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.
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.
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.
answered yesterday
AcccumulationAcccumulation
1,68626
$endgroup$
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.
answered yesterday
AcccumulationAcccumulation
1,68626
answered yesterday
AcccumulationAcccumulation
1,68626
answered yesterday
AcccumulationAcccumulation
1,68626
answered yesterday
AcccumulationAcccumulation
1,68626
1,68626
add a comment |
add a comment |
Benaya Trabelsi is a new contributor. Be nice, and check out our Code of Conduct.
Benaya Trabelsi is a new contributor. Be nice, and check out our Code of Conduct.
Benaya Trabelsi is a new contributor. Be nice, and check out our Code of Conduct.
Benaya Trabelsi is a new contributor. Be nice, and check out our Code of Conduct.
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-hypothesis-testing
$begingroup$
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