two types of coins, decide which type it is based on 100 flips Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)how 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
Strange behaviour of Check
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two types of coins, decide which type it is based on 100 flips
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)how 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 Apr 11 at 10:16
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 Apr 11 at 10:16
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
Apr 11 at 10:22
$begingroup$
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
Apr 11 at 10:36
$begingroup$
"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
Apr 11 at 12:40
$begingroup$
"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
Apr 11 at 12:44
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 Apr 11 at 10:16
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 Apr 11 at 10:16
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 Apr 11 at 10:16
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 Apr 11 at 10:16
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 Apr 11 at 10:16
Benaya TrabelsiBenaya Trabelsi
111
asked Apr 11 at 10:16
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
Apr 11 at 10:22
$begingroup$
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
Apr 11 at 10:36
$begingroup$
"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
Apr 11 at 12:40
$begingroup$
"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
Apr 11 at 12:44
add a comment |
$begingroup$
Can you lay out your calculations?
$endgroup$
– bi_scholar
Apr 11 at 10:22
$begingroup$
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
Apr 11 at 10:36
$begingroup$
"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
Apr 11 at 12:40
$begingroup$
"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
Apr 11 at 12:44
$begingroup$
Can you lay out your calculations?
$endgroup$
– bi_scholar
Apr 11 at 10:22
$begingroup$
Can you lay out your calculations?
$endgroup$
– bi_scholar
Apr 11 at 10:22
$begingroup$
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
Apr 11 at 10:36
$begingroup$
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
Apr 11 at 10:36
$begingroup$
"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
Apr 11 at 12:40
$begingroup$
"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
Apr 11 at 12:40
$begingroup$
"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
Apr 11 at 12:44
$begingroup$
"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
Apr 11 at 12:44
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 Apr 11 at 11:13
bi_scholarbi_scholar
42813
$endgroup$
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 Apr 11 at 11:15
gunesgunes
7,5161316
$endgroup$
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 Apr 11 at 15:36
AcccumulationAcccumulation
1,68626
$endgroup$
add a comment |
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3 Answers
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active
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3 Answers
3
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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 Apr 11 at 11:13
bi_scholarbi_scholar
42813
$endgroup$
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 Apr 11 at 11:13
bi_scholarbi_scholar
42813
$endgroup$
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 Apr 11 at 11:13
bi_scholarbi_scholar
42813
$endgroup$
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 Apr 11 at 11:13
bi_scholarbi_scholar
42813
answered Apr 11 at 11:13
bi_scholarbi_scholar
42813
answered Apr 11 at 11:13
bi_scholarbi_scholar
42813
answered Apr 11 at 11:13
bi_scholarbi_scholar
42813
42813
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 Apr 11 at 11:15
gunesgunes
7,5161316
$endgroup$
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 Apr 11 at 11:15
gunesgunes
7,5161316
$endgroup$
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 Apr 11 at 11:15
gunesgunes
7,5161316
$endgroup$
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 Apr 11 at 11:15
gunesgunes
7,5161316
answered Apr 11 at 11:15
gunesgunes
7,5161316
answered Apr 11 at 11:15
gunesgunes
7,5161316
answered Apr 11 at 11:15
gunesgunes
7,5161316
7,5161316
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 Apr 11 at 15:36
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 Apr 11 at 15:36
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 Apr 11 at 15:36
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 Apr 11 at 15:36
AcccumulationAcccumulation
1,68626
answered Apr 11 at 15:36
AcccumulationAcccumulation
1,68626
answered Apr 11 at 15:36
AcccumulationAcccumulation
1,68626
answered Apr 11 at 15:36
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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$begingroup$
Can you lay out your calculations?
$endgroup$
– bi_scholar
Apr 11 at 10:22
$begingroup$
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
Apr 11 at 10:36
$begingroup$
"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
Apr 11 at 12:40
$begingroup$
"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
Apr 11 at 12:44