Independent coin tosses, Hypothesis testing
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I have 10 independent coin tosses with probability p for heads.
I'm testing the null hypothesis
$H_0: p=1/2$
Versus the alternative
$H_1 : p > 1/2 $
$W$ is the number of heads in the 10 coin tosses.
We accept $H_0$ if $W leq c$ else reject $H_0$
Decide c so that $alpha = 0.05$
I started off saying $W$ is $binomial(10,½)$ so the probability of type I errors is:
$P(Type I Error)=P(Reject H_0 mid H_0) = P(RejectH_0 mid p=½) = P(W >cmid p=½)=sumlimits_{k=c+1}^{infty} P(W=k)$
$And because of W being binomial(10,½)
$=sumlimits_{k=c+1}^{10} 0.5^k 0.5^{10-k}={(10-c)}/1024 $
And so because of us wanting $alpha =0.05$ we choose a c so that
$(10-c)/1024 leq alpha $
So I get
$c=10-alpha*1024=-41.2$
Which seems wrong to me, where did i go wrong in my thought process? Because if c=-41.2 then i would never be able to say the coin is fair and accept the null hypothesis
probability statistics hypothesis-testing
asked Nov 22 at 20:33
Winther
227
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up vote
0
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I have 10 independent coin tosses with probability p for heads.
I'm testing the null hypothesis
$H_0: p=1/2$
Versus the alternative
$H_1 : p > 1/2 $
$W$ is the number of heads in the 10 coin tosses.
We accept $H_0$ if $W leq c$ else reject $H_0$
Decide c so that $alpha = 0.05$
I started off saying $W$ is $binomial(10,½)$ so the probability of type I errors is:
$P(Type I Error)=P(Reject H_0 mid H_0) = P(RejectH_0 mid p=½) = P(W >cmid p=½)=sumlimits_{k=c+1}^{infty} P(W=k)$
$And because of W being binomial(10,½)
$=sumlimits_{k=c+1}^{10} 0.5^k 0.5^{10-k}={(10-c)}/1024 $
And so because of us wanting $alpha =0.05$ we choose a c so that
$(10-c)/1024 leq alpha $
So I get
$c=10-alpha*1024=-41.2$
Which seems wrong to me, where did i go wrong in my thought process? Because if c=-41.2 then i would never be able to say the coin is fair and accept the null hypothesis
probability statistics hypothesis-testing
asked Nov 22 at 20:33
Winther
227
2
Check probabilities for binomial distrbution.
– NCh
Nov 22 at 23:19
Using the discrete distribution BINOM(10,.5) you can't get exactly $alpha = 0.05.$ In R: $P(X ge 9|p=1/2) approx 0.011,$ computed assum(dbinom(9:10, 10, 1/2))which returns 0.01074219 and $P(X ge 8|p = 1/2) approx 0.055)$ with codesum(dbinom(8{10, 10, 1/2)), which returns 0.0546875. // Exact normal computations on a calculator are feasible, You can try to sweep the discreteness under the rug by using a normal approximation, but that doesn't change the truth.
– BruceET
Nov 23 at 0:42
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I have 10 independent coin tosses with probability p for heads.
I'm testing the null hypothesis
$H_0: p=1/2$
Versus the alternative
$H_1 : p > 1/2 $
$W$ is the number of heads in the 10 coin tosses.
We accept $H_0$ if $W leq c$ else reject $H_0$
Decide c so that $alpha = 0.05$
I started off saying $W$ is $binomial(10,½)$ so the probability of type I errors is:
$P(Type I Error)=P(Reject H_0 mid H_0) = P(RejectH_0 mid p=½) = P(W >cmid p=½)=sumlimits_{k=c+1}^{infty} P(W=k)$
$And because of W being binomial(10,½)
$=sumlimits_{k=c+1}^{10} 0.5^k 0.5^{10-k}={(10-c)}/1024 $
And so because of us wanting $alpha =0.05$ we choose a c so that
$(10-c)/1024 leq alpha $
So I get
$c=10-alpha*1024=-41.2$
Which seems wrong to me, where did i go wrong in my thought process? Because if c=-41.2 then i would never be able to say the coin is fair and accept the null hypothesis
probability statistics hypothesis-testing
asked Nov 22 at 20:33
Winther
227
I have 10 independent coin tosses with probability p for heads.
I'm testing the null hypothesis
$H_0: p=1/2$
Versus the alternative
$H_1 : p > 1/2 $
$W$ is the number of heads in the 10 coin tosses.
We accept $H_0$ if $W leq c$ else reject $H_0$
Decide c so that $alpha = 0.05$
I started off saying $W$ is $binomial(10,½)$ so the probability of type I errors is:
$P(Type I Error)=P(Reject H_0 mid H_0) = P(RejectH_0 mid p=½) = P(W >cmid p=½)=sumlimits_{k=c+1}^{infty} P(W=k)$
$And because of W being binomial(10,½)
$=sumlimits_{k=c+1}^{10} 0.5^k 0.5^{10-k}={(10-c)}/1024 $
And so because of us wanting $alpha =0.05$ we choose a c so that
$(10-c)/1024 leq alpha $
So I get
$c=10-alpha*1024=-41.2$
Which seems wrong to me, where did i go wrong in my thought process? Because if c=-41.2 then i would never be able to say the coin is fair and accept the null hypothesis
probability statistics hypothesis-testing
probability statistics hypothesis-testing
asked Nov 22 at 20:33
Winther
227
asked Nov 22 at 20:33
Winther
227
edited Nov 22 at 20:39
asked Nov 22 at 20:33
Winther
227
asked Nov 22 at 20:33
Winther
227
asked Nov 22 at 20:33
Winther
227
227
2
Check probabilities for binomial distrbution.
– NCh
Nov 22 at 23:19
Using the discrete distribution BINOM(10,.5) you can't get exactly $alpha = 0.05.$ In R: $P(X ge 9|p=1/2) approx 0.011,$ computed assum(dbinom(9:10, 10, 1/2))which returns 0.01074219 and $P(X ge 8|p = 1/2) approx 0.055)$ with codesum(dbinom(8{10, 10, 1/2)), which returns 0.0546875. // Exact normal computations on a calculator are feasible, You can try to sweep the discreteness under the rug by using a normal approximation, but that doesn't change the truth.
– BruceET
Nov 23 at 0:42
add a comment |
2
Check probabilities for binomial distrbution.
– NCh
Nov 22 at 23:19
Using the discrete distribution BINOM(10,.5) you can't get exactly $alpha = 0.05.$ In R: $P(X ge 9|p=1/2) approx 0.011,$ computed assum(dbinom(9:10, 10, 1/2))which returns 0.01074219 and $P(X ge 8|p = 1/2) approx 0.055)$ with codesum(dbinom(8{10, 10, 1/2)), which returns 0.0546875. // Exact normal computations on a calculator are feasible, You can try to sweep the discreteness under the rug by using a normal approximation, but that doesn't change the truth.
– BruceET
Nov 23 at 0:42
2
2
Check probabilities for binomial distrbution.
– NCh
Nov 22 at 23:19
Check probabilities for binomial distrbution.
– NCh
Nov 22 at 23:19
Using the discrete distribution BINOM(10,.5) you can't get exactly $alpha = 0.05.$ In R: $P(X ge 9|p=1/2) approx 0.011,$ computed as
sum(dbinom(9:10, 10, 1/2)) which returns 0.01074219 and $P(X ge 8|p = 1/2) approx 0.055)$ with code sum(dbinom(8{10, 10, 1/2)), which returns 0.0546875. // Exact normal computations on a calculator are feasible, You can try to sweep the discreteness under the rug by using a normal approximation, but that doesn't change the truth.– BruceET
Nov 23 at 0:42
Using the discrete distribution BINOM(10,.5) you can't get exactly $alpha = 0.05.$ In R: $P(X ge 9|p=1/2) approx 0.011,$ computed as
sum(dbinom(9:10, 10, 1/2)) which returns 0.01074219 and $P(X ge 8|p = 1/2) approx 0.055)$ with code sum(dbinom(8{10, 10, 1/2)), which returns 0.0546875. // Exact normal computations on a calculator are feasible, You can try to sweep the discreteness under the rug by using a normal approximation, but that doesn't change the truth.– BruceET
Nov 23 at 0:42
add a comment |
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2
Check probabilities for binomial distrbution.
– NCh
Nov 22 at 23:19
Using the discrete distribution BINOM(10,.5) you can't get exactly $alpha = 0.05.$ In R: $P(X ge 9|p=1/2) approx 0.011,$ computed as
sum(dbinom(9:10, 10, 1/2))which returns 0.01074219 and $P(X ge 8|p = 1/2) approx 0.055)$ with codesum(dbinom(8{10, 10, 1/2)), which returns 0.0546875. // Exact normal computations on a calculator are feasible, You can try to sweep the discreteness under the rug by using a normal approximation, but that doesn't change the truth.– BruceET
Nov 23 at 0:42