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How likely is sample A and sample B is from distribution C?
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$begingroup$
Let's say I have a sample A: [0,0,0,1]
and another sample B: [2,0,5,10,100,3,2,6]
I would like to know the probability that A and B are both picked from the same population C.
I tried applying a hypothesis test, but it gives me a p value of approx. 0.39 and I think it should be clear that it's very unlikely that both samples are from the same distribution.
probability hypothesis-testing distributions p-value multivariate-analysis
New contributor
$endgroup$
add a comment |
$begingroup$
Let's say I have a sample A: [0,0,0,1]
and another sample B: [2,0,5,10,100,3,2,6]
I would like to know the probability that A and B are both picked from the same population C.
I tried applying a hypothesis test, but it gives me a p value of approx. 0.39 and I think it should be clear that it's very unlikely that both samples are from the same distribution.
probability hypothesis-testing distributions p-value multivariate-analysis
New contributor
$endgroup$
$begingroup$
I'm guessing you used a pooled 2-sample t test, which is not a good choice here because sample sizes are small, 100 is a far outlier, and sample variances are hugely different. But your intuition that these data are not likely to have come from the same population is correct.
$endgroup$
– BruceET
6 hours ago
$begingroup$
As phrased the question (which contains a request for a probability), appears to be framed as a Bayesian problem. I expect that a Bayesian analysis is likely not the OP's intent, but if answers talk about hypothesis tests they should also discuss what question those answer (in place of what the question asks).
$endgroup$
– Glen_b♦
1 hour ago
add a comment |
$begingroup$
Let's say I have a sample A: [0,0,0,1]
and another sample B: [2,0,5,10,100,3,2,6]
I would like to know the probability that A and B are both picked from the same population C.
I tried applying a hypothesis test, but it gives me a p value of approx. 0.39 and I think it should be clear that it's very unlikely that both samples are from the same distribution.
probability hypothesis-testing distributions p-value multivariate-analysis
New contributor
$endgroup$
Let's say I have a sample A: [0,0,0,1]
and another sample B: [2,0,5,10,100,3,2,6]
I would like to know the probability that A and B are both picked from the same population C.
I tried applying a hypothesis test, but it gives me a p value of approx. 0.39 and I think it should be clear that it's very unlikely that both samples are from the same distribution.
probability hypothesis-testing distributions p-value multivariate-analysis
probability hypothesis-testing distributions p-value multivariate-analysis
New contributor
New contributor
New contributor
asked 8 hours ago
Franc WeserFranc Weser
113 bronze badges
113 bronze badges
New contributor
New contributor
$begingroup$
I'm guessing you used a pooled 2-sample t test, which is not a good choice here because sample sizes are small, 100 is a far outlier, and sample variances are hugely different. But your intuition that these data are not likely to have come from the same population is correct.
$endgroup$
– BruceET
6 hours ago
$begingroup$
As phrased the question (which contains a request for a probability), appears to be framed as a Bayesian problem. I expect that a Bayesian analysis is likely not the OP's intent, but if answers talk about hypothesis tests they should also discuss what question those answer (in place of what the question asks).
$endgroup$
– Glen_b♦
1 hour ago
add a comment |
$begingroup$
I'm guessing you used a pooled 2-sample t test, which is not a good choice here because sample sizes are small, 100 is a far outlier, and sample variances are hugely different. But your intuition that these data are not likely to have come from the same population is correct.
$endgroup$
– BruceET
6 hours ago
$begingroup$
As phrased the question (which contains a request for a probability), appears to be framed as a Bayesian problem. I expect that a Bayesian analysis is likely not the OP's intent, but if answers talk about hypothesis tests they should also discuss what question those answer (in place of what the question asks).
$endgroup$
– Glen_b♦
1 hour ago
$begingroup$
I'm guessing you used a pooled 2-sample t test, which is not a good choice here because sample sizes are small, 100 is a far outlier, and sample variances are hugely different. But your intuition that these data are not likely to have come from the same population is correct.
$endgroup$
– BruceET
6 hours ago
$begingroup$
I'm guessing you used a pooled 2-sample t test, which is not a good choice here because sample sizes are small, 100 is a far outlier, and sample variances are hugely different. But your intuition that these data are not likely to have come from the same population is correct.
$endgroup$
– BruceET
6 hours ago
$begingroup$
As phrased the question (which contains a request for a probability), appears to be framed as a Bayesian problem. I expect that a Bayesian analysis is likely not the OP's intent, but if answers talk about hypothesis tests they should also discuss what question those answer (in place of what the question asks).
$endgroup$
– Glen_b♦
1 hour ago
$begingroup$
As phrased the question (which contains a request for a probability), appears to be framed as a Bayesian problem. I expect that a Bayesian analysis is likely not the OP's intent, but if answers talk about hypothesis tests they should also discuss what question those answer (in place of what the question asks).
$endgroup$
– Glen_b♦
1 hour ago
add a comment |
1 Answer
1
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$begingroup$
You don't say what kind of hypothesis test you used.
Doing inference on such small samples as these is always
going to be difficult. However, a nonparametric Kolmogorov-Smirnov test (in R) does reject the null hypothesis that these
two samples were randomly sampled from the same population.
There is a warning message that (on account of the ties), the P-value is not exact, but 0.034 seems sufficiently smaller than 0.05 to say that we can reject at the 5% level.
x1 = c(0,0,0,1)
x2 = c(2,0,5,10,100,3,2,6)
ks.test(x1, x2)
Two-sample Kolmogorov-Smirnov test
data: x1 and x2
D = 0.875, p-value = 0.0337
alternative hypothesis: two-sided
Warning message:
In ks.test(x1, x2) : cannot compute exact p-value with ties
Similar data without ties gives a 'cleaner' test--rejecting the null hypothesis with no warning messages.
y1 = c(.01, .02, .03, .9)
y2 = c(2,0,5,10,100,3,2.1,6)
ks.test(y1, y2)
Two-sample Kolmogorov-Smirnov test
data: y1 and y2
D = 0.875, p-value = 0.0202
alternative hypothesis: two-sided
Another possible test is the two-sample Wilcoxon (rank sum test). Its distribution theory is also somewhat disturbed by ties, but it does find a significant difference between your two samples. Looking just at the P-value, we have:
wilcox.test(x1,x2)$p.val
[1] 0.02434338
Warning message:
In wilcox.test.default(x1, x2) :
cannot compute exact p-value with ties
$endgroup$
add a comment |
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$begingroup$
You don't say what kind of hypothesis test you used.
Doing inference on such small samples as these is always
going to be difficult. However, a nonparametric Kolmogorov-Smirnov test (in R) does reject the null hypothesis that these
two samples were randomly sampled from the same population.
There is a warning message that (on account of the ties), the P-value is not exact, but 0.034 seems sufficiently smaller than 0.05 to say that we can reject at the 5% level.
x1 = c(0,0,0,1)
x2 = c(2,0,5,10,100,3,2,6)
ks.test(x1, x2)
Two-sample Kolmogorov-Smirnov test
data: x1 and x2
D = 0.875, p-value = 0.0337
alternative hypothesis: two-sided
Warning message:
In ks.test(x1, x2) : cannot compute exact p-value with ties
Similar data without ties gives a 'cleaner' test--rejecting the null hypothesis with no warning messages.
y1 = c(.01, .02, .03, .9)
y2 = c(2,0,5,10,100,3,2.1,6)
ks.test(y1, y2)
Two-sample Kolmogorov-Smirnov test
data: y1 and y2
D = 0.875, p-value = 0.0202
alternative hypothesis: two-sided
Another possible test is the two-sample Wilcoxon (rank sum test). Its distribution theory is also somewhat disturbed by ties, but it does find a significant difference between your two samples. Looking just at the P-value, we have:
wilcox.test(x1,x2)$p.val
[1] 0.02434338
Warning message:
In wilcox.test.default(x1, x2) :
cannot compute exact p-value with ties
$endgroup$
add a comment |
$begingroup$
You don't say what kind of hypothesis test you used.
Doing inference on such small samples as these is always
going to be difficult. However, a nonparametric Kolmogorov-Smirnov test (in R) does reject the null hypothesis that these
two samples were randomly sampled from the same population.
There is a warning message that (on account of the ties), the P-value is not exact, but 0.034 seems sufficiently smaller than 0.05 to say that we can reject at the 5% level.
x1 = c(0,0,0,1)
x2 = c(2,0,5,10,100,3,2,6)
ks.test(x1, x2)
Two-sample Kolmogorov-Smirnov test
data: x1 and x2
D = 0.875, p-value = 0.0337
alternative hypothesis: two-sided
Warning message:
In ks.test(x1, x2) : cannot compute exact p-value with ties
Similar data without ties gives a 'cleaner' test--rejecting the null hypothesis with no warning messages.
y1 = c(.01, .02, .03, .9)
y2 = c(2,0,5,10,100,3,2.1,6)
ks.test(y1, y2)
Two-sample Kolmogorov-Smirnov test
data: y1 and y2
D = 0.875, p-value = 0.0202
alternative hypothesis: two-sided
Another possible test is the two-sample Wilcoxon (rank sum test). Its distribution theory is also somewhat disturbed by ties, but it does find a significant difference between your two samples. Looking just at the P-value, we have:
wilcox.test(x1,x2)$p.val
[1] 0.02434338
Warning message:
In wilcox.test.default(x1, x2) :
cannot compute exact p-value with ties
$endgroup$
add a comment |
$begingroup$
You don't say what kind of hypothesis test you used.
Doing inference on such small samples as these is always
going to be difficult. However, a nonparametric Kolmogorov-Smirnov test (in R) does reject the null hypothesis that these
two samples were randomly sampled from the same population.
There is a warning message that (on account of the ties), the P-value is not exact, but 0.034 seems sufficiently smaller than 0.05 to say that we can reject at the 5% level.
x1 = c(0,0,0,1)
x2 = c(2,0,5,10,100,3,2,6)
ks.test(x1, x2)
Two-sample Kolmogorov-Smirnov test
data: x1 and x2
D = 0.875, p-value = 0.0337
alternative hypothesis: two-sided
Warning message:
In ks.test(x1, x2) : cannot compute exact p-value with ties
Similar data without ties gives a 'cleaner' test--rejecting the null hypothesis with no warning messages.
y1 = c(.01, .02, .03, .9)
y2 = c(2,0,5,10,100,3,2.1,6)
ks.test(y1, y2)
Two-sample Kolmogorov-Smirnov test
data: y1 and y2
D = 0.875, p-value = 0.0202
alternative hypothesis: two-sided
Another possible test is the two-sample Wilcoxon (rank sum test). Its distribution theory is also somewhat disturbed by ties, but it does find a significant difference between your two samples. Looking just at the P-value, we have:
wilcox.test(x1,x2)$p.val
[1] 0.02434338
Warning message:
In wilcox.test.default(x1, x2) :
cannot compute exact p-value with ties
$endgroup$
You don't say what kind of hypothesis test you used.
Doing inference on such small samples as these is always
going to be difficult. However, a nonparametric Kolmogorov-Smirnov test (in R) does reject the null hypothesis that these
two samples were randomly sampled from the same population.
There is a warning message that (on account of the ties), the P-value is not exact, but 0.034 seems sufficiently smaller than 0.05 to say that we can reject at the 5% level.
x1 = c(0,0,0,1)
x2 = c(2,0,5,10,100,3,2,6)
ks.test(x1, x2)
Two-sample Kolmogorov-Smirnov test
data: x1 and x2
D = 0.875, p-value = 0.0337
alternative hypothesis: two-sided
Warning message:
In ks.test(x1, x2) : cannot compute exact p-value with ties
Similar data without ties gives a 'cleaner' test--rejecting the null hypothesis with no warning messages.
y1 = c(.01, .02, .03, .9)
y2 = c(2,0,5,10,100,3,2.1,6)
ks.test(y1, y2)
Two-sample Kolmogorov-Smirnov test
data: y1 and y2
D = 0.875, p-value = 0.0202
alternative hypothesis: two-sided
Another possible test is the two-sample Wilcoxon (rank sum test). Its distribution theory is also somewhat disturbed by ties, but it does find a significant difference between your two samples. Looking just at the P-value, we have:
wilcox.test(x1,x2)$p.val
[1] 0.02434338
Warning message:
In wilcox.test.default(x1, x2) :
cannot compute exact p-value with ties
edited 7 hours ago
answered 7 hours ago
BruceETBruceET
9,5581 gold badge8 silver badges24 bronze badges
9,5581 gold badge8 silver badges24 bronze badges
add a comment |
add a comment |
Franc Weser is a new contributor. Be nice, and check out our Code of Conduct.
Franc Weser is a new contributor. Be nice, and check out our Code of Conduct.
Franc Weser is a new contributor. Be nice, and check out our Code of Conduct.
Franc Weser is a new contributor. Be nice, and check out our Code of Conduct.
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$begingroup$
I'm guessing you used a pooled 2-sample t test, which is not a good choice here because sample sizes are small, 100 is a far outlier, and sample variances are hugely different. But your intuition that these data are not likely to have come from the same population is correct.
$endgroup$
– BruceET
6 hours ago
$begingroup$
As phrased the question (which contains a request for a probability), appears to be framed as a Bayesian problem. I expect that a Bayesian analysis is likely not the OP's intent, but if answers talk about hypothesis tests they should also discuss what question those answer (in place of what the question asks).
$endgroup$
– Glen_b♦
1 hour ago