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Hypothesis testing using binomial distributions
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TSR360
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#1
Question: A dice is rolled 5 times and 3 sixes are observed. Test at the 5% significance level if the dice is biased.H_0: p = 1/6 (dice is unbiased)H_1: p > 1/6 (dice is biased towards 6)
Signifiance level: 5% for X ~ B(5, 1/6) where X = #. of sixes obtainedP(X >= 3) = 1 - [P(X = 0) + P(X = 1) + P(X = 2)] = 0.0350.035 < 0.05 so H_0 is rejected in favour of H_1.
According to the textbook, this question is actually a two-tailed test and H_1 is p ≠ 1/6: http://prntscr.com/u0bq4r
If that's the case, shouldn't the w/o include the probabilities for each tail? I'm so confused.....
Signifiance level: 5% for X ~ B(5, 1/6) where X = #. of sixes obtainedP(X >= 3) = 1 - [P(X = 0) + P(X = 1) + P(X = 2)] = 0.0350.035 < 0.05 so H_0 is rejected in favour of H_1.
According to the textbook, this question is actually a two-tailed test and H_1 is p ≠ 1/6: http://prntscr.com/u0bq4r
If that's the case, shouldn't the w/o include the probabilities for each tail? I'm so confused.....
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RDKGames
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#2
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#2
(Original post by TSR360)
Question: A dice is rolled 5 times and 3 sixes are observed. Test at the 5% significance level if the dice is biased.H_0: p = 1/6 (dice is unbiased)H_1: p > 1/6 (dice is biased towards 6)
Signifiance level: 5% for X ~ B(5, 1/6) where X = #. of sixes obtainedP(X >= 3) = 1 - [P(X = 0) + P(X = 1) + P(X = 2)] = 0.0350.035 < 0.05 so H_0 is rejected in favour of H_1.
According to the textbook, this question is actually a two-tailed test and H_1 is p ≠ 1/6: http://prntscr.com/u0bq4r
If that's the case, shouldn't the w/o include the probabilities for each tail? I'm so confused.....
Question: A dice is rolled 5 times and 3 sixes are observed. Test at the 5% significance level if the dice is biased.H_0: p = 1/6 (dice is unbiased)H_1: p > 1/6 (dice is biased towards 6)
Signifiance level: 5% for X ~ B(5, 1/6) where X = #. of sixes obtainedP(X >= 3) = 1 - [P(X = 0) + P(X = 1) + P(X = 2)] = 0.0350.035 < 0.05 so H_0 is rejected in favour of H_1.
According to the textbook, this question is actually a two-tailed test and H_1 is p ≠ 1/6: http://prntscr.com/u0bq4r
If that's the case, shouldn't the w/o include the probabilities for each tail? I'm so confused.....
You only need to care about both tails at once when you are finding the critical region, or finding the actual sig level, etc..
Last edited by RDKGames; 1 year ago
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TSR360
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#3
(Original post by RDKGames)
But 3 is on the upper-tail of the distribution, so you need to test in the upper-tail.
You only need to care about both tails at once when you are finding the critical region, or finding the actual sig level, etc..
But 3 is on the upper-tail of the distribution, so you need to test in the upper-tail.
You only need to care about both tails at once when you are finding the critical region, or finding the actual sig level, etc..
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#4
(Original post by TSR360)
That makes sense, but why isn't H_1: p < 1/6?
That makes sense, but why isn't H_1: p < 1/6?
It's asking you to test whether a die is biased. It does not specify whether this is in favour or against rolling a six.
Hence it's a two tailed test
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#5
(Original post by RDKGames)
You need to interpret the context carefully.
It's asking you to test whether a die is biased. It does not specify whether this is in favour or against rolling a six.
Hence it's a two tailed test
You need to interpret the context carefully.
It's asking you to test whether a die is biased. It does not specify whether this is in favour or against rolling a six.
Hence it's a two tailed test
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#6
(Original post by TSR360)
But isn't it implied or else why would they mention 3 rolls being a six? What if it was four 4s instead?
But isn't it implied or else why would they mention 3 rolls being a six? What if it was four 4s instead?
If it was four 4s then it would still be a two tailed test. You would still test in the upper tail because 4 is greater than the mean of X ~ B(4,1/6) [which implies it is in the upper-tail of the distribution!]
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#7
(Original post by RDKGames)
No it is not implied.
If it was four 4s then it would still be a two tailed test. You would still test in the upper tail because 4 is greater than the mean of X ~ B(4,1/6) [which implies it is in the upper-tail of the distribution!]
No it is not implied.
If it was four 4s then it would still be a two tailed test. You would still test in the upper tail because 4 is greater than the mean of X ~ B(4,1/6) [which implies it is in the upper-tail of the distribution!]
Also, if you're only supposed to care about both tails for critical regions/sig levels, how would you answer this question http://prntscr.com/u0e66n ?
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#8
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#8
(Original post by TSR360)
That makes sense.
Also, if you're only supposed to care about both tails for critical regions/sig levels, how would you answer this question http://prntscr.com/u0e66n ?
That makes sense.
Also, if you're only supposed to care about both tails for critical regions/sig levels, how would you answer this question http://prntscr.com/u0e66n ?
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#9
(Original post by RDKGames)
Test in the upper-tail. See whether under H0 which is p=0.5, we have P(X>9) < 0.025 where X ~ B(10,0.5)
Test in the upper-tail. See whether under H0 which is p=0.5, we have P(X>9) < 0.025 where X ~ B(10,0.5)
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#10
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#10
(Original post by TSR360)
The textbook answer is different where both tail ends are taken into account: http://prntscr.com/u0pvb1
The textbook answer is different where both tail ends are taken into account: http://prntscr.com/u0pvb1
Together they are less than the significance level 0.05.
BUT this is exactly the same as simply looking at P(X>9) and seeing that it is less than 0.025. This shortcut is due to the symmetry of the binomial distribution; P(X>9) = P(X<1).
Last edited by RDKGames; 1 year ago
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TSR360
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#11
(Original post by RDKGames)
Unbiased claim means either in favour of government or not. The textbook took the probability P(X>9) and added it to P(X<1) which is the reflection in the other tail.
Together they are less than the significance level 0.05.
BUT this is exactly the same as simply looking at P(X>9) and seeing that it is less than 0.025. This shortcut is due to the symmetry of the binomial distribution; P(X>9) = P(X<1).
Unbiased claim means either in favour of government or not. The textbook took the probability P(X>9) and added it to P(X<1) which is the reflection in the other tail.
Together they are less than the significance level 0.05.
BUT this is exactly the same as simply looking at P(X>9) and seeing that it is less than 0.025. This shortcut is due to the symmetry of the binomial distribution; P(X>9) = P(X<1).
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#12
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#12
(Original post by TSR360)
If I wanted to answer my original question using the method of adding both tail end probabilities, how would I do that?
If I wanted to answer my original question using the method of adding both tail end probabilities, how would I do that?
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