Hypothesis testing using binomial distributions

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] = 0.0350.035 < 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.....

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..

That makes sense, but why isn't H_1: p < 1/6?

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

But isn't it implied or else why would they mention 3 rolls being a six? What if it was four 4s instead?

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!]

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 ?

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)

The textbook answer is different where both tail ends are taken into account: http://prntscr.com/u0pvb1

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).

If I wanted to answer my original question using the method of adding both tail end probabilities, how would I do that?