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Edexcel S2: Hypothesis testing



What have you tried so far? :h:
Original post by SeanFM
What have you tried so far? :h:


All the stuff my teacher taught me:

H0: p = 0.3
H1: p 0.3

Under H0:
X ~ B(30, 0.3)

And then look for values from tables that are less than 0.05% and more than 99.95%

After that I'm stuck ( I may be missing something or be looking in the wrong direction I'm not sure)
Original post by AlphaArgonian
All the stuff my teacher taught me:

H0: p = 0.3
H1: p 0.3

Under H0:
X ~ B(30, 0.3)

And then look for values from tables that are less than 0.05% and more than 99.95%

After that I'm stuck ( I may be missing something or be looking in the wrong direction I'm not sure)


Correct up to that point. So what values did you find?
Original post by SeanFM
Correct up to that point. So what values did you find?


P(X≤2) = 0.0021

P(X≤18) = 0.9998

Now what? This is what confuses me :frown:
Original post by AlphaArgonian
P(X≤2) = 0.0021

P(X≤18) = 0.9998

Now what? This is what confuses me :frown:


Remember that you want it as close to 0.01, so as close to 0.005 on either side.

Since P(X≤2) is less than 0.005, you have to look at the other side of 0.005 (which'll be P(X≤3) and then choose whichever one is closest to 0.005.

The other tail is slightly trickier as you're looking for P(X≥c) to be as close to 0.005 as possible, but you're reading P(X≤x) off the table. If P(X≤18) = 0.9998, using just that, can you get an inequality of the form P(X≥c) and a corresponding probability?
Original post by SeanFM
Remember that you want it as close to 0.01, so as close to 0.005 on either side.

Since P(X≤2) is less than 0.005, you have to look at the other side of 0.005 (which'll be P(X≤3) and then choose whichever one is closest to 0.005.

The other tail is slightly trickier as you're looking for P(X≥c) to be as close to 0.005 as possible, but you're reading P(X≤x) off the table. If P(X≤18) = 0.9998, using just that, can you get an inequality of the form P(X≥c) and a corresponding probability?


P(X≥17)? Regardless of the answer, this makes a lot more sense now, thank you very much!!! I have a chance for an A now cheers!
Original post by AlphaArgonian
P(X≥17)? Regardless of the answer, this makes a lot more sense now, thank you very much!!! I have a chance for an A now cheers!


Not quite, this bit is important as well to get the right critical region.

What is 1 - P(X≤18)? (Not the numerical value of it in this case, but what it represents).

Eg if I have 5 cubes in a bag, either blue or green, and the probability of getting blue is x, then 1-x represents...
Original post by SeanFM
Not quite, this bit is important as well to get the right critical region.

What is 1 - P(X≤18)? (Not the numerical value of it in this case, but what it represents).

Eg if I have 5 cubes in a bag, either blue or green, and the probability of getting blue is x, then 1-x represents...


Everything apart from the observed value like P(A)'
Original post by AlphaArgonian
Everything apart from the observed value like P(A)'


Correct.

Using that it's a discrete distribution, let's look at the question and choose a binomial distribution with n = 30, then what's the complement of (X less than or equal to 18)?

(i.e 1-P(X less than or equal to 18) is P(...?...))
Original post by SeanFM
Correct.

Using that it's a discrete distribution, let's look at the question and choose a binomial distribution with n = 30, then what's the complement of (X less than or equal to 18)?

(i.e 1-P(X less than or equal to 18) is P(...?...))

Greater than or equal to 19?
Original post by AlphaArgonian
Greater than or equal to 19?


Correct :h: so you know that P(X> greater than or equal to 19) = .... so you need to look at P(X > greater than or equal to...) and compare them and see which is closer to ....
Original post by SeanFM
Correct :h: so you know that P(X> greater than or equal to 19) = .... so you need to look at P(X > greater than or equal to...) and compare them and see which is closer to ....


Ahhhh ok thanks a bunch!
Hey, I know it's not the same topic but for continuous uniform distributions, does anyone known if we need to be able to derive the formula for var(x) or do we just need to know the formula and apply it? Thank you :smile:
Reply 14
Original post by Melanierobertson
Hey, I know it's not the same topic but for continuous uniform distributions, does anyone known if we need to be able to derive the formula for var(x) or do we just need to know the formula and apply it? Thank you :smile:


Just know the formula and apply it unless explicitly asked to prove it, in which case it's just a standard integration.
Original post by Zacken
Just know the formula and apply it unless explicitly asked to prove it, in which case it's just a standard integration.


Thankyou :smile:
Reply 16
Original post by Melanierobertson
Thankyou :smile:


You're welcome.

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