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# S2 Hypothesis Testing (Critical Values)

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1. Finding an upper critical value
Test 5% level
Assume X~B(20,0.25) is true
P(X>=r)=<0.05
r=7 -> 1-0.8982=0.1018
r=8 -> 1-0.9591=0.0409 =<0.05

I forgot S2.
Why is 9 the CV?
2. (Original post by AlmostNotable)
Finding an upper critical value
Test 5% level
Assume X~B(20,0.25) is true
P(X>=r)=<0.05
r=7 -> 1-0.8982=0.1018
r=8 -> 1-0.9591=0.0409 =<0.05

I forgot S2.
Why is 9 the CV?
Because it is the first value or r s.t P(X>=r) < 0.05 is satisfied.
3. (Original post by Zacken)
Because it is the first value or r s.t P(X>=r) < 0.05 is satisfied.
When r=8, P(>=r)=0.0409<0.05
4. (Original post by AlmostNotable)
When r=8, P(>=r)=0.0409<0.05
Remember that the tables give you values of P(X=<x), so if you're looking at P(X=<r) then 1 - P(X=<r) = 1 - P(X>r)*** = (1 - P(X >= r+1) as it is a discrete distribution, so if it's greater than r then it's greater or equal to r+1 as there are no values between r and r+1.

*** Remember that P(X=<r) + P(X>r), where you can read that as probability of something being less than or equal to r, or greater than r, covers all real numbers so the sum = 1.
5. (Original post by AlmostNotable)
When r=8, P(>=r)=0.0409<0.05

P(X >= r) <= 0.05 is what you use to determine your c.v.

But P(X >= r) = 1 - P(X <= r - 1)

SO you want

1- P(X <= r-1) <= 0.05

Remember that binomial tables are cumulative and the distribution is discrete.

So r-1 = 8 is the largest value that satisfies the inequality, so r = 9.
6. (Original post by Zacken)

P(X >= r) <= 0.05 is what you use to determine your c.v.

But P(X >= r) = 1 - P(X <= r - 1)

SO you want

1- P(X <= r-1) <= 0.05

Remember that binomial tables are cumulative and the distribution is discrete.

So r-1 = 8 is the largest value that satisfies the inequality, so r = 9.
Oh yeah its discrete.

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