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Help With This Maths Problem

i have this hypothesis testing question: ive dont the first half but not sure on what to do for the second (iii) and (iv)

1) sweets called "scruffies" are sold in packs of 18
they come in a variety of colours, and market research shows red is the most populat. they are packed randomly and on average 25% are red

(i) find p of no more than 6 red
X-B(18,0.25)
p(X≤6) = 0.86101...

(ii) find p of exactly 4
p(X=4) = 0.21298...

(iii) 1 in 10 contain 19 sweets instead of 18. what is the p of exactly 4 now


(iv) to inscrease sales manufacturer sais increase production of red. (assume contail 18 per pack). pack contains 8 scruffies. test at the 5% significance level if this supports the claim

h0= p=0.25
h1= p≠ 0.25
assume h0 to be true, X-B(18,0.25)


How do you do (iii) and (iv). help would be great in the right direction for these
For part iii
Work out P(18 scruffies and X=4)
work out P(19 scruffies and X=4)
Add the probabilities together
Hope that helps 👍
Let me know if you need a bit more guidance
(edited 3 years ago)
Reply 2
Original post by jamiet0185
For part iii
Work out P(18 scruffies and X=4)
work out P(19 scruffies and X=4)
Add the probabilities together
Hope that helps 👍
Let me know if you need a bit more guidance

Cheers, that confirms along the lines I was working

I have another question in the same set i was working through which is attached as question 2 if you can help. would be much appriciated

this is what i've done so far
(i) because of the random sample being only 30 young people 40% of this would be around 12. this means around 12 young people of the 30 selected would fail the test

(ii) X-B(30,0.4)
P(X=12) = 0.1473
Find the probability of each of the young people passing from 1 - 30 and then find the most common amount of people

(iii) let p be probability of failing fitness test


Thanks
Original post by Harvey2019
Cheers, that confirms along the lines I was working

I have another question in the same set i was working through which is attached as question 2 if you can help. would be much appriciated

this is what i've done so far
(i) because of the random sample being only 30 young people 40% of this would be around 12. this means around 12 young people of the 30 selected would fail the test

(ii) X-B(30,0.4)
P(X=12) = 0.1473
Find the probability of each of the young people passing from 1 - 30 and then find the most common amount of people

(iii) let p be probability of failing fitness test


Thanks

Are you Ok with writing out the null and alternative hypothesis?
Then you will want to work out the probability of X being greater than or equal to 12. This is then compared with the significance level and your conclusion made.
See if you can get that far and then I'll help with the critical region bit if you need 👍
Reply 4
Original post by jamiet0185
Are you Ok with writing out the null and alternative hypothesis?
Then you will want to work out the probability of X being greater than or equal to 12. This is then compared with the significance level and your conclusion made.
See if you can get that far and then I'll help with the critical region bit if you need 👍

h0= 0.4

h1 0.4
Assume h0 to be true, X-B(20,0.4)

p(X≥12) = 1-P(X≤11)
= 0.0565...

think thats right
Original post by Harvey2019
h0= 0.4

h1 0.4
Assume h0 to be true, X-B(20,0.4)

p(X≥12) = 1-P(X≤11)
= 0.0565...

think thats right


Yep that's correct 👍
Reply 6
Original post by jamiet0185
Yep that's correct 👍

what do you do for the region. i know you have to state the critical region and the values that you will use to exclude the null hypothesis. what do i do here?
for the conclusion part 0.0565 (<0.05) which isnt a significant result
Original post by Harvey2019
what do you do for the region. i know you have to state the critical region and the values that you will use to exclude the null hypothesis. what do i do here?
for the conclusion part 0.0565 (<0.05) which isnt a significant result


Use the tables to find:
The highest value of a such that p (X smaller than or equal to a) < 0.025
This gives you the upper limit for the lower tail critical region
See if you can find the lower limit for the upper tail critical region
Then you're done 👍
Reply 8
Original post by jamiet0185
Use the tables to find:
The highest value of a such that p (X smaller than or equal to a) < 0.025
This gives you the upper limit for the lower tail critical region
See if you can find the lower limit for the upper tail critical region
Then you're done 👍

Cool, thanks for your help.
Might able to continue doing the other questions i'm working though now
Thank
Original post by Harvey2019
Cool, thanks for your help.
Might able to continue doing the other questions i'm working though now
Thank

No problem at all 😀
If you come across anything else you're struggling with I'm more than happy to help!
Ive done most of this question but not sure on the last part. I guess its similar to the last on here.

(i) expected containing vouchers = 6 packets
X-B(30,0.2)
p(X=6) = 0.17945

(ii) at least 1
p(X≥1) = 1-p(X≤0)
= 0.9987

(iii)
let p be probability of voucher
h0=0.2
h1≠0.2

(iv)
significance level 5%
assume h0 to be true X-B(12, 0.2)
Original post by Harvey2019
Ive done most of this question but not sure on the last part. I guess its similar to the last on here.

(i) expected containing vouchers = 6 packets
X-B(30,0.2)
p(X=6) = 0.17945

(ii) at least 1
p(X≥1) = 1-p(X≤0)
= 0.9987

(iii)
let p be probability of voucher
h0=0.2
h1≠0.2

(iv)
significance level 5%
assume h0 to be true X-B(12, 0.2)

For the lower tail to be empty , P(X=0)<0.025
Can you take it from there?
Original post by jamiet0185
For the lower tail to be empty , P(X=0)<0.025
Can you take it from there?

yep. got ya
Reply 13
@Harvey2019 for your first post, for part iv), the alternative hypotheses should be: p > 0.25,

as you are testing if the proportion of the red variety has increased.
(edited 3 years ago)
Original post by simon0
@Harvey2019 for your first post, for part iv), the alternative hypotheses should be: p > 0.25,

as you are testing if the proportion of the red variety has increased.

thanks

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