# Help With This Maths Problem

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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

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

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#2

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

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

Last edited by jamiet0185; 5 months ago

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(Original post by

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

**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

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

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#4

(Original post by

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

**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

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 👍

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(Original post by

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 👍

**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 👍

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

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#6

(Original post by

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

**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

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(Original post by

Yep that's correct 👍

**jamiet0185**)Yep that's correct 👍

for the conclusion part 0.0565 (<0.05) which isnt a significant result

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#8

(Original post by

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

**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

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 👍

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(Original post by

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 👍

**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 👍

Might able to continue doing the other questions i'm working though now

Thank

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#10

(Original post by

Cool, thanks for your help.

Might able to continue doing the other questions i'm working though now

Thank

**Harvey2019**)Cool, thanks for your help.

Might able to continue doing the other questions i'm working though now

Thank

If you come across anything else you're struggling with I'm more than happy to help!

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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)

(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)

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#12

(Original post by

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)

**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)

Can you take it from there?

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(Original post by

For the lower tail to be empty , P(X=0)<0.025

Can you take it from there?

**jamiet0185**)For the lower tail to be empty , P(X=0)<0.025

Can you take it from there?

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#14

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.

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

Last edited by simon0; 5 months ago

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(Original post by

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.

**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.

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