Statistics 72 - How was it?

[Zephyro]

Easiest paper I ever saw till date tbh. Expecting 49 or 50/50. How about everyone else?

 

[kirashinagami]

I cannot do the 7 marks, pity me

 

[Zephyro]

last question? Took me like 3 pages to solve it but got it in the end.

 

[redapple20]

I couldnt prove it either got near though

 

[Zephyro]

how much time do I have to wait before writing the method for it (q7 last part)? I guess mine was right but cant be 100% sure, so I want to check with:

if you got the answer you were suppose to prove its right. No point posting the solution as people know whether they got it or not by checking the final answer against the prove.

 

[Zephyro]

well yeah but its the first time a question of this type came so it would be good to know that the method was right...

you found the two rej regions and found the prob after that, right? It's actually come in previous papers but in different words so people didnt make out the question.

 

[Saiyan]

I found the rejection region of no. 7 but still could not prove the probability. Was the question about one-tail or two-tail test? I don't know why but my answer kept on coming 0.9. The others went well though. I am disappointed with 7!

 

[kirashinagami]

And the bathing question is also funny lol

 

[qassim]

It was pretty easy, last question was bleh though . hahaha lol, at first i thought they meant they bathed together xD

 

[redapple20]

Variance had to b changed in 7

 

[Echelon94]

you found the two rej regions and found the prob after that, right? It's actually come in previous papers but in different words so people didnt make out the question.

I found the rejection region of no. 7 but still could not prove the probability. I don't know why but my answer kept on coming 0.9.

CV z = ±1.645, old µ = 2, σ^2 = 1.85, n=200. Using these values, find the Acceptance Region. It'll come out to be 1.842<x̄<2.158.​

Type 2 means accepting H0 when H0 is false. So that means, now, using the new µ = 2.12, calculate P(1.842<x̄<2.158).​

P(Type 2) = P(1.842<x̄<2.158)​

= Φ(0.395) - Φ(-2.891)​

= 0.6536 - 0.001920​

= 0.6517 ≈ 0.652​

 

[Zephyro]

CV z = ±1.645, old µ = 2, σ^2 = 1.85, n=200. Using these values, find the Acceptance Region. It'll come out to be 1.842<x̄<2.158.​

Type 2 means accepting H0 when H0 is false. So that means, now, using the new µ = 2.12, calculate P(1.842<x̄<2.158).​

P(Type 2) = P(1.842<x̄<2.158)​

= Φ(0.395) - Φ(-2.891)​

= 0.6536 - 0.001920​

= 0.6517 ≈ 0.652​

Did the same except I found type 1 first and did 1 - P of Type 1 to find type 2. My final answer was 0.6524, rounded off to 3 significant figures 0.652

 

[kirashinagami]

CV z = ±1.645, old µ = 2, σ^2 = 1.85, n=200. Using these values, find the Acceptance Region. It'll come out to be 1.842<x̄<2.158.​

Type 2 means accepting H0 when H0 is false. So that means, now, using the new µ = 2.12, calculate P(1.842<x̄<2.158).​

P(Type 2) = P(1.842<x̄<2.158)​

= Φ(0.395) - Φ(-2.891)​

= 0.6536 - 0.001920​

= 0.6517 ≈ 0.652​

Oh my, its so easy, scumbag brain

 

[redapple20]

Was bacteria Q poiss0n?

 

[Zephyro]

Was bacteria Q poiss0n?

First part of it was poisson. Last was normal. You can do the first one aswell with normal so marks will be given if anyone did it that way.

What did you people write for why the prob density function might not be reliable?

 

[Saiyan]

CV z = ±1.645, old µ = 2, σ^2 = 1.85, n=200. Using these values, find the Acceptance Region. It'll come out to be 1.842<x̄<2.158.​

Type 2 means accepting H0 when H0 is false. So that means, now, using the new µ = 2.12, calculate P(1.842<x̄<2.158).​

P(Type 2) = P(1.842<x̄<2.158)​

= Φ(0.395) - Φ(-2.891)​

= 0.6536 - 0.001920​

= 0.6517 ≈ 0.652​

Thanks a lot. Now I know where did I got that wrong. I found the critical regions with the new mean 2.12. But we had to do with 2! Oh darn! Do you think I will lose the whole 7 marks? My process was right though.

 

[Saiyan]

First part of it was poisson. Last was normal. You can do the first one aswell with normal so marks will be given if anyone did it that way.

What did you people write for why the prob density function might not be reliable?

I wrote that for probability density function to be reliable, the demand needs to be a continuous variable but the demand is always discrete as demand can never be in decimals. So it is not realistic.

 

[redapple20]

Was the ans to part 1 of 3 something like 0.069?
in the last Q part a was it reject h o?

 

[redapple20]

And what did we have to do with that flour Q how did we find prob that 20 kg not enuff?

 

[Saiyan]

Was the ans to part 1 of 3 something like 0.069?
in the last Q part a was it reject h o?

Can't remember that but I did not get any prob like that.