statistics A2 level: post doubts here

[Ranzis]

Hey bro m posting my doubts can u help me plz!??

Didn't I explain it?
Remember what the definition of Type 1 error is, then apply to the situation like I did.
"Type 1 error is the probability that you reject the claim that it is 20% red for sugar-coated chocolate beans when they actually are 20%." That's pretty much the answer.

View attachment 17960

Man, I didn't think you needed this much help. This is pretty simple.
Ho = null hypothesis
Ha = Alternative hypothesis

Ho: mean = 21.2
Ha: mean does not equal 21.2

So test the example they give you, 19.4 words mean in 90 sentences.
X~N(21.2, 7.3^2)
P(X<19.4)
= P(Z< (19.4-21.2)/(7.3/sqrt 90))
= P(Z<-2.339)

Since this is two tailed, use the z value for .975, which is 1.960.
|-2.339| > 1.960, therefore -2.339 is in the rejection region, so therefore:
Reject Ho, the mean sentence length is not the same as the author's to a 95% confidence interval.

Type I error with reference to the question is:
(Actually, it's not the probability, it's just straight out) Saying that the mean sentence length is not the same as the author, when it actually is the same.

Type I error is just calculated by the rejection region probability, which is 1-0.95 = 0.05

 

[shanky631]

Ranzis thanks bro, but how do u know in a two tail whether to check X<19.4 or X>19.4

 

[Ranzis]

Ranzis thanks bro, but how do u know in a two tail whether to check X<19.4 or X>19.4

If it's below the mean use <19.4. If above use >19.4

 

[parthrocks]

If it's below the mean use <19.4. If above use >19.4

hey i didnt get thata can u be more precise and explain me!plz

 

[Ranzis]

hey i didnt get thata can u be more precise and explain me!plz

He wants to know when to use x>19.4 or x<19.4 when finding the rejection region. You can tell, because if you think of it as on a side of the mean, you need to find the rejection region for that side.
Hard to explain, hit the books on when to use them if you're still unsure.

 

[parthrocks]

He wants to know when to use x>19.4 or x<19.4 when finding the rejection region. You can tell, because if you think of it as on a side of the mean, you need to find the rejection region for that side.
Hard to explain, hit the books on when to use them if you're still unsure.

Hey bro ohhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhh ok!!I got that...I know too difficult to explain in words!!bt referring to book would make it clear!!and say hey i have some serious doubts!!need to clarify them asap!!r u der for some while!!!//till when then!reply in a conversation!

 

[parthrocks]

For this you need to realise that these are discrete values.
The Type I error is at most 0.1.
Null hypothesis is when p = 0.5.
Alternative hypothesis is p does not equal to 0.5. Therefore this is a two tailed test.

X~B(10, 0.5)
P(X=0) = 0.000976562
P(X=1) = 0.009765625
P(X=2) = 0.0439453125

Since it is two tailed, we are calculating up to 0.05.
To find the acceptance region for the test, we need to find what values of x do not fall into the rejection region.
P(X=0) is lower than 0.05.
P(X=<1) is lower than 0.05 (add P(X=1) and P(X=0) together.
but P(X=<2) is higher than 0.05. This means that when X=<2 it is not rejected as being not biased to a 0.1 significance level.

Therefore the acceptance region of the test would be P(X=>2), as any values below that fall into the rejection region.

For the second part of calculating a Type II error.
Type II error is the probability of accepting the null hypothesis when it is actually false.
Ignore that 'when it is actually false part', it always confused me. Just calculate the probability of accepting the null hypothesis, ie. not in the rejection region. The rejection region was P(X<2).
So the acceptance region for accepting the null hypothesis is P(X=>2).

Using the new probability, X~B(10, 0.7), calculate P(X=>2).
ie. 1 - P(x<2), which is 1- P(X=1) - P(X=0)
=0.9998563141 = 1.00.
Okay, something's wrong here.
What paper was this and the question?
I probably made a error in calculation somewhere.

So we can use any two tailed test or 1 tail test bt when using 2 tail 10% to be taken and when using one tail 5% to be taken!!so m i correct?

 

[Kumkum]

Please help me with question 2.
I know how to do it but got confused when the standard deviation was divided by 60.why was it done?

 

[Ranzis]

Please help me with question 2.
I know how to do it but got confused when the standard deviation was divided by 60.why was it done?

The mean was averaged from 60 samples. So divide s.d by sqrt 60.
Also, everyone else who wants help should actually post here, and should actually show some form of working, otherwise I'm just going to ignore.
I don't want to do whole problems when you could look at the mark scheme.

 

[shanky631]

The mean was averaged from 60 samples. So divide s.d by sqrt 60.
Also, everyone else who wants help should actually post here, and should actually show some form of working, otherwise I'm just going to ignore.
I don't want to do whole problems when you could look at the mark scheme.

in the above question, we divide by 60 due to central limit theorem???

 

[Ranzis]

in the above question, we divide by 60 due to central limit theorem???

Yeah, I guess.

 

[parthrocks]

in the above question, we divide by 60 due to central limit theorem???

Hey bro then why is CC not applied!!!
as it should then be 1/2n and so 1/2(60)=1/120 we needs to be deucted from 2.8 and soo!!!in the marking scheme they have not included cc but so do they allow us???
shanky631
Ranzis
Kumkum

 

[parthrocks]

in the above question, we divide by 60 due to central limit theorem???

ya thats becos its n is more than 30!

 

[parthrocks]

hELP ME!!
Ranzis
shanky631
Kumkum

 

[shanky631]

Hey bro then why is CC not applied!!!
as it should then be 1/2n and so 1/2(60)=1/120 we needs to be deucted from 2.8 and soo!!!in the marking scheme they have not included cc but so do they allow us???
shanky631
Ranzis
Kumkum

it should be applied. cc is a must when converting binomial to normal.

 

[parthrocks]

it should be applied. cc is a must when converting binomial to normal.

hey bt here its central limit theorem applying from binomial to normal thats why!!i suppose???
and hey also ya its a must!!!!
bt here as lamda is greater than 30 so thats why they we can take 1/2n and ....so on
and not just 1/2

 

[parthrocks]

it should be applied. cc is a must when converting binomial to normal.

Do u want that notes??which i showed u that day???npw?

 

[parthrocks]

it should be applied. cc is a must when converting binomial to normal.

Bt if they had given us X~N(12,1/4) or some what like that
then use of central limit theorem is not required and we dont take 60 into consideration then!!as just what is required is normal disctribution formula......
and here as original distribution in binomial to convert to normal and then APplying CC is absolutely thinkable!!

 

[parthrocks]

One more silly doubt friends
shanky631
Kumkum
Ranzis

 

[parthrocks]

Ranzis