statistics 2: post your doubts and solve mine! lol

[redapple20]

A type 1 error means rejecting someones claim ie ho when the claim was true! Here the claim is HO=5.2 this was correct but was rejected
so type 1 error is ... the mean number of white blood cells were 5.2 but we rejected them. Now for da probability. Here's wat ma sir told me
Its less 5.2 the alternative hypotheses that is we reject 5.2 and say it has decreased .prob starts from 0. So wekeep taking out prob using poissn
And keep checking whether each ans is less than 0.1 our sig level.we ail stop wen our prob exceeds 0.1. Hope u liked my story

 

[kagome]

But 1.8 is for 1 player everly Rogers. Its now asking about the whole team!

They don't mention for one player. Read the question again. It says "goals per match by everly rovers".

 

[kagome]

A type 1 error means rejecting someones claim ie ho when the claim was true! Here the claim is HO=5.2 this was correct but was rejected
so type 1 error is ... the mean number of white blood cells were 5.2 but we rejected them. Now for da probability. Here's wat ma sir told me
Its less 5.2 the alternative hypotheses that is we reject 5.2 and say it has decreased .prob starts from 0. So wekeep taking out prob using poissn
And keep checking whether each ans is less than 0.1 our sig level.we ail stop wen our prob exceeds 0.1. Hope u liked my story

Actually I still dunno how to obtain the answer. Can u plz do it step by step plz..??

 

[redapple20]

Am I supposed to laugh or cry?

 

[redapple20]

Actually I still dunno how to obtain the answer. Can u plz do it step by step plz..??

e to the power -5.2 *(5.2) to the power zero divided by 0factorial= 5.51*10 TO the power minus three smaller than o.1 @ x=0 then take out prob @ x=1 e to the power -5.2*(5.2)^1 divided by 1 factorial = o.o28 add this to prob @ 0 = o.o342 still less than o.1 then @ 2. e to the power -5.2 *(5.2)^2 divided by 2 factorial =o.o745 add this to prob @ 0,1= o.o342= o.108 exceeds 0.1 so we cant take this value highest value before this is o.o34 and tats the answer!

 

[ALstudentd]

Hi,
Can anyone please give me a summary of Hypothesis Testing and how to use it in different situations (like with Poisson, Binomial, Proportions in a population etc.). I find this topic very challenging and would appreciate any help

 

[redapple20]

Tel u what I'm gonna fail s2!

 

[kirashinagami]

Hi,
Can anyone please give me a summary of Hypothesis Testing and how to use it in different situations (like with Poisson, Binomial, Proportions in a population etc.). I find this topic very challenging and would appreciate any help

Just do at least 5 past papers with ms and er.. You will get the feeling , seriously

 

[gudboy01]

A fair coin is tossed 5 times and the number of heads is recorded.
i) The number of heads is doubled and denoted by the random variable y.State the variance of Y.

 

[gudboy01]

Weights of men follow normal distribution with mean 71 kg and standard deviation 7 kg and women follow with mean 57 kg and standard deviation 5 kg.The total weight of 5 men and 2 women chosen randomly is denoted by X kg.
i)show that var(X)=295
=> Here while calculating variance why do we hab to do 5*7 square+2*5square instead of 5 sq. * 7 sq.+2sq. * 5sq???plzzzz make me understand clearly......

 

[gudboy01]

hey could you please help me with some s2 doubts.Can't seem to understand q2 part i of 2009 72.Please reply whenever you can.
thanks

hey man..i am taking s2..u r talking abt m/j 09 or o/n 09??

 

[gudboy01]

Can u help with me with M/J 2010 71 Q7(i). How to find probability of type 1 error? Thx

null hypothesis = 502
alternative hypothesis < 502
At 10% significance level
- P(0)=e^-5.2=0.00552
- P(1)=e^(-5.2)*5.2= 0.02869
P(0)+P(1) = 0.00552+0.02869=0.03421<0.1(10%) [so u hab to calculate further]
- P(2)= (e^(-5.2)*5.2 square)/2!=0.07458
P(0)+P(1)+P(2)= 0.03421+0.07458=0.10879>0.1
So 2 is rejected and P(type I error)= P(X<2) becoz P(0)+(1) is less than significance level....

 

[2pac]

hey man..i am taking s2..u r talking abt m/j 09 or o/n 09??

oct 2009

 

[2pac]

Weights of men follow normal distribution with mean 71 kg and standard deviation 7 kg and women follow with mean 57 kg and standard deviation 5 kg.The total weight of 5 men and 2 women chosen randomly is denoted by X kg.
i)show that var(X)=295
=> Here while calculating variance why do we hab to do 5*7 square+2*5square instead of 5 sq. * 7 sq.+2sq. * 5sq???plzzzz make me understand clearly......

well your supposed to square only fractions and coefficients when related to variance.Here its just 5xmen and 2xwomen so squaring is not needed as as if you add the variance of men 5 times,it'll be the same value as multiplied.Had the question been that the variance of men is twice of women's then squaring would be applicable as the initial condition is 2 x some give variance value.Hope you understood.

 

[gudboy01]

oct 2009

its given dat lamda=1.27 per hour. but in part i) it is asked for the period of 5 hours so for part i) lamda becomes 6.35 per 5 hrs(5x1.27). Now u hab to calculate more than 1 unwanted email so P(X>1)=?
P(X>1)=1-P(X=0,1)
=1-[e^-6.35(1+6.35)]
=0.987
THATS IT...

 

[2pac]

its given dat lamda=1.27 per hour. but in part i) it is asked for the period of 5 hours so for part i) lamda becomes 6.35 per 5 hrs(5x1.27). Now u hab to calculate more than 1 unwanted email so P(X>1)=?
P(X>1)=1-P(X=0,1)
=1-[e^-6.35(1+6.35)]
=0.987
THATS IT...

okay this is embarassing,I mistyped the q number,meant q4 part i.so sorry

 

[redapple20]

This is terrible...

 

[gudboy01]

okay this is embarassing,I mistyped the q number,meant q4 part i.so sorry

ohh its ok..
in that que. its given dat p=5 and i think u r confused in significance level..its given dat type I error is at most 0.1 which means 10% significance level.
as to get acceptance region sum of prob sud b more than 0.1 and here we go for test.
P(0)=10C0*0.5^10 = 0.00098 which is also equals P(10) [its sud b checked with smallest, P(0), and largest, P(10), wen p=q and in this que p=q=0.5]
P(1)=10C1*0.5*0.5^9 = 0.00977 which is also equals P(9) [check by calculating P(9)]
now lets check P(O)+P(1)+P(10)+P(9)= 0.00098x2+0.00977x2 = 0.02150 which is less than 0.1
again we go for P(2)= 10C2*0.5^2*0.5^8 = 0.04395 which also equals P(8)
now again lets check P(0)+P(1)+P(2)+P(8)+P(9)+P(10)= 0.02150+0.04395x2 = 0.1094 which is greater than 0.1
Hence acceptance region is 2 greater than or equal to X less than or equal to 8...
Hope u understood....

 

[ALstudentd]

I have a few doubts, could someone please explain them to me

M/J 2011 p73
Q2 part i) Firstly, isn't the approximation from binomial to normal taken as (n*p, n*p*(1-p))? The examiner report states that it is (p, [p*(1-p)]/n). Please clarify this.
and secondly, from the question, I concluded that it is one-tailed, as the question mentions bars of soap being undersized, however, the z value used in the MS was for a two-tailed test ( 1.645 was used, not 1.282 for the 90% C.I.) :/

and in questions like Q3 part i), i don't know which proportions to use in my binomial calculation ( in this case 0.15 and 0.85 were the original proportions, but i used 2/30 and 28/30 )

 

[2pac]

ohh its ok..
in that que. its given dat p=5 and i think u r confused in significance level..its given dat type I error is at most 0.1 which means 10% significance level.
as to get acceptance region sum of prob sud b more than 0.1 and here we go for test.
P(0)=10C0*0.5^10 = 0.00098 which is also equals P(10) [its sud b checked with smallest, P(0), and largest, P(10), wen p=q and in this que p=q=0.5]
P(1)=10C1*0.5*0.5^9 = 0.00977 which is also equals P(9) [check by calculating P(9)]
now lets check P(O)+P(1)+P(10)+P(9)= 0.00098x2+0.00977x2 = 0.02150 which is less than 0.1
again we go for P(2)= 10C2*0.5^2*0.5^8 = 0.04395 which also equals P(8)
now again lets check P(0)+P(1)+P(2)+P(8)+P(9)+P(10)= 0.02150+0.04395x2 = 0.1094 which is greater than 0.1
Hence acceptance region is 2 greater than or equal to X less than or equal to 8...
Hope u understood....

when do we consider this method of having P of 0=P of 10 and so on.Obviously it'll be only when p and q have same probability but still,maybe perhaps only when the h1 is not equal to?