ECMT1010 - Revision Questions (1 Viewer)

stazi

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sarevok said:
Just going through past quiz possible questions, any one know what the 'modal class' of a histogram is?
the heaviest populated class. Think 'mode' in statistics.
 

stazi

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ressul said:
Anyone knows how to do Q11, I); Q21,h); Q48 and the full process for Q57c)?
q11 has been done for you
q21 h is a misprint. I'm assuming it means g) not c)
 

stazi

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yvonne_710 said:
can i ask, how to calculate marginal distribution and coniditional distribution?thanks~!!!!!!!!
This wasn't explained that clearly.

Let's suppose
X
0 .25
1 .35
2 .40

The marginal distribution of X looks like:
P(X=0) = .25
P(X=1) = .35
P(X=2) = .40

Conditional probability is obtained using this formula:
P(A|B)=P(A&B)/P(B)
 

yvonne_710

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1Time4thePpl said:
This wasn't explained that clearly.

Let's suppose
X
0 .25
1 .35
2 .40

The marginal distribution of X looks like:
P(X=0) = .25
P(X=1) = .35
P(X=2) = .40

Conditional probability is obtained using this formula:
P(A|B)=P(A&B)/P(B)
i almost understand what is marginal distribution, but for conditional distribution..... :confused:
 

yvonne_710

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ressul said:
Why do u make S *1.01?
bcz u need to pay extra one month, without 1.01, there are only 35 months, u need pay the 36th month ..
 

stazi

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yvonne_710 said:
i almost understand what is marginal distribution, but for conditional distribution..... :confused:
condition distribution just uses the condition probability formula.

So it'd be
P(X=1|Y=1)=
P(X=2|Y=1)=
P(X=3|Y=1)=
etc.
 

Rorix

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yvonne_710 said:
i almost understand what is marginal distribution, but for conditional distribution..... :confused:


your total possibilities has been reduced because you limit it to the cases where Y=1 - or whatever the condition happens to be. e.g. if P(Y=1)=.4 and P(Y=0)=.6 [marginal distribution] then your new probabilities will all be /.4, so P(X=1 and Y=1)=.2 will be .5 in the condition distribution

As P(A|B) = P(A and B)/P(B) as stas said
 

sarevok

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Hmm...for these test the slope questions, don't we have to make some sort of assumption? Like something about iid? I think Tig said something about this in consultation but notes aren't really clear.
 

ressul

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Question

See Q33 C):
y(hat)=41.6667-2/3X
H0: B1=0; H1: B1<0
In the answer, its conclusion is Don't Rej H0
Does it mean that the slop coefficient is NOT negative?
But in the equation, the slop IS negative, which is -2/3.
Is my thinking right? How should we explain this?
 

stazi

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did we ever get to the bottom of using a p-value hypothesis test on a t-stat?
I understand Z stat but tstat escapes me.
 

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It's not a probelm cause of the assumptions you use for p values, something like P(s) > 5 and P(1-s) > 5 or whetever it is assume normality and the t distribution is obselete. This raises another question, when actually using T distribution do you assume normality or what? What do you actually state to justfy using T stats cause if you assume normality why not just use the z tables. And another thing for hypothesis testing when can you use the normal tables, I have heard something like when n>100 use z tables but the CLT is for n>30, and if you can't use the z tables ie n<30 what do you say in your assumptions again for justifying the use of a t stat, n<30 therefore assume t'ially distributed. If anyone can clarify I would be grateful.

Ps I don't believe in paragraphs.
 

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hey i was wondering how to do the confidence questions where they give you the sum of the stastics. like q13, 18 etc...

how can u find the confidence interval without the variance!!
 

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Without the variance you use the t critical values! (I think) Wait that might be population variation vs sample variance. You may have to calculate it yourself if possible or it's a proportion qs perhaps.
 

yvonne_710

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ressul said:
See Q33 C):
y(hat)=41.6667-2/3X
H0: B1=0; H1: B1<0
In the answer, its conclusion is Don't Rej H0
Does it mean that the slop coefficient is NOT negative?
But in the equation, the slop IS negative, which is -2/3.
Is my thinking right? How should we explain this?
its nearly same like q2(e).............
 

stazi

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jpr333 said:
It's not a probelm cause of the assumptions you use for p values, something like P(s) > 5 and P(1-s) > 5 or whetever it is assume normality and the t distribution is obselete. This raises another question, when actually using T distribution do you assume normality or what? What do you actually state to justfy using T stats cause if you assume normality why not just use the z tables. And another thing for hypothesis testing when can you use the normal tables, I have heard something like when n>100 use z tables but the CLT is for n>30, and if you can't use the z tables ie n<30 what do you say in your assumptions again for justifying the use of a t stat, n<30 therefore assume t'ially distributed. If anyone can clarify I would be grateful.

Ps I don't believe in paragraphs.
n<30: Assume normal distribution. Can use t statt
30<n<130: Use CLT. Say that t is a robust estimator
n>130: Use Z table. Use CLT.
 

stazi

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monkey187 said:
hey i was wondering how to do the confidence questions where they give you the sum of the stastics. like q13, 18 etc...

how can u find the confidence interval without the variance!!
q13:
You're given the sum of X: 56. So the mean = 56/n =56/10
You're given the sum of X^2: 360

To find the variance remember the formula:
E(X^2)-[E(X)]^2

So it'd be 360-5.6^2

And don't forget to square root it. We don't need the variance. We need the standard deviation.
 

sarevok

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From my notes,

Given that population is normal, and when o is uknown, use t-stat.
n<15 do not use t-stat unless given that population is normal.
15<n<40 use t-stat unless strong skewness in population.
n>40 use t-stat even if there is strong skewness.
n>120 t distribution approximates normal distribution, use z tables.
 

stazi

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sarevok said:
From my notes,

Given that population is normal, and when o is uknown, use t-stat.
n<15 do not use t-stat unless given that population is normal.
15<n<40 use t-stat unless strong skewness in population.
n>40 use t-stat even if there is strong skewness.
n>120 t distribution approximates normal distribution, use z tables.
What i posted is what an exam marker told my tutor before our last tutorial.
The text book also says this.
 

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