ECMT quiz 2 (1 Viewer)

[stazi]

can someone explain how they got the mean and stdev from:
1:2000-u/SD=2.33(Z value from normal table)
2:6000-u/SD=1.88(As above)

 

[ToO LaZy ^*]

1.88 = (6000 - u) / o

multiply both sides by o and we get:
1.88o = 6000 - u

solve for u

(1) u = 6000 - 1.88o

similarly with the other one

(2) u = 2000 + 2.33o........... (+) because Z = -2.33

solve simultaneously
6000 - 1.88o = 2000 + 2.33o

solve for o...then sub the value for o into either equation 1 or 2

 

[stazi]

ive just gotta pray that we dont get that one. i dont know how simultaneous equations are done. they unarouse me.

 

[stazi]

edit: im silly dont mind me, read q 3d as an interval one.

 

[stazi]

did anyone get a different answer to absolution in 3.d.?
I got:
Assume normality using CLT, as N>30

Z= X - u
------
... o/root n

=(12 - 10)/(5/root40)

=2.53 (2d.p.)

P(X>12) = 1-P(Z<12)

=0.0057

 

[sarevok]

3e)

Statistic: Z = Ps - P/Sqrt(p(1-p)/n)

np = 200(0.1) = 20
n(1-p_ = 200(0.90) = 180

Reason for use of statistic:
*Dealing with sampling distribution of the proportion
*np and n(1-p) are greater than or equal to 5 and so the normal distribution can be used to approximate the binomial distribution

Z1 = 0.15 - 0.1/SQRT(0.1(1-0.1)/200) = 2.36 (2dp)
Z2 = 0.25 - 0.1/SQRT(0.1(1-0.1)/200) = 7.07 (2dp)

The second Z value is massive and I don't know why, but I'm sure that's the right method

Btw I emailed my tutor and we will only be given the normal table for positive Z's

 

[ToO LaZy ^*]

yeah i got that
EDIT: for 3d)

 

[stazi]

edit: just saw sarevoks solution

 

[Lainee]

on another note..someone said that 2b) was 1 and 99..can't remember who
that answer is incorrect i'm affraid.
the answer posted by absolution (pg 2) is the correct answer

Er yeah... that was wrong, I doubled checked my answers. Sorry about that.

 

[Lainee]

3e)

Statistic: Z = Ps - P/Sqrt(p(1-p)/n)

np = 200(0.1) = 20
n(1-p_ = 200(0.90) = 180

Reason for use of statistic:
*Dealing with sampling distribution of the proportion
*np and n(1-p) are greater than or equal to 5 and so the normal distribution can be used to approximate the binomial distribution

Z1 = 0.15 - 0.1/SQRT(0.1(1-0.1)/200) = 2.36 (2dp)
Z2 = 0.25 - 0.1/SQRT(0.1(1-0.1)/200) = 7.07 (2dp)

The second Z value is massive and I don't know why, but I'm sure that's the right method

Btw I emailed my tutor and we will only be given the normal table for positive Z's

Yeah, I got that too for 3(e). Just assume that P(Z<7.07)=0.9999 I suppose.

 

[sarevok]

er, nevermind. made a stupid error.

 

[stazi]

working for f. can anyone just confirm?

Sample mean =22500/60 = 375
Assume normality, using the CLT, as the sample size >30
1) 375+ t.05,59 * 109.42/sqrt60
=375 + 1.671 (14.13)
=398.61

2) = 351.39

Therefore 90% CI = 351.39, 398.61kWh

 

[stazi]

could someone please do the working out for 3 (g)
Can't seem to find notes for this sort of thing anywhere

 

[ToO LaZy ^*]

working for f. can anyone just confirm?

Sample mean =22500/60 = 375
Assume normality, using the CLT, as the sample size >30
1) 375+ t.05,59 * 109.42/sqrt60
=375 + 1.671 (14.13)
=398.61

2) = 351.39

Therefore 90% CI = 351.39, 398.61kWh

:uhhuh: :uhhuh: ??

 

[sarevok]

that's what i got

3h) the confidence interval for the proportion question

Statistic:
Z = ps - p/Sqrt(p(1-p)/n)

Statistic is sampling distribution of the proportion and has a binomial distribution.

Why this statistic?
-As dealing with sampling distribution of the proportion.
-np = 27 and n(1-p) = 73. np and n(1-p) are greater than or equal to 5 so normal distribution can be used to approximate binomial dsitribution.

Constructing CI:
Critical value of Z: + or - 1.96

Formula used for CI: ps + or - Z * SQRT(ps(1-ps)/n)

Lower lim:
0.27 - 1.96 * Sqrt(0.27(1-0.27)/100) = 0.183

Upper lim:
0.27 + 1.96 Sqrt(0.27(1-0.27)/100 = 0.36

Confidence interval:

18.3% < u < 36%

 

[stazi]

too lazy, are u on msn. cant see u there. speak to me.

 

[stazi]

sarevok - any chance of working out for 3g?

 

[ToO LaZy ^*]

that's what i got

3h) the confidence interval for the proportion question

Statistic:
Z = ps - p/Sqrt(p(1-p)/n)

Statistic is sampling distribution of the proportion and has a binomial distribution.

Why this statistic?
-As dealing with sampling distribution of the proportion.
-np = 27 and n(1-p) = 73. np and n(1-p) are greater than or equal to 5 so normal distribution can be used to approximate binomial dsitribution.

Constructing CI:
Critical value of Z: + or - 1.96

Formula used for CI: ps + or - Z * SQRT(ps(1-ps)/n)

Lower lim:
0.27 - 1.96 * Sqrt(0.27(1-0.27)/100) = 0.183

Upper lim:
0.27 + 1.96 Sqrt(0.27(1-0.27)/100 = 0.36

Confidence interval:

18.3% < u < 36%

yep
got the same answer

 

[sarevok]

zzzzzzzzzzzzzzzzzzzz

 

[sarevok]

here is a scan from my old 2u maths book for those who do not know how to do simultaneous equations (just a coupel of examples)

quality of the scan hopefully not too low