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hongco1990

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Qmb pass

Hi all,
Im a bit lost in this exercise (from PASS Kevin-Fiona). Hope you guys will help me out with this. Ta.

Population is normally distributed. Standard deviation of 20mins. How large a sample if wanna achieve 90% confidence within 1min?

I know we gotta use the sample size formula, but does the difference in time (20mins and 1min) matter in this case?
 
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mr.nice.guy

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honestly, forget about that pass exserice, try starting the QMB project
 

hongco1990

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lol, we gonna have in tut test nxt wk, and i dun wanna screw evrything out. I worked it out anyway

btw, I emailed Sasan yesterday about the project concern as my fckng Chinese tutor didnt know anything about project discussion which was designed in wk 7 tut. He said I could go to his/tutor in charge's consultation for help. So I think if you have any trouble with the instruction sheet itself, you should go and ask. Dont just wait until the last minute.
 

runnable

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Do you have the answer?

(2(1.645 x20))/1 squared

to give 4329.64... 4330?
 
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mr.nice.guy

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lol, we gonna have in tut test nxt wk, and i dun wanna screw evrything out. I worked it out anyway

btw, I emailed Sasan yesterday about the project concern as my fckng Chinese tutor didnt know anything about project discussion which was designed in wk 7 tut. He said I could go to his/tutor in charge's consultation for help. So I think if you have any trouble with the instruction sheet itself, you should go and ask. Dont just wait until the last minute.
btw, wif our quiz 3 .. they say its topics from week 1 - 8, but would the topics be mostly from lectures 6-8? corz there ain't not much point quizing us the same shit again..
Btw whaat were the topics between weeks 6-8
 

hongco1990

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@ mr.nice.guy: I think mostly based on wk 6-8 lec, which include normal distribution, converting binomial to normal distribution and hypothesis testing. Of course the knowledge covered in previous lecs from wk1-5 can be tested as well, but I think they wont test us directly on that.

@ runnable: I think Aronawa21's answer should be right. As in this week tut solution on Vista, they multiplied ((standard deviation * z alpha/2) / B) by 2 I think pretty much because the question is "the width is no more than 2 units". If in the question we are given "within 2 units", so we dont need to multiply by 2 at all. So the formula for this PASS question will be the same as in text book:
n = [(std deviation * z alpha/2) / B] squared
 

mr.nice.guy

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@ mr.nice.guy: I think mostly based on wk 6-8 lec, which include normal distribution, converting binomial to normal distribution and hypothesis testing. Of course the knowledge covered in previous lecs from wk1-5 can be tested as well, but I think they wont test us directly on that.

@ runnable: I think Aronawa21's answer should be right. As in this week tut solution on Vista, they multiplied ((standard deviation * z alpha/2) / B) by 2 I think pretty much because the question is "the width is no more than 2 units". If in the question we are given "within 2 units", so we dont need to multiply by 2 at all. So the formula for this PASS question will be the same as in text book:
n = [(std deviation * z alpha/2) / B] squared
tanks
 

mr.nice.guy

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Re: Qmb pass

Hi all,
Im a bit lost in this exercise (from PASS Kevin-Fiona). Hope you guys will help me out with this. Ta.

Population is normally distributed. Standard deviation of 20mins. How large a sample if wanna achieve 90% confidence within 1min?

I know we gotta use the sample size formula, but does the difference in time (20mins and 1min) matter in this case?
yo i got a question as well:
week 8 pass nick and nick and james :
In a certain community, 60% of the residents are in favour of building a new library. A random sample of 200 ppl is taken. Find the probability that 100 or less of these ppl favour the construction of the library.

Ans: 0.0025
 

chewy123

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Re: Qmb pass

I don't think the answer is correct. 0.0025 is wayyy to small when you think about it.

Presumably the question is about correctional factors. The mean is just 200x0.6=120
The standard deviation is root npq. that is, root(200x0.6x0.4) which gives 48.
Then use standard transformation NOTE X=99.5 NOT 100 due to correction stuff.
P(z<-0.427) = 0.5 - 0.1628= 0.3372

Methink that's the answer, but i could be wrong.....:(
 

runnable

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Re: Qmb pass

chewy: npq gives variance, not s.d so the s.d is actually root 48

I got the ans to be 0.0025 with X=100.5, not 99.5
 
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runnable

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The conversion is

X< or equal to 100
so X < 100.5 in normal approximation.
 

chewy123

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+1 runnable

sigh sigh, i screwed my qmb test so severely, sigh sigh~~
 

whoisurdaddy

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A lot of people got questions with 2 different samples, have we even learnt this?!
 

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