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Post ECMT Quiz 2 Feedback here. Plz. (1 Viewer)
- Thread starter what971
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Yeah, 7 and 8 are the only ones really that I've had to look through slowly. It's not that hard.
Q7.a. is about the normal approximation to the sample proportion.
p= 0.1 (the proportion of the entire population) and n=200 (sample size tested)... and now we need to find Pr(0.15≤p-hat≤0.25). p-hat is indicative of the sample proportion.
So that you can normally approximate...
find the mean (mu) = p = 0.1
find the variance (sigma^2) = p(1-p)/ n = 0.1x0.9/200 = 0.00045
And then prove that CLT applies...
-that n≥30 (which it does, n=200) and that
0< p +(or)- 3√p(1-p)/n < 1 ... that std. deviations thing. The text book has written it clearer than i can type
.
so √p(1-p)/n = √0.00045 = 0.021213203 = 0.02 (2.dp)
therefore 3 X 0.02= 0.06...
p +(or)- 3√p(1-p)/n = 0.1 +(or)- 0.06
= 0.04, 0.16 ... which are both between 0 and 1. Therefore the CLT applies.
now find Pr(0.15≤p-hat≤0.25) ... standardise both 0.15 and 0.25 using the formula... z= p-hat - p/ standard deviation
so for 0.15 it would be...
z= 0.15- 0.1/0.02 = 2.5 ... then calculate probabilities as usual...
Q7.a. is about the normal approximation to the sample proportion.
p= 0.1 (the proportion of the entire population) and n=200 (sample size tested)... and now we need to find Pr(0.15≤p-hat≤0.25). p-hat is indicative of the sample proportion.
So that you can normally approximate...
find the mean (mu) = p = 0.1
find the variance (sigma^2) = p(1-p)/ n = 0.1x0.9/200 = 0.00045
And then prove that CLT applies...
-that n≥30 (which it does, n=200) and that
0< p +(or)- 3√p(1-p)/n < 1 ... that std. deviations thing. The text book has written it clearer than i can type
.so √p(1-p)/n = √0.00045 = 0.021213203 = 0.02 (2.dp)
therefore 3 X 0.02= 0.06...
p +(or)- 3√p(1-p)/n = 0.1 +(or)- 0.06
= 0.04, 0.16 ... which are both between 0 and 1. Therefore the CLT applies.
now find Pr(0.15≤p-hat≤0.25) ... standardise both 0.15 and 0.25 using the formula... z= p-hat - p/ standard deviation
so for 0.15 it would be...
z= 0.15- 0.1/0.02 = 2.5 ... then calculate probabilities as usual...
I'm sorry it's taking me so long to explain .
7.b. is different- this is how i understand it.
"if the true proportion of defects in a production run of 25000 items is 3.5%"... this is population info... so p= 0.035, N (population) = 25 000 . I don't think N=25 000 is relevant again in the question.
The point of drawing the sample- proportion (p- hat) is to test that claim that
p≤0.5 (population). so in the sample proportion; n=500.
Pr(p- hat>0.5)... test it like that.
and you calculate the same as in 7a, the probability that quality control rejects the production run in error...
z= 0.5- p/ √p(1-p)/n
= 0.5- 0.035 / √0.035(1-0.035)/25 000 ... and so on.
.7.b. is different- this is how i understand it.
"if the true proportion of defects in a production run of 25000 items is 3.5%"... this is population info... so p= 0.035, N (population) = 25 000 . I don't think N=25 000 is relevant again in the question.
The point of drawing the sample- proportion (p- hat) is to test that claim that
p≤0.5 (population). so in the sample proportion; n=500.
Pr(p- hat>0.5)... test it like that.
and you calculate the same as in 7a, the probability that quality control rejects the production run in error...
z= 0.5- p/ √p(1-p)/n
= 0.5- 0.035 / √0.035(1-0.035)/25 000 ... and so on.
and finally Q8...
a, mu=mean= 500 (population)
(sample info)
Xbar= 505.2, standard deviation (s)= 16
and n=150
you compare the results of the population mean and the sample mean. the population mean is 500... but the sample mean, which is 505.2, suggests that it is greater than 500 (505.2). So the question is asking you to find- the pvalues (probability) of the mean being greater than 500, less than 500, or not equal to 500. so you test the probability that xbar (the sample mean) will actually be greater/smaller than 505.2 (if you think about how the sample is representative of the population... how it's mean is indicative of the population mean)
Xbar, the sample mean, is a random variable with mean mu (as in the population) and variance (sigma^2/n)... which they've given you in the question.
Before you find the p- values (the probabilities) you need to show that CLT applies. If the Xs are independent and identically distributed (which we just assume they are, there's no way to prove)... and n≥30 (which it is, n=150) then it does and you can move on with the question... these are just the assumptions/considerations that you need to show at the beginning.
So find probability that mu>500...
by CLT Xbar ~ N(mu, (sigma^2/n) )
Xbar ~ N(500, 16^2/150)
so standardise as normal.
z= xbar- mu/ sigma/ √n ... where sigma/√n is standard deviation=16/√150=1.31(2.d.p)
z= 505.2-500/1.31 = 3.98 (2.d.p) and this is the z-value you use throughout Q8a.
so for i, Pr(mu>500) = Pr(z>3.98) = whatever... this is therefore your p-value.
ii, Pr(mu<500) = Pr(z<3.98) = ...
iii, since the claim is that mu is not equal to 500 at all, multiply the p-value/percentage you got in the above Qs (i and ii) by TWO this is the probability/ pvalue.
8b is the exact same thing.
I'm so sorry if this isn't very clear, but it's very hard to describe through type. I learnt from the textbook... it's muchhh better explained there, I promise. Good luck for your quiz .
a, mu=mean= 500 (population)
(sample info)
Xbar= 505.2, standard deviation (s)= 16
and n=150
you compare the results of the population mean and the sample mean. the population mean is 500... but the sample mean, which is 505.2, suggests that it is greater than 500 (505.2). So the question is asking you to find- the pvalues (probability) of the mean being greater than 500, less than 500, or not equal to 500. so you test the probability that xbar (the sample mean) will actually be greater/smaller than 505.2 (if you think about how the sample is representative of the population... how it's mean is indicative of the population mean)
Xbar, the sample mean, is a random variable with mean mu (as in the population) and variance (sigma^2/n)... which they've given you in the question.
Before you find the p- values (the probabilities) you need to show that CLT applies. If the Xs are independent and identically distributed (which we just assume they are, there's no way to prove)... and n≥30 (which it is, n=150) then it does and you can move on with the question... these are just the assumptions/considerations that you need to show at the beginning.
So find probability that mu>500...
by CLT Xbar ~ N(mu, (sigma^2/n) )
Xbar ~ N(500, 16^2/150)
so standardise as normal.
z= xbar- mu/ sigma/ √n ... where sigma/√n is standard deviation=16/√150=1.31(2.d.p)
z= 505.2-500/1.31 = 3.98 (2.d.p) and this is the z-value you use throughout Q8a.
so for i, Pr(mu>500) = Pr(z>3.98) = whatever... this is therefore your p-value.
ii, Pr(mu<500) = Pr(z<3.98) = ...
iii, since the claim is that mu is not equal to 500 at all, multiply the p-value/percentage you got in the above Qs (i and ii) by TWO this is the probability/ pvalue.
8b is the exact same thing.
I'm so sorry if this isn't very clear, but it's very hard to describe through type. I learnt from the textbook... it's muchhh better explained there, I promise. Good luck for your quiz
.
Last edited:
what971... please see my correction to Q8!!! They've given us the standard deviation of the sample rather than the population, use that. Not 16^2/150... because it's NOT sigma. It's s... you don't need to change it! So Xbar ~ N(500, 16) NOT Xbar ~ N(500, 16^2/150) !!!
I'm glad that they didn't ask you. You see, they asked me... and i divided by s only in standardisation. But apparently I'll only lose one mark if my workings correct... and that might be the only mark i lose... this quiz was much better than the last.I heard acct was hard though :| like a 70mrk question at the end?