Angels1234
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Hiya ,

https://postimg.cc/image/e5m7bbgpp/

https://postimg.cc/image/osg0gwowt/

I’m confused on Part C , how is this question different from asking about 90% confidence interval . Also where has n gone in the formula they have used in solution bank when working out the answer . Also in part E the formula on the mark scheme uses n as 100 . I am aware that N is equal to 100 for the unleaded petrol in 1990 but when we use the formula we are using the population mean for 1989 so how do we know that the n value in the formula is the same in 1989 as it is a 1990 ..which is hundred. Also in the formula it sample mean- population mean so why don’t we do the population mean of the 1990 rather than the population mean of 1989. I know we are comparing 1990 and 1989 but isn’t there supposed to be some kind of consistency within the formula
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Gregorius
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(Original post by Angels1234)
Hiya ,

https://postimg.cc/image/e5m7bbgpp/

https://postimg.cc/image/osg0gwowt/

I’m confused on Part C , how is this question different from asking about 90% confidence interval . Also where has n gone in the formula they have used in solution bank when working out the answer . Also in part E the formula on the mark scheme uses n as 100 . I am aware that N is equal to 100 for the unleaded petrol in 1990 but when we use the formula we are using the population mean for 1989 so how do we know that the n value in the formula is the same in 1989 as it is a 1990 ..which is hundred. Also in the formula it sample mean- population mean so why don’t we do the population mean of the 1990 rather than the population mean of 1989. I know we are comparing 1990 and 1989 but isn’t there supposed to be some kind of consistency within the formula
You use confidence intervals to estimate the precision of an estimate - for example, if you were estimating the mean of the underlying population using the mean of a sample. Here you are working with the distribution of the population - which you are told.

For part (e), you are told what the population mean for 1989 is, and you use a sample of size 100 to estimate what the population mean for 1990 is. So you are doing a test of whether the sample mean is consistent with a known population mean.
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Gregorius
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Also, we looked at this question not long ago right here!
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Angels1234
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(Original post by Gregorius)
Also, we looked at this question not long ago right here!
I still don’t understand why where the 10 has gone and why we don’t use it
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Gregorius
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(Original post by Angels1234)
I still don’t understand why where the 10 has gone and why we don’t use it
I can't see a 10 in any of the working - do you mean 100? Let's go through your original question about part (e) in a bit more detail.

Also in part E the formula on the mark scheme uses n as 100 . I am aware that N is equal to 100 for the unleaded petrol in 1990 but when we use the formula we are using the population mean for 1989 so how do we know that the n value in the formula is the same in 1989 as it is a 1990 ..which is hundred.
You are told what the population mean (and standard deviation, by implication) are for unleaded petrol in 1989. You are not told these figures for 1990, but you are told what the mean of a sample of 100 is drawn from the 1990 population. You are asked to use this evidence to see whether the sample mean from 1990 is consistent with the population mean from 1989. So you are comparing a random quantity from 1990 with given mean and standard deviation that involves the sample size, with a fixed and known constant, the population mean from 1989.

Also in the formula it sample mean- population mean so why don’t we do the population mean of the 1990 rather than the population mean of 1989. I know we are comparing 1990 and 1989 but isn’t there supposed to be some kind of consistency within the formula
Precisely because you are comparing the known population mean from 1989 with the sample mean from 1990 (which estimates the population mean from 1990).
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Angels1234
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(Original post by Gregorius)
Precisely because you are comparing the known population mean from 1989 with the sample mean from 1990 (which estimates the population mean from 1990).
This made everything a whole lot clearer . Thank you . I was talking about part C when I said 10, I should of been more clear . I thought the sample size was 100 so I meant root N was 10 but I can’t see any mention of n in the formula in part C so I was confused why this is ?
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Gregorius
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(Original post by Angels1234)
This made everything a whole lot clearer . Thank you . I was talking about part C when I said 10, I should of been more clear . I thought the sample size was 100 so I meant root N was 10 but I can’t see any mention of n in the formula in part C so I was confused why this is ?
Ah, right! In part (c) you're not estimating anything about the mean of the population - so you're not working with the distribution of the mean (that has the square root of 100 in it), you're working directly on the population distribution.
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Angels1234
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(Original post by Gregorius)
Ah, right! In part (c) you're not estimating anything about the mean of the population - so you're not working with the distribution of the mean (that has the square root of 100 in it), you're working directly on the population distribution.
So is this formula that they’ve used nothing about confidence intervals ,because I can see it follows the same kind of format but obviously there is no N in the solution . I haven’t seen this type of formula before so I’m still struggling to understand where the N has gone. I think I kind of understand what you’re talking about ..are you basically saying that the question is not talking about sales from the sample ( sample mean of leaded petrol isn’t given anyway) but we just looking at the population mean so we dont include N as if we did we would be looking at a sample . Are we basically trying to find an interval for which the true mean of the leaded petrol would lie and is the formula on solution bank something I need to learn for the exam?
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Gregorius
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(Original post by Angels1234)
So is this formula that they’ve used nothing about confidence intervals
Correct. You're doing calculations on the actual population distribution for leaded petrol.

,because I can see it follows the same kind of format but obviously there is no N in the solution . I haven’t seen this type of formula before so I’m still struggling to understand where the N has gone. I think I kind of understand what you’re talking about ..are you basically saying that the question is not talking about sales from the sample ( sample mean of leaded petrol isn’t given anyway) but we just looking at the population mean so we dont include N as if we did we would be looking at a sample .
Yes. You are trying to find an interval within which 90% of the population lies.

Are we basically trying to find an interval for which the true mean of the leaded petrol would lie and is the formula on solution bank something I need to learn for the exam?
In part (c), you're not trying to infer anything about the mean - you are already told what the population mean is. You are trying to find an interval in which 90% of the population lies.
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