S3 help , finding limits but not confidence intervals Watch

Angels1234
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Hi ,

https://postimg.cc/image/8kki2vr3h/

I’m stuck on part A of the question . I’m struggling to understand why we can’t use confidence intervals . The mark scheme uses the normal formula for confidence intervals of 95% but without the involvement of n . Why is this so ?

Thank you
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Pangol
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(Original post by Angels1234)
Hi ,

https://postimg.cc/image/8kki2vr3h/

I’m stuck on part A of the question . I’m struggling to understand why we can’t use confidence intervals . The mark scheme uses the normal formula for confidence intervals of 95% but without the involvement of n . Why is this so ?

Thank you
The question is asking about the interval within which 95% of the individual orchids lie. In a way, this is the same as finding a confidence interval, but that wording is reserved for when you are finding an interval within which the mean of the population is likely to lie.

If we call the distribution of the heights of the orchids X, then what you have to work out is the value of a for which P(-a < X < a) = 0.95. This is just a standard normal distribution question - no need for samples. The problem that you have is that you do not know the value of the population mean, which you would need to find the exact value of a. But you can use your sample mean as an estimate for the population mean, which won't give you a perfect answer, but that is why you are asked for an estimate of the limits rather than the actual limits.
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Angels1234
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(Original post by Pangol)
The question is asking about the interval within which 95% of the individual orchids lie. In a way, this is the same as finding a confidence interval, but that wording is reserved for when you are finding an interval within which the mean of the population is likely to lie.

If we call the distribution of the heights of the orchids X, then what you have to work out is the value of a for which P(-a < X < a) = 0.95. This is just a standard normal distribution question - no need for samples. The problem that you have is that you do not know the value of the population mean, which you would need to find the exact value of a. But you can use your sample mean as an estimate for the population mean, which won't give you a perfect answer, but that is why you are asked for an estimate of the limits rather than the actual limits.
Ohhh okay , so are you saying that in this type of question it’s different from confidence intervals in the sense we are looking at the individual orchids in the sample provided (only 1 sample )rather than looking at the entire population? But it’s similar in the sense that we are looking for a 95 % probability. So are we still looking at the sample mean provided for the one sample provided and trying to find the intervals between the sample mean that give a 95% probability. Or are we not looking at the sample mean at all?
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Pangol
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(Original post by Angels1234)
Ohhh okay , so are you saying that in this type of question it’s different from confidence intervals in the sense we are looking at the individual orchids in the sample provided (only 1 sample )rather than looking at the entire population? But it’s similar in the sense that we are looking for a 95 % probability. So are we still looking at the sample mean provided for the one sample provided and trying to find the intervals between the sample mean that give a 95% probability. Or are we not looking at the sample mean at all?
The question is asking you to estimate the limits within which 95% of the heights of the orchids in the population lie within. To do this properly, you would need the population mean and standard deviation. But you only have the population standard deviation, so you have to use the sample mean as the next best thing. This is why your answer will only be an estimate.

This is therefore just a standard normal distribution question, equivalent to "find a when P(-a < X < a) = 0.95". It is about the population. The only connection to the sample is that you have to use it so that you have an estimated value for the population mean.

In (b) you use the one sample that you have in order to find a confidence interval for the population mean. This does include using the sample mean again, yes, but it is for a different purpose. There is still no guarantee that the real population mean is inside the confidence interva, of course.
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Angels1234
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(Original post by Pangol)
The question is asking you to estimate the limits within which 95% of the heights of the orchids in the population lie within. To do this properly, you would need the population mean and standard deviation. But you only have the population standard deviation, so you have to use the sample mean as the next best thing. This is why your answer will only be an estimate.

This is therefore just a standard normal distribution question, equivalent to "find a when P(-a < X < a) = 0.95". It is about the population. The only connection to the sample is that you have to use it so that you have an estimated value for the population mean.

In (b) you use the one sample that you have in order to find a confidence interval for the population mean. This does include using the sample mean again, yes, but it is for a different purpose. There is still no guarantee that the real population mean is inside the confidence interva, of course.
Okay so are you basically saying that we are trying to work out what the interval is but to do this we would need the population mean (if they gave the pop mean would I just do that plus or minus 1.6449 to get the 95% interval ).. but because they don’t give population mean we have to use a sample mean to try and estimate the population mean . So am I allowed to think of it as doing the standard confidence interval procedure using the sample mean (xbar) as an estimate for the population mean but as sample size is one then the root n part of the formala would be one so the formula becomes xbar +\- 1.6449 x sigma

One question I have is that if we are still using the sample mean of one to try and get the 95% interval for the true population mean then isn’t there no real difference between part A and B except the fact that on this occasion there is only one sample ?

Thanks for your time and help
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Pangol
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(Original post by Angels1234)
Okay so are you basically saying that we are trying to work out what the interval is but to do this we would need the population mean (if they gave the pop mean would I just do that plus or minus 1.6449 to get the 95% interval ).. but because they don’t give population mean we have to use a sample mean to try and estimate the population mean . So am I allowed to think of it as doing the standard confidence interval procedure using the sample mean (xbar) as an estimate for the population mean but as sample size is one then the root n part of the formala would be one so the formula becomes xbar +\- 1.6449 x sigma
Yes - but I think it is a bit unhelpful to talk about "a sample of size 1", because I think that this is where the confusion is arrising. The method does look the same as if you work on this basis, but this is really nothing to do with a sample. If I gave you a normal distribution X with a known mean and variance, you would be able to work out the value of a such that P(-a < X < a) = 0.95 in exactly the way that you describe, but there is no need to make any reference to samples. You can do this just with your knowledge of the normal distribution.

The only reason that the sample in this question comes into this is that you don't know the population mean, so you need to estimate it using the given sample.

(Original post by Angels1234)
One question I have is that if we are still using the sample mean of one to try and get the 95% interval for the true population mean then isn’t there no real difference between part A and B except the fact that on this occasion there is only one sample ?
I would say that they are about different things, although the actual methods are quite similar. Part (a) is about finding an interval within which 95% of the whole distribution lies. Part (b) is about finding an intervial within which the population mean lies, with 95% probability.
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Angels1234
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(Original post by Pangol)
I would say that they are about different things, although the actual methods are quite similar. Part (a) is about finding an interval within which 95% of the whole distribution lies. Part (b) is about finding an intervial within which the population mean lies, with 95% probability.
Oh yeah that makes more sense now . Sorry for all these questions I do apologise I just don’t want to get confused in the exam

I think I defo see the difference in the questions now. So if part A is looking for an interval where 95 percent of the data lies then why do we need to involve the sample mean ?
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Pangol
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(Original post by Angels1234)
I think I defo see the difference in the questions now. So if part A is looking for an interval where 95 percent of the data lies then why do we need to involve the sample mean ?
(Original post by Pangol)
The only reason that the sample in this question comes into this is that you don't know the population mean, so you need to estimate it using the given sample.
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