Retsek
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I'm familiar with the simple straightforward mean hypothesis testing but I have no idea how to do this one, thanks in advance
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Sir Cumference
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(Original post by Retsek)
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I'm familiar with the simple straightforward mean hypothesis testing but I have no idea how to do this one, thanks in advance
What do you mean by straightforward mean hypothesis testing? Are you familiar with the distribution of the sample mean?
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Retsek
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(Original post by Notnek)
What do you mean by straightforward mean hypothesis testing? Are you familiar with the distribution of the sample mean?
I might not have all the terminology down, is that like how the variance of the sample mean is the variance of the population divided by the size of the sample (n)
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Sir Cumference
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(Original post by Retsek)
I might not have all the terminology down, is that like how the variance of the sample mean is the variance of the population divided by the size of the sample (n)
Yes that's right. Have you ever done a hypothesis test that involves the sample mean? Are you just stuck on Q25 or can you not do the ones above it also?
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Retsek
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(Original post by Notnek)
Yes that's right. Have you ever done a hypothesis test that involves the sample mean? Are you just stuck on Q25 or can you not do the ones above it also?
Just 25, it's different in the way it's worded and I felt like it didn't give you all the necessary info, although my friends said it was just like the others, I would show the others so you could figure out what I mean but it was on a test and I just snapped a picture at the end
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Sir Cumference
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(Original post by Retsek)
Just 25, it's different in the way it's worded and I felt like it didn't give you all the necessary info, although my friends said it was just like the others, I would show the others so you could figure out what I mean but it was on a test and I just snapped a picture at the end
Hmm just had another look at the question and there's one thing I need to have a think about but it's getting late. I'm still not a stats expert yet

I can have another look tomorrow but if you need an answer now I recommend starting a new thread so that it's unanswered and someone else can help. You may not get an answer tonight though.
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Retsek
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(Original post by Notnek)
Hmm just had another look at the question and there's one thing I need to have a think about but it's getting late. I'm still not a stats expert yet

I can have another look tomorrow but if you need an answer now I recommend starting a new thread so that it's unanswered and someone else can help. You may not get an answer tonight though.
No worries, I'm not in any rush to figure it out, thanks for your help though
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Sir Cumference
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(Original post by Retsek)
No worries, I'm not in any rush to figure it out, thanks for your help though
Please remind me tomorrow if I forget and no one else has replied. I want to have another look at this.
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RDKGames
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(Original post by Retsek)
I'm familiar with the simple straightforward mean hypothesis testing but I have no idea how to do this one, thanks in advance
I haven't done any of this 'significance level' and 'hypothesis testing' at all, but I've done confidence intervals at A-Level (a while ago) which might be useful here but don't take my word for it since I could be wrong somewhere.

Here's what I think:

The significance level is 5% which is 2.5% on either side of the distribution hence our z value is 1.96.

Then we see whether we have the interval 19.78 \pm 1.96 \cdot \left( \dfrac{\sigma}{\sqrt{10}} \right) entirely below 20, so we test whether 19.78 + 1.96 \cdot \dfrac{\sigma}{\sqrt{10}} < 20

If so, we can say with 95% confidence that \mu < 20 hence accept the claim.

We can say that \sigma^2 \approx 0.811 since the sample variance is an unbiased estimate of the population variance so we can finish it off.
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Sir Cumference
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(Original post by RDKGames)
I haven't done any of this 'significance level' and 'hypothesis testing' at all, but I've done confidence intervals at A-Level (a while ago) which might be useful here but don't take my word for it since I could be wrong somewhere.

Here's what I think:

The significance level is 5% which is 2.5% on either side of the distribution hence our z value is 1.96.

Then we see whether we have the interval 19.78 \pm 1.96 \cdot \left( \dfrac{\sigma}{\sqrt{10}} \right) entirely below 20, so we test whether 19.78 + 1.96 \cdot \dfrac{\sigma}{\sqrt{10}} < 20

If so, we can say with 95% confidence that \mu < 20 hence accept the claim.

We can say that \sigma^2 \approx 0.811 since the sample variance is an unbiased estimate of the population variance so we can finish it off.
I'm not sure I fully understand what you've done since I don't know much about confidence intervals. Also I'm not sure this is a hypothesis test?

But what you said about the sample variance being an unbiased estimate of the population variance makes sense. So Retsek I'm thinking you can do this like a standard sample mean hypothesis test with

\sigma^2 =  0.811

So the sample mean \bar{X} will be normally distributed with

\bar{X} \sim N\left(\mu, \frac{0.811}{10}\right)

and H_0 : \mu = 20, H_1 : \mu < 20.

But I'm not sure though because I haven't come across this type of problem at A Level.

Tagging ghostwalker who may be able to confirm.
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(Original post by Notnek)
I'm not sure I fully understand what you've done since I don't know much about confidence intervals. Also I'm not sure this is a hypothesis test?
It's essentially constructing an interval around the sample mean which says that we are 95% sure that the population mean \mu is in there somewhere, hence the condition on the upper bound being < 20.

But as I said, I could be wrong and this wouldn't be what's expected in the new spec most likely, so this is just me throwing ideas around to hopefully help the more appropriate approaches come about

The answer I found:
Spoiler:
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For what it's worth, I found that we should reject the claim that \mu &lt; 20.

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Pepperpeople
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I agree with Notnek with his choice of hypotheses and the distribution of the sample mean. I believe you then, assuming the null hypothesis is true, find the probability that the sample mean is less than or equal to 19.78. If this probability is less than 0.05, then you reject the null hypothesis as it is a highly unlikely result if the sample mean is in fact 20.
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ghostwalker
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(Original post by RDKGames)
We can say that \sigma^2 \approx 0.811 since the sample variance is an unbiased estimate of the population variance so we can finish it off.
(Original post by Notnek)
I

Tagging ghostwalker who may be able to confirm.
:holmes: I think we have a couple of issues here:

The sample variance is not an unbiased estimate of the population variance.

Rather \displaystyle s^2=\frac{n}{n-1}\sigma_x^2, for your unbiased estimator.

Where \sigma_x^2 is the sample variance.

Also we are dealing with a small sample from a population of unknown variance and mean, albeit normal, so we need the t-distribution.

Edited to bring in line with Edexcel terminology.
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Sir Cumference
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(Original post by ghostwalker)
:holmes: I think we have a couple of issues here:

The sample variance, s^2, is not an unbiased estimate of the population variance.

Rather \displaystyle \hat{\sigma}^2=\frac{n}{n-1}s^2, for your unbiased estimator.

Also we are dealing with a small sample from a population of unknown variance and mean, albeit normal, so we need the t-distribution.
Ah this is further maths stuff which is why I don't know it yet

I was trying to apply A Level maths methods because I was assuming it wasn't further maths for some reason.
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old_engineer
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(Original post by ghostwalker)
:holmes: I think we have a couple of issues here:

The sample variance, s^2, is not an unbiased estimate of the population variance.

Rather \displaystyle \hat{\sigma}^2=\frac{n}{n-1}s^2, for your unbiased estimator.

Also we are dealing with a small sample from a population of unknown variance and mean, albeit normal, so we need the t-distribution.
Hmmm might be some terminology issues here. I'm not putting myself forward as a stats expert, but the Edexcel (2008 syllabus) line on this is that, if we denote X as a sample, then Var(X) is a biased estimator for the population variance, but S^2(X) is an unbiased estimator for the population variance. In Edexcel-speak, S^2 = Var(X) x n/(n - 1), where, again, Var(X) is the the sample variance (calculated as the mean of the sample squares minus the square of the sample mean).
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GCSE2016Troop
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(Original post by Notnek)
Ah this is further maths stuff which is why I don't know it yet

I was trying to apply A Level maths methods because I was assuming it wasn't further maths for some reason.
Isn’t this S2 not further maths?
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(Original post by old_engineer)
Hmmm might be some terminology issues here.
Could well be.

I've always used \sigma^2 for the population, and s^2 for the sample.
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Sir Cumference
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(Original post by GCSE2016Troop)
Isn’t this S2 not further maths?
Maybe it's in S2 I don't know. I only know new spec stats and this would be further maths.
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GCSE2016Troop
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(Original post by Notnek)
Maybe it's in S2 I don't know. I only know new spec stats and this would be further maths.
Yeah it’s S2 hypothesis testing on old spec
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Sir Cumference
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(Original post by GCSE2016Troop)
Yeah it’s S2 hypothesis testing on old spec
I just had a look and it seems like this is S3 for Edexcel. I don't know what exam board you're using.
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