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Normal Mean Hypothesis Testing

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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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Reply 1
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?
Reply 2
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)
Reply 3
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?
Reply 4
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
Reply 5
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 :smile:

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.
Reply 6
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 :smile:

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
Reply 7
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.
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 zz value is 1.961.96.

Then we see whether we have the interval 19.78±1.96(σ10)19.78 \pm 1.96 \cdot \left( \dfrac{\sigma}{\sqrt{10}} \right) entirely below 2020, so we test whether 19.78+1.96σ10<2019.78 + 1.96 \cdot \dfrac{\sigma}{\sqrt{10}} < 20

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

We can say that σ20.811\sigma^2 \approx 0.811 since the sample variance is an unbiased estimate of the population variance so we can finish it off.
(edited 6 years ago)
Reply 9
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 zz value is 1.961.96.

Then we see whether we have the interval 19.78±1.96(σ10)19.78 \pm 1.96 \cdot \left( \dfrac{\sigma}{\sqrt{10}} \right) entirely below 2020, so we test whether 19.78+1.96σ10<2019.78 + 1.96 \cdot \dfrac{\sigma}{\sqrt{10}} < 20

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

We can say that σ20.811\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

σ2=0.811\sigma^2 = 0.811

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

XˉN(μ,0.81110)\bar{X} \sim N\left(\mu, \frac{0.811}{10}\right)

and H0:μ=20H_0 : \mu = 20, H1:μ<20H_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.
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 μ\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 :smile:

The answer I found:

Spoiler

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.
(edited 6 years ago)
Original post by RDKGames

We can say that σ20.811\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 s2=nn1σx2\displaystyle s^2=\frac{n}{n-1}\sigma_x^2, for your unbiased estimator.

Where σx2\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.
(edited 6 years ago)
Reply 13
Original post by ghostwalker
:holmes: I think we have a couple of issues here:

The sample variance, s2s^2, is not an unbiased estimate of the population variance.

Rather σ^2=nn1s2\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 :smile:

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

The sample variance, s2s^2, is not an unbiased estimate of the population variance.

Rather σ^2=nn1s2\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).
Original post by Notnek
Ah this is further maths stuff which is why I don't know it yet :smile:

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?
Original post by old_engineer
Hmmm might be some terminology issues here.


Could well be.

I've always used σ2\sigma^2 for the population, and s2s^2 for the sample.
Reply 17
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.
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
Reply 19
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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