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shreya_2003
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#1
for part b
H0: p=27.5
H1: p>27.5
N~X(27.5,3/root60)
Im not sure what the
N> would be to test it against
in the previous trial i got the mean as 28 so would it would it be (N>28) and comapare that to 5%
H0: p=27.5
H1: p>27.5
N~X(27.5,3/root60)
Im not sure what the
N> would be to test it against
in the previous trial i got the mean as 28 so would it would it be (N>28) and comapare that to 5%
Last edited by shreya_2003; 1 year ago
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username1732133
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#2
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#2
(Original post by shreya_2003)
for part b
H0: p=27.5
H1: p>27.5
N~X(27.5,3/root60)
Im not sure what the
N> would be to test it against
for part b
H0: p=27.5
H1: p>27.5
N~X(27.5,3/root60)
Im not sure what the
N> would be to test it against
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Blue112225
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#3
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#3
(Original post by shreya_2003)
for part b
H0: p=27.5
H1: p>27.5
N~X(27.5,3/root60)
Im not sure what the
N> would be to test it against
in the previous trial i got the mean as 28 so would it would it be (N>28) and comapare that to 5%
for part b
H0: p=27.5
H1: p>27.5
N~X(27.5,3/root60)
Im not sure what the
N> would be to test it against
in the previous trial i got the mean as 28 so would it would it be (N>28) and comapare that to 5%
Then test the mean that you have observed as your test statistic against that distribution
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mqb2766
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#4
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#4
(Original post by shreya_2003)
for part b
H0: p=27.5
H1: p>27.5
N~X(27.5, 3/root60)
Im not sure what the
N> would be to test it against
for part b
H0: p=27.5
H1: p>27.5
N~X(27.5, 3/root60)
Im not sure what the
N> would be to test it against
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shreya_2003
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#5
(Original post by Blue112225)
You have to say that X̄ (the mean of your sample) is normally distributed as: X̄~N(28, 10.24/60)
Then test the mean that you have observed as your test statistic against that distribution
You have to say that X̄ (the mean of your sample) is normally distributed as: X̄~N(28, 10.24/60)
Then test the mean that you have observed as your test statistic against that distribution
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mqb2766
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#6
(Original post by shreya_2003)
where did you get the 10.24 from ?
where did you get the 10.24 from ?
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Blue112225
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#7
(Original post by shreya_2003)
where did you get the 10.24 from ?
where did you get the 10.24 from ?
(does give a nice std deviation though)
Last edited by Blue112225; 1 year ago
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shreya_2003
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#8
(Original post by mqb2766)
Ir would probably be better to upload your working for part a) then write down the hypothesis test clearly?
Ir would probably be better to upload your working for part a) then write down the hypothesis test clearly?
SD= root (47654.4/60 -28^2)= 3.2
b) normal distributed X~N(28, 3^2) new variance = root(3^/60)
test stat
H0: p=28
H1=p>28
testing for (X>27.5) to which i got 0.9016 which seems way too big to compare to 0.05
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#9
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#9
(Original post by shreya_2003)
a) mean = 1680/60 = 28
SD= root (47654.4/60 -28^2)= 3.2
b) normal distributed X~N(28, 3^2) new variance = root(3^/60)
test stat
H0: p=28
H1=p>28
testing for (X>27.5) to which i got 0.9016 which seems way too big to compare to 0.05
a) mean = 1680/60 = 28
SD= root (47654.4/60 -28^2)= 3.2
b) normal distributed X~N(28, 3^2) new variance = root(3^/60)
test stat
H0: p=28
H1=p>28
testing for (X>27.5) to which i got 0.9016 which seems way too big to compare to 0.05
Last edited by mqb2766; 1 year ago
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#10
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#10
(Original post by mqb2766)
That is not correct. The new time trial has mean 28 and std dev 3/sqrt(60)?
That is not correct. The new time trial has mean 28 and std dev 3/sqrt(60)?
Also, keep using the exact values
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shreya_2003
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#11
(Original post by mqb2766)
That is not correct. The new time trial has mean 28 and std dev 3/sqrt(60)?
That is not correct. The new time trial has mean 28 and std dev 3/sqrt(60)?
Variance before was a typo it was meant to be 3^2
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#12
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#12
(Original post by shreya_2003)
if i root the (3^2/60) variance it gives SD of 3/root60
if i root the (3^2/60) variance it gives SD of 3/root60
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shreya_2003
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#13
(Original post by Blue112225)
So when you write that X̄ is distributed normally you have to use the variance (3.2)^2/60, rather than std dev
So when you write that X̄ is distributed normally you have to use the variance (3.2)^2/60, rather than std dev
So its X~b(28,root(3^2/60))
Then writing it with the for the SD is (28,3/root60)?
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#14
(Original post by shreya_2003)
Ok so when i first write it i use variance
So its X~b(28,root(3^2/60))
Then writing it with the for the SD is (28,3/root60)?
Ok so when i first write it i use variance
So its X~b(28,root(3^2/60))
Then writing it with the for the SD is (28,3/root60)?
For the random variable Y, normally distributed with mean μ and standard deviation σ, you would write it as:
Y ~ N(μ,σ^2)
In this specific case:
X̄~N(28, 10.24/60)
Your σ^2 is (3.2/sqrt(60))^2 = 10.24/60
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shreya_2003
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#15
(Original post by Blue112225)
Sorry - I think I've confused you.
For the random variable Y, normally distributed with mean μ and standard deviation σ, you would write it as:
Y ~ N(μ,σ^2)
In this specific case:
X̄~N(28, 10.24/60)
Your σ^2 is (3.2/sqrt(60))^2 = 10.24/60
Sorry - I think I've confused you.
For the random variable Y, normally distributed with mean μ and standard deviation σ, you would write it as:
Y ~ N(μ,σ^2)
In this specific case:
X̄~N(28, 10.24/60)
Your σ^2 is (3.2/sqrt(60))^2 = 10.24/60
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#16
(Original post by shreya_2003)
But for part b it says the SD for this trial is 3?
But for part b it says the SD for this trial is 3?
They're two different distributions. In addition, when you compare or test the means, you dvide the population std dev by sqrt(n), so 3/sqrt(60), which is different again. You assume the sigma part is known. Thsi represents the std dev of the mean estimates of sample size n, so a third distribution..
Last edited by mqb2766; 1 year ago
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shreya_2003
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#17
(Original post by mqb2766)
What does your textbook or notes say about how you estimate and compare a population mean? The population std dev is 3, whch is different from the estimate or sample std dev in part a, 3.2.
They're two different distributions. In addition, when you compare or test the means, you assume you dvide the population std dev by sqrt(n), so 3/sqrt(60), which is different again. You assume the sigma part is known in the sample distribution.
What does your textbook or notes say about how you estimate and compare a population mean? The population std dev is 3, whch is different from the estimate or sample std dev in part a, 3.2.
They're two different distributions. In addition, when you compare or test the means, you assume you dvide the population std dev by sqrt(n), so 3/sqrt(60), which is different again. You assume the sigma part is known in the sample distribution.
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#18
(Original post by shreya_2003)
To do them separately and compare them?
To do them separately and compare them?
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shreya_2003
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#19
(Original post by mqb2766)
Not sure what you mean. Do you understand what youre trying to do or not? Do you need a tutorial or look over your notes or ..
Not sure what you mean. Do you understand what youre trying to do or not? Do you need a tutorial or look over your notes or ..
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#20
(Original post by shreya_2003)
A tutorial please as i tried applying my notes but im still very confused.
A tutorial please as i tried applying my notes but im still very confused.
https://www.cimt.org.uk/projects/mep...fstats_ch3.pdf
https://onlinestatbook.com/2/tests_o...ngle_mean.html
https://madasmaths.com/archive/maths...stribution.pdf
There are other more detailed explanations, just google.
As a brief comment on the above (in addition to the link/textbook/..., not replace reading it), I think the confusion arises from what normal distribution you're really comparing. You have
1) The given or assumed population of cyclists doing time trials. The mean is 27.5 and the std dev is 3 and its normal. N(27.5, 3^2)
2) Conducting one experiment, you get one data set size n. The mean of this sample is 28 and the std dev is 3.2 and its approximately normal, as its a finite data set. The data samples in this experiment are approximately N(28, 3.2^2). or 28 is an estimate of the mean and 3.2 is an estimate of the standard deviation in the underlying distribution.
3) Generally you conduct lots of experiments and collect lots of mean estimates, where each mean estimate is calculated from one experiment of sample size n. The std dev of this distribution of mean estimates is 3/sqrt(n), assuming the population std dev is given. You're forming the distribution of mean estimates. The mean estimate in item 2) is one sample from this distribution. The mean of this distribution is not known you're trying to test whether its different or not from the assumed/given population mean.
i
So you have one sample (mean) estimate 28 and the std dev of the mean estimate distribution from which its drawn is 3/sqrt(60). Youre testing to see if the given population mean is significantly different to 28.
In doing this comparison, you have to use the population std dev/sqrt(n) rather than the estimated std dev/sqrt(n) to make the ""normal stats work". Otherwise you're looking at different distributions and it gets more messy.
Last edited by mqb2766; 1 year ago
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