Edexcel A level Maths Specification question

Does anyone know if we need to learn how to calculate the sample standard deviation for the Stats and Mechanics exam, as Jack Brown does in this video: https://www.youtube.com/watch?v=ts9SWUiFn6k . I haven't come across this version once in the textbook so I'm just wondering.

Reply 1

_gcx

Study Forum Helper

Using

n - 1

is Bessel's correction, and is only relevant when you're using a sample to estimate the variance/standard deviation of the whole population, (because using

n

, on average, gives an underestimate, and

n - 1

on average, gives the right value for variance) as explained by the video. You do not cover anything about biased/unbiased estimators in single maths, so you don't need to worry about this. I think it will be accepted in the exam, because some calculators use it.

(edited 5 years ago)

Reply 2

dont know it

Original post by _gcx

Using

n - 1

is Bessel's correction, and is only relevant when you're using a sample to estimate the variance/standard deviation of the whole population, (because using

n

, on average, gives an underestimate, and

n - 1

Thanks but in a question like this: https://www.youtube.com/watch?v=aQF7L3TCI_U I'm kinda confused, would this ever come up?

Reply 3

_gcx

Study Forum Helper

Original post by dont know it

Thanks but in a question like this: https://www.youtube.com/watch?v=aQF7L3TCI_U I'm kinda confused, would this ever come up?

Ah, you would realistically use

n - 1

in that situation as you want (an unbiased estimate of) the population variance. However, the specification does say that they expect you to use

\displaystyle \sqrt{\frac{S_{xx}} n}

for sd but using

n - 1

is acceptable. So you'd be fine with either.

(edited 5 years ago)

Reply 4

dont know it

Original post by _gcx

Ah, you would realistically use

n - 1

in that situation as you want (an unbiased estimate of) the population variance. However, the specification does say that they expect you to use

\displaystyle \sqrt{\frac{S_{xx}} n}

for sd but using

n - 1

is acceptable. So you'd be fine with either.

Why do you have to divide by 8 in this question after finding the sample variance? You've found your sample variance to be 54.786, isn't that all you need?

Reply 5

_gcx

Study Forum Helper

Original post by dont know it

Why do you have to divide by 8 in this question after finding the sample variance? You've found your sample variance to be 54.786, isn't that all you need?

It's a result you need to know, that if you have

n

independent random variables

X_1, X_2, \ldots, X_n

each with distribution

N(\mu, \sigma^2)

, then their mean

\bar X

has distribution

\displaystyle N\left(\mu, \frac{\sigma^2} n\right)

. We know that if the null hypothesis is true, then

\bar X \sim N\left(\mu, \frac{\sigma^2} n\right)

, and we want to check whether our observation

\bar X

is consistent with this, by either finding a critical region appropriate to our significance level or checking p-values.

Reply 6

begbie68

no. you've answered this yourself in your OP. https://www.youtube.com/watch?v=ts9SWUiFn6k

Original post by dont know it

Why do you have to divide by 8 in this question after finding the sample variance? You've found your sample variance to be 54.786, isn't that all you need?

Reply 7

begbie68

strangely, confusingly (and probably wrongly) this was the formula given for Standard Deviation by MEI in the old S1 spec. However, the MS would (usually) allow sqrt(

\sigma

^2/n)

Original post by _gcx

Using

n - 1

is Bessel's correction, and is only relevant when you're using a sample to estimate the variance/standard deviation of the whole population, (because using

n

, on average, gives an underestimate, and

n - 1

(edited 5 years ago)

Reply 8

dont know it

Original post by begbie68

no. you've answered this yourself in your OP. https://www.youtube.com/watch?v=ts9SWUiFn6k

Ah so when we find the variance initially, we are finding the variance of the 8 samples. To find the variance of the sample means, we must divide this by the number of samples.