Maths 3A/3B

[anotherworld2b]

I was wondering what particular formulas I would need to know. I'm quite sure what the formula is for p or p hat

[RuiAce]

I was wondering what particular formulas I would need to know. I'm quite sure what the formula is for p or p hat

Your previous question involved a confidence interval. That formula is one you need to know.

There isn't a rule of thumb as to how to actually find \(p\) or \(\hat{p}\). That will depend entirely on the question given to you.

It just so happens that all of your previous questions involved the binomial distribution, so the method of finding \(p\) and \(\hat{p}\) are pretty much the same.
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Also, you can show this via explicit computation, but in general if you want a larger percent confidence interval, the interval will be wider. The intuitive explanation to this is that if you want to be more confident (that your true value lies in the confidence interval), you have to relax your restriction instead. By relaxing your interval, you're really just making it larger.

[anotherworld2b]

Your previous question involved a confidence interval. That formula is one you need to know.

There isn't a rule of thumb as to how to actually find \(p\) or \(\hat{p}\). That will depend entirely on the question given to you.

It just so happens that all of your previous questions involved the binomial distribution, so the method of finding \(p\) and \(\hat{p}\) are pretty much the same.
_________________________________________________

Also, you can show this via explicit computation, but in general if you want a larger percent confidence interval, the interval will be wider. The intuitive explanation to this is that if you want to be more confident (that your true value lies in the confidence interval), you have to relax your restriction instead. By relaxing your interval, you're really just making it larger.

I see
I was wondering does E[^p) stand for the mean of p hat?
E[^p]=E[P/100]
E[p] = mean of population portion?

I was wondering is it always 1/100 x E[p] x 100?
=1/100E[P]
=1/100 ×(100×13/18)
=13\18

[RuiAce]

I see
I was wondering does E[^p) stand for the mean of p hat?
E[^p]=E[P/100]
E[p] = mean of population portion?

I was wondering is it always 1/100 x E[p] x 100?
=1/100E[P]
=1/100 ×(100×13/18)
=13\18

Well basically yeah. For the binomial distribution this does tend to be the case.

And I'm fairly sure yes, in general the mean of the sample is the same as the mean of the population proportion.

[anotherworld2b]

I am having trouble with commenting on the results for part c. I'm not quite sure what to compare and focus on

[Shadowxo]

I am having trouble with commenting on the results for part c. I'm not quite sure what to compare and focus on

You wouldn't normally have to answer a question as ambiguous as this, but things you could mention are
- p^ is quite different to p. Reasons for this discrepancy could include
-Small sample size (only 240)
-Not a random sample - it's from "a particular region in the country" so doesn't represent the population as a whole

[anotherworld2b]

Hi I was wondering if I could get help with question 14

[anotherworld2b]

And help with part d for this question too please

[jamonwindeyer]

Hi I was wondering if I could get help with question 14

The first bit is literally just substituting the maximum and minimum values from the expression of \(0.241\pm0.06\).

As for the second bit, it's been a year since I've done confidence intervals, but pretty sure this is the job for a large scale confidence interval?



It might also be the t-distribution since variance is unknown - Sorry, I'm pretty rusty on this. Whatever style interval you have been using will likely work again, you'll be taking the 0.06 from the expression and putting it equal to the expression being added/subtracted to get the endpoints of your interval. You'll be able to draw values out of there I'm fairly certain

Second one should be similar, your answer for (c) would have endpoints that are more than 3% from the mean. You want them to be 3%, that is, the endpoints need to be \(\bar{x}\pm0.03\bar{x}\). Find the value for \(n\) that achieves this with everything else the same

[anotherworld2b]

Can I have help with part a and b?

[RuiAce]

Can I have help with part a and b?

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[anotherworld2b]

Thank you for your help
Can I also have help with this question

[RuiAce]

c) is now quite easily doable because it's just plugging in numbers to find the expected value

[anotherworld2b]

Thank you 
I was also hoping to get help with this question please

[RuiAce]

Thank you 
I was also hoping to get help with this question please

All you need to do here is to add the areas of the rectangles