anonymous
  • anonymous
Help with confidence interval problem (screenshot attached)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
@ganeshie8 @tkhunny @jim_thompson5910 X)
anonymous
  • anonymous
Would I just simply use\[\huge \text{Z}_\alpha=\frac{\bar{x}-\mu}{\sigma}\]and solve for \(\mu\)?
ganeshie8
  • ganeshie8
whats the z* value for 99% confidence level ?

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anonymous
  • anonymous
-2.33
anonymous
  • anonymous
Oh, so it'd just simply have to be greater than 2.33?
ganeshie8
  • ganeshie8
|dw:1449638671166:dw|
ganeshie8
  • ganeshie8
that table says \(Z^{\star} = 2.576\) ?
anonymous
  • anonymous
Ah. On the chart I have it's between -2.32 and -2.33
ganeshie8
  • ganeshie8
Make sure you're not looking at single tail sheet
anonymous
  • anonymous
Wouldn't it be single tailed though? Because it only mentioned that it's greater than certain value.
ganeshie8
  • ganeshie8
Oh, thats right !
anonymous
  • anonymous
I just used that \(\alpha=0.01\)
anonymous
  • anonymous
Ok haha I thought I was going crazy. Where can I go from here with this information?
ganeshie8
  • ganeshie8
find the margin of error by multiplying "standard error" and "Z*"
anonymous
  • anonymous
Okay, I got -0.8621
ganeshie8
  • ganeshie8
margin of error = 0.37*2.33 = 0.86
anonymous
  • anonymous
Yay X)
ganeshie8
  • ganeshie8
margin of error is always positive
anonymous
  • anonymous
Ah, okay.
ganeshie8
  • ganeshie8
|dw:1449639292614:dw|
ganeshie8
  • ganeshie8
subtract margin of error from the mean to get the lower bound of true mean
anonymous
  • anonymous
Alright, I have 134.5279.
ganeshie8
  • ganeshie8
looks good to me!
anonymous
  • anonymous
Just curious, why is it the lower bound?
anonymous
  • anonymous
Having a hard time in general with this material with picturing what's going on
ganeshie8
  • ganeshie8
That is due to central limit theorem
ganeshie8
  • ganeshie8
what does CLT say about the "sampling distribution" ?
ganeshie8
  • ganeshie8
let me ask this question what is a sampling distribution ?
anonymous
  • anonymous
The probability of a statistic based on a sample from a population, yeah?
anonymous
  • anonymous
We can use it to make inferences about the population directly from the sample
ganeshie8
  • ganeshie8
"sampling distribution" as the name says, it is a distribution
ganeshie8
  • ganeshie8
distribution of what ?
anonymous
  • anonymous
I'm not sure :S
ganeshie8
  • ganeshie8
suppose you want to know the average height of ppl in your country
ganeshie8
  • ganeshie8
it is impossible to collect data of each and everyone in your country
anonymous
  • anonymous
True, that makes sense. So we try to take multiple samples right?
ganeshie8
  • ganeshie8
so you just select "n" people randombly and take their average height
ganeshie8
  • ganeshie8
Now is the real question
ganeshie8
  • ganeshie8
How confident are you about your sample ?
ganeshie8
  • ganeshie8
How do you know that the "average height in your sample" is same as the "average height of all the people in your country" ?
anonymous
  • anonymous
Definitely not 100% because you don't know ALL of the heights
ganeshie8
  • ganeshie8
right, that is where we use central limit theorem
anonymous
  • anonymous
Which means that we can approximate the sample as normal, right?
ganeshie8
  • ganeshie8
what do you mean the sampel is normal ?
anonymous
  • anonymous
I mean, it will follow a normal distribution
ganeshie8
  • ganeshie8
CLT talks about this : Suppose you have one sample of size \(n\) with average height = \(\overline{x}\) and standard deviation of \(\sigma\). Now take multiple samples of the same size \(n\) and find the average height in each of those samples.
ganeshie8
  • ganeshie8
You have a collection of average height from samples. CLT says that this collection of average height from samples is a normal distribution
anonymous
  • anonymous
Ahhh, I guess I was kinda close X)
anonymous
  • anonymous
So how does that lower bound come into play here?
ganeshie8
  • ganeshie8
As you can see, there is a very less chance for the average height of your single sampel to be same as the average height of the total population
ganeshie8
  • ganeshie8
Would you agree that there are more chances for the population mean to lie somewhere around your sample mean ?
ganeshie8
  • ganeshie8
Using confidence intervals, we give some range of values around the sample mean in which the population mean lies with certain confidence
ganeshie8
  • ganeshie8
Let me ask you another question If we want to be more confident about the population mean to lie in our interval, do we need a wider interval or a shorter interval ?
anonymous
  • anonymous
We'd want a wider interval, correct?
ganeshie8
  • ganeshie8
Yes, CLT also gives you the \(\overline{x}\) and \(\sigma\) of the sampling distribution : 1) mean of sampling distribution is same as the mean of a sample 2) standard deviation of sampling distribution depends on size of sample : \(\dfrac{\sigma}{\sqrt{n}}\)
ganeshie8
  • ganeshie8
Once you have the "sampling" distribution, you can find the confidence intervals using zscores
anonymous
  • anonymous
Ohh, I think I'm getting the idea now
ganeshie8
  • ganeshie8
Under normal curve, you know that 99% of the area lies after the zscore of -2.33 therefore, to say that the true population mean is greater than some X with 99% confidence, you need to take the observation(X) corresponding to the zscore of -2.33
anonymous
  • anonymous
Yeah that makes sense now!
ganeshie8
  • ganeshie8
single tailed 99% confidence interval looks as below : \[(\overline{x}-2.33*\dfrac{\sigma}{\sqrt{n}},~~\infty)\]
ganeshie8
  • ganeshie8
double tailed 99% confidence interval looks as below : \[(\overline{x}-2.576*\dfrac{\sigma}{\sqrt{n}},~~\overline{x}+2.576*\dfrac{\sigma}{\sqrt{n}})\]
anonymous
  • anonymous
Ahh, it's starting to look more familiar now :P
ganeshie8
  • ganeshie8
Either case, the interval captures 99% of the area under the normal curve
anonymous
  • anonymous
That's why it's only \(\alpha\) for a single tail and \(\alpha/2\) for a double tail.
anonymous
  • anonymous
Hehe, that makes more sense now. Thank you!! X)

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