I think this is about the Central Limit Theorem

• whitejac
In summary, the engineer is measuring a quantity q and takes n measurements to estimate its value, assuming a random error in each measurement. The engineer reports the average of the measurements and wants to know how many measurements are needed to be 95% sure that the final error is less than 0.1 units. Using the Central Limit Theorem and the formula Zn = (Mx - μ) / (σ / √n), it is determined that n should be between 1000 and 2000.
whitejac

Homework Statement

An engineer is measuring a quantity q. It is assumed that there is a random error in each measurement, so the engineer will take n measurements and reports the average of the measurements as the estimated value of q. Specifically, if Yi is the value that is obtained in the i'th measurement, we assume that

Yi=q+Xi,
where Xi is the error in the ith measurement. We assume that Xi's are i.i.d. with EXi=0 and Var(Xi)=4 units. The engineer reports the average of measurements

Mn = (Y1 + ... Yn) / n

How
many measurements does the engineer need to make until he is 95% sure that the final error is less than 0.1 units? In other words, what should the value of n be such that
P(q−0.1≤Mn≤q+0.1)≥0.95?

Homework Equations

Central Limit Theorem states:
Zn = (Mx - μ) / (σ / √n)

The Attempt at a Solution

So this is the formula I chose to use. It seems like a simple variable swap, but my problem is pulling n out.

P(y1 ≤ Y ≤ y2)
= P( (y1 - nμ) / (σ√n) ≤ ((X1 +... Xn) - nμ) / (σ√n)) ≤ (y2 - nμ) / (σ√n)

Then this would give ((q + 0.1) / (2√n)) - ((q - 0.1) / (2√n)) = Φ-1 (0.95) = 1.64

combining this would give:(q/2√n) + (0.1 / 2√n) - (q / 2√n)- (0.1 / 2√n) = 0.2 / 2√n = 1/√n = 1.64 ⇒ n = (1/1.64)2 = 0.37

Now... this is not an integer, and makes absolutely no sense. I am semi-confident in my process, but I think I may have made too many assumptions about the central limit theorem.

Why do you change from ##\ \sigma\over\sqrt n\ ## to ##\ \sigma\sqrt n\ ## ?

It seems your n is used for two different purposes as well

...

BvU said:
Why do you change from ##\ \sigma\over\sqrt n\ ## to ##\ \sigma\sqrt n\ ## ?

It seems your n is used for two different purposes as well

...

When you multiply by n/n you get σ√n

I would call that multiplying by n, not multiplying by n/n

whitejac said:
P(y1 ≤ Y ≤ y2)
= P( (y1 - nμ) / (σ√n) ≤ ((X1 +... Xn) - nμ) / (σ√n)) ≤ (y2 - nμ) / (σ√n)

Then this would give ((q + 0.1) / (2√n)) - ((q - 0.1) / (2√n)) = Φ-1 (0.95) = 1.64
Check it anyway !

whitejac said:
Central Limit Theorem states:
Zn = (Mx - μ) / (σ / √n)
It has to be n/n because Mx is (X1 +...+Xn)/n.

In my extrapolation of the formula you quoted, the y1 = q - 0.1, nμ = 0 and σ = √4
Which portion did I misinterpret/use twice?

The "Then" part, where you divide by n

Another comment: Your 1.64 looks one-sided to me. And your notation asks more from the reader than is reasonable; teachers shouldn't allow that ! (there is no using n twice; I got confused -- sorry)

I end up with an awful lot of measurements required, which is rather to be expected if you start out with such a large variance !

By the way, even "Var(Xi)=4 units" is confusing. Shouldn't it be Var(Xi)=4 units2 ?More (admittedly nitpicking)
Central Limit Theorem states: Zn = (Mx - μ) / (σ / √n) is normally distributed with standard deviation 1

Matter of completeness: for some readers Zn is something else: an impedance or whatever.

Last edited:
BvU said:
By the way, even "Var(Xi)=4 units" is confusing. Shouldn't it be Var(Xi)=4 units2 ?
Sorry, it ought to be. My book isn't the best despite my professor's claim. I guess they're assuming "units" is general enough.
BvU said:
The "Then" part, where you divide by n
Maybe I'm not looking at it the same way you are, I don't divide by n. I multiply by n to get y1 and y2 = q-0.1, q+0.1, respectively. This is consistent with examples in my book, I think. Are you saying I mistook these values as y1,y2 when they were in fact y1 - nμ, y2 + nμ?

I apologize for my not showing as many steps, or in the best way. It's much neater on my paper. I'll upload it when i get home if it remains unclear.

No need to apologize, I'm trying to help.
whitejac said:
P(y1 ≤ Y ≤ y2)
= P( (y1 - nμ) / (σ√n) ≤ ((X1 +... Xn) - nμ) / (σ√n)) ≤ (y2 - nμ) / (σ√n)

Then this would give ((q + 0.1) / (2√n)) - ((q - 0.1) / (2√n)) = Φ-1 (0.95) = 1.64
In the top line you have y1 for nq, in the bottom line you have q only. You only divide the numerator by n.

The 1.64 is for 5% in one tail, you want 2.5% in each tail.

--

 The more I read it, the more nonsensical it becomes:

P(y1 ≤ Y ≤ y2) = P( (y1 - nμ) / (σ√n) ≤ ((X1 +... Xn) - nμ) / (σ√n)) ≤ (y2 - nμ) / (σ√n)
What does it say here ?

Last edited:
whitejac said:
Sorry, it ought to be. My book isn't the best despite my professor's claim. I guess they're assuming "units" is general enough.

Maybe I'm not looking at it the same way you are, I don't divide by n. I multiply by n to get y1 and y2 = q-0.1, q+0.1, respectively. This is consistent with examples in my book, I think. Are you saying I mistook these values as y1,y2 when they were in fact y1 - nμ, y2 + nμ?

I apologize for my not showing as many steps, or in the best way. It's much neater on my paper. I'll upload it when i get home if it remains unclear.

I suggest you go back to square one. You need to figure out how to make a 95% probability interval (centered at the mean) come out with a length of 0.2 (i.e., from -0.1 to + 0.1). In terms of the unit normal distribution, how many standard deviations wide would that be? Then carefully figure out what that means in terms of n. When I do it I get a value of n between 1000 and 2000, but I will not state the precise value.

1. What is the Central Limit Theorem?

The Central Limit Theorem (CLT) is a statistical concept that states that the average of a large number of independent and identically distributed variables will follow a normal distribution. In simpler terms, it means that the more data you have, the closer your sample mean will be to the true population mean.

2. Why is the Central Limit Theorem important?

The CLT is important because it allows us to make inferences about a population based on a sample. This is essential in many fields of science, as it is often impossible or impractical to collect data from an entire population. The CLT also forms the basis for many statistical tests and models.

3. What are the assumptions of the Central Limit Theorem?

The main assumptions of the CLT are that the sample size is large enough (typically at least 30 observations), the observations are independent and identically distributed, and there is no significant skewness or outliers in the data. Violating these assumptions can lead to incorrect conclusions.

4. How is the Central Limit Theorem used in practice?

The CLT is used in a variety of ways in scientific research. For example, it is used to calculate confidence intervals and p-values, test hypotheses, and estimate population parameters. It is also used in quality control processes to monitor and improve product quality.

5. Are there any limitations to the Central Limit Theorem?

While the CLT is a powerful tool, it does have some limitations. It assumes that the underlying population follows a normal distribution, which may not always be the case. It also requires a sufficiently large sample size to accurately estimate the population parameters. Additionally, it may not be appropriate to use the CLT for small or skewed data sets.

• Calculus and Beyond Homework Help
Replies
10
Views
2K
• Calculus and Beyond Homework Help
Replies
1
Views
2K
• General Math
Replies
22
Views
3K
• Linear and Abstract Algebra
Replies
1
Views
2K
• Calculus and Beyond Homework Help
Replies
10
Views
6K
• Set Theory, Logic, Probability, Statistics
Replies
4
Views
1K
• Calculus and Beyond Homework Help
Replies
1
Views
2K
• Calculus and Beyond Homework Help
Replies
3
Views
2K
• Calculus and Beyond Homework Help
Replies
5
Views
6K
• Calculus and Beyond Homework Help
Replies
1
Views
6K