Statistics:minimizing an interval for a standard normal distribution

In summary: WV0aXZlLCBpbnN1bWVkIHJlcG9ydHMgYW5kIGJlIGNoYW5nZWQgb2YgY29udGVudCBub3QgaW5jbHVkZWQgaW4gcCA8IDEsMjE0KSwgc2hvdyB0aGF0IHRoZSBtZXRob2QgZ2l2ZSBhbmQgW1gtaW4gaW5kZXggZnVuY3Rpb24gb2YgdGhlIGRpc3RyaWJ1dGlvbiBmdW5jdGlvbil
  • #1
k3k3
78
0

Homework Statement


Generalize: For arbitrary 0 < p < 1, show that the method giving a and b produces the minimum length interval.

Hint: It might be helpful to use local extrema for the inverse function of the distribution function.


Homework Equations


The method is is talking about is locating the z scores using (1-p)/2 and [1-(1-p)/2]


The Attempt at a Solution


Let a be the area on the tail end of the distribution not included in p
Let b be the other end so that

a+b=1-p and b=1-p-a

Then the points A and B are the end points of the interval containing p.

B-A = (F^-1)(p+a)-(F^-1)(a)

This is where I am stuck. I know f(y). So d/dy(F(y))=f(y) and then (f^-1)'(y)=1/(f'(f^-1)(y))

I am not sure how to proceed.
 
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  • #2
k3k3 said:

Homework Statement


Generalize: For arbitrary 0 < p < 1, show that the method giving a and b produces the minimum length interval.

Hint: It might be helpful to use local extrema for the inverse function of the distribution function.


Homework Equations


The method is is talking about is locating the z scores using (1-p)/2 and [1-(1-p)/2]


The Attempt at a Solution


Let a be the area on the tail end of the distribution not included in p
Let b be the other end so that

a+b=1-p and b=1-p-a

Then the points A and B are the end points of the interval containing p.

B-A = (F^-1)(p+a)-(F^-1)(a)

This is where I am stuck. I know f(y). So d/dy(F(y))=f(y) and then (f^-1)'(y)=1/(f'(f^-1)(y))

I am not sure how to proceed.

It is helpful to assume the normal distribution is standard (mean = 0, variance = 1). Do you see why you can do that without loss of generality? Now, if F(z) is the standard normal cdf, you want to minimize b-a, subject to the constraint F(b) - F(a) = p. Hint: Lagrange multipliers.

RGV
 

FAQ: Statistics:minimizing an interval for a standard normal distribution

What is a standard normal distribution?

A standard normal distribution is a specific type of probability distribution that has a mean of 0 and a standard deviation of 1. It is often used in statistics to model continuous data that follows a symmetric, bell-shaped curve.

Why is it important to minimize the interval for a standard normal distribution?

Minimizing the interval for a standard normal distribution is important because it allows for more precise estimates of probabilities and confidence intervals. By reducing the interval, we can increase the accuracy of our statistical analysis and make more informed decisions.

How can an interval be minimized for a standard normal distribution?

An interval for a standard normal distribution can be minimized by increasing the sample size, reducing the level of confidence, or using a more accurate estimation method such as the central limit theorem. Additionally, using a more precise measurement or reducing the variability in the data can also help minimize the interval.

What is the relationship between interval width and confidence level for a standard normal distribution?

The interval width and confidence level for a standard normal distribution have an inverse relationship. This means that as the confidence level increases, the interval width also increases. Similarly, as the confidence level decreases, the interval width decreases.

How do you interpret the interval for a standard normal distribution?

The interval for a standard normal distribution can be interpreted as the range of values within which the true population mean is likely to fall. For example, if the 95% confidence interval is 1.5 to 2.5, we can say with 95% confidence that the true population mean falls within this range. It is important to note that this interpretation is probabilistic and not absolute.

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