Finding the cutoff Xmin where we reject the null hypothesis

[t.kirschner99]

Homework Statement

A random sample of X1, X2, · · · , Xn is taken from a population of values that is modeled by the following probability density function:

f(x_i; β) = e^-(x_i-β), x_i≥β

Suppose we test the hypothesis:

H_o: β = 1 , H_A: β > 1

a) Suppose you are to test the above hypothesis based on a random sample of n = 10 data points and regulating the probability of committing Type I Error to be 0.05. State the values of X_min in that would indicate that β > 1.

Homework Equations

f(X_min;β) = ne^{-n(X_min - β)}, X_min ≥ β

The Attempt at a Solution

Getting stuck at the end due to the negative natural log situation that develops... was wondering if anyone would be able to point out the mistake I have made. The scribble at the end there ended up giving me a negative "c" value, which does not make sense as it is less than β.

Thanks!

 

[tnich]

Homework Statement

Suppose you are to test the above hypothesis based on a random sample of n = 10 data points and regulating the probability of committing Type I Error to be 0.05. State the values of X_min in that would indicate that β > 1.

H_o: β = 1 , H_A: β > 1

Homework Equations

f(X_min;β) = ne^{-n(X_min - β)}, X_min ≥ β

The Attempt at a Solution

View attachment 224175

Getting stuck at the end due to the negative natural log situation that develops... was wondering if anyone would be able to point out the mistake I have made. The scribble at the end there ended up giving me a negative "c" value, which does not make sense as it is less than β.

Thanks!

It would be nice to have the full problem statement. What is the hypothesis you are testing?
As to your negative c value, -ln(x) is not necessarily negative. It depends on the value of x.

 

[tnich]

It would be nice to have the full problem statement. What is the hypothesis you are testing?
As to your negative c value, -ln(x) is not necessarily negative. It depends on the value of x.

Oh, OK, I see that $c= 1 + \frac {ln(.05)} {10} - 1$ would be negative.

 

[t.kirschner99]

It would be nice to have the full problem statement. What is the hypothesis you are testing?
As to your negative c value, -ln(x) is not necessarily negative. It depends on the value of x.

I have added the whole problem to the original post. The Xmin formula was given to us in the question.

 

[tnich]

I have added the whole problem to the original post. The Xmin formula was given to us in the question.

I can't read your solution photo, so let's start at the beginning. What plan are you following in solving this problem?

 

[t.kirschner99]

I can't read your solution photo, so let's start at the beginning. What plan are you following in solving this problem?

Sorry about the quality of the photo. I'll include a new photo at the bottom of this post.

For my strategy in this problem, the question asked for me to find the cutoff value using the Type I error of 0.05. Thus, I started with the Type I error definition which is...

α = P(Reject Ho | Ho is true)

The first part of the above probability statement is using the Xmin, while the last part is using the hypothesis given to us, therefore the completed probability statement is:

0.05 = P(X_min < c | β = 1)

And from there, I did the work outlined in the photo below:

If the photo is still unreadable, i'll go ahead and type it up.

EDIT: I'll say that the embedded photo is a lot more blurry than the one on the imgur link. Here is the direct link to it: https://imgur.com/a/Dj4yN

 

[tnich]

Sorry about the quality of the photo. I'll include a new photo at the bottom of this post.

For my strategy in this problem, the question asked for me to find the cutoff value using the Type I error of 0.05. Thus, I started with the Type I error definition which is...

α = P(Reject Ho | Ho is true)

The first part of the above probability statement is using the Xmin, while the last part is using the hypothesis given to us, therefore the completed probability statement is:

0.05 = P(X_min < c | β = 1)

And from there, I did the work outlined in the photo below:

View attachment 224229

If the photo is still unreadable, i'll go ahead and type it up.

EDIT: I'll say that the embedded photo is a lot more blurry than the one on the imgur link. Here is the direct link to it: https://imgur.com/a/Dj4yN

In the lines you crossed out, you took $ln(e^{-10c} + \frac {0.05} {e^{10}})$ and got $-10c+ln(\frac {0.05} {e^{10}})$. Do you see why that is not correct?

Here something that would simplify your derivation: What is $\frac d{dx} e^{-n(x-β)}$?

Another useful fact in problems like this is,
for $x_i$ iid and $x_{min} \equiv \begin{matrix} minimum\\i \end{matrix}x_i$,
$P(x_{min}>c)=\prod_{i=1}^n P(x_i>c)=P^n(x>c)$.

This makes it easy to get the cdf of $x_{min}$.

 

[t.kirschner99]

In the lines you crossed out, you took $ln(e^{-10c} + \frac {0.05} {e^{10}})$ and got $-10c+ln(\frac {0.05} {e^{10}})$. Do you see why that is not correct?

Here something that would simplify your derivation: What is $\frac d{dx} e^{-n(x-β)}$?

Another useful fact in problems like this is,
for $x_i$ iid and $x_{min} \equiv \begin{matrix} minimum\\i \end{matrix}x_i$,
$P(x_{min}>c)=\prod_{i=1}^n P(x_i>c)=P^n(x>c)$.

This makes it easy to get the cdf of $x_{min}$.

a) Yes, ln(x+y) does not equal ln(x) + ln(y), only ln(xy) = ln(x) + ln(y). Think that scribble on the bottom was my brain losing grasp of reality lol.
b) $\frac d{dx} e^{-n(x-β)}$ = $-n e^{-n(x-β)}$. Therefore, the anti-derivative is $\frac {-1}{n} e^{-n(x-β)}$.
c) Before getting to the last part of your response, I realized that I may have over simplified (leading to a wrong simplification) in my original response. I went ahead and redid it without separating the β term from the integrand and got an answer of 1.005129 for c. Unfortunately, I don't have the answer, but it is a value that is greater than 1.

Now my question for the $P(x_{min}>c)=\prod_{i=1}^n P(x_i>c)=P^n(x>c)$ term you stated. Is it necessary to use that in the question or is the answer attainable by evaluating the anti-derivative above between the bounds [1,c]? I would assume if you use the equation you stated above, you would have to assume that α = P(X_min > c | β=1)?

Thanks for your help with this!

 

[tnich]

I went ahead and redid it without separating the β term from the integrand and got an answer of 1.005129 for c.

That is the correct answer.

Now my question for the $P(x_{min}>c)=\prod_{i=1}^n P(x_i>c)=P^n(x>c)$ term you stated. Is it necessary to use that in the question or is the answer attainable by evaluating the anti-derivative above between the bounds [1,c]? I would assume if you use the equation you stated above, you would have to assume that α = P(X_min > c | β=1)?

No, this is not required to work the problem since you already have $f(x_i; β) = e^{-(xi-β)}$. It would give you an alternate way to work the problem. If you start with $α = P(X_{min} \leq c | β=1)$, then $1-α = P(X_{min} > c | β=1)$.

 

[t.kirschner99]

That is the correct answer.No, this is not required to work the problem since you already have $f(x_i; β) = e^{-(xi-β)}$. It would give you an alternate way to work the problem. If you start with $α = P(X_{min} \leq c | β=1)$, then $1-α = P(X_{min} > c | β=1)$.

Thanks for the help on this question!

 

[tnich]

State the values of Xmin in that would indicate that β > 1.

I went ahead and redid it without separating the β term from the integrand and got an answer of 1.005129 for c.

Actually, this is not quite the correct answer, the question asks for "the values of Xmin in that would indicate that β > 1". So you will need to state an interval of values for c.