Finding the cutoff Xmin where we reject the null hypothesis

  • Thread starter t.kirschner99
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In summary: This equation can be rearranged to get,\frac {dx}{dt} = -βx_i+nx_{min}which is the equation you were trying to solve for ##c##.
  • #1
t.kirschner99
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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(xi; β) = e-(xi-β), xi≥β

Suppose we test the hypothesis:

Ho: β = 1 , HA: 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 Xmin in that would indicate that β > 1.

Homework Equations


f(Xmin;β) = ne-n(Xmin - β), Xmin ≥ β

The Attempt at a Solution



1Z2PqSL.jpg


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!
 

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  • #2
t.kirschner99 said:

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 Xmin in that would indicate that β > 1.

Ho: β = 1 , HA: β > 1

Homework Equations


f(Xmin;β) = ne-n(Xmin - β), Xmin ≥ 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.
 
  • #3
tnich said:
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.
 
  • #4
tnich said:
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.
 
  • #5
t.kirschner99 said:
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?
 
  • #6
tnich said:
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(Xmin < c | β = 1)

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

KuCpOD4.jpg


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
 

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  • #7
t.kirschner99 said:
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(Xmin < 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}##.
 
  • #8
tnich said:
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(Xmin > c | β=1)?

Thanks for your help with this!
 
  • #9
t.kirschner99 said:
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(Xmin > 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)##.
 
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  • #10
tnich said:
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!
 
  • #11
State the values of Xmin in that would indicate that β > 1.
t.kirschner99 said:
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.
 

Related to Finding the cutoff Xmin where we reject the null hypothesis

1. What is the significance of finding the cutoff Xmin in rejecting the null hypothesis?

The cutoff Xmin is a critical value used in hypothesis testing to determine whether to reject the null hypothesis. It represents the minimum value of a test statistic that is required to reject the null hypothesis. If the test statistic falls below this cutoff value, we fail to reject the null hypothesis.

2. How is the cutoff Xmin calculated?

The cutoff Xmin is calculated based on the level of significance chosen for the hypothesis test. It is determined using statistical tables or software programs, and it corresponds to a specific p-value. The lower the level of significance, the higher the cutoff Xmin value will be.

3. What factors determine the value of the cutoff Xmin?

The value of the cutoff Xmin is primarily determined by the level of significance chosen for the hypothesis test. Other factors that can influence the cutoff Xmin include the sample size, the type of test being performed, and the assumptions made about the data.

4. What happens when the test statistic falls below the cutoff Xmin?

If the test statistic falls below the cutoff Xmin, it means that the results are not statistically significant and we fail to reject the null hypothesis. This suggests that there is not enough evidence to support the alternative hypothesis and we must accept the null hypothesis as the best explanation for the data.

5. Can the cutoff Xmin value change for different hypothesis tests?

Yes, the cutoff Xmin value can change for different hypothesis tests. It is based on the specific parameters and assumptions of the test being performed. For example, a t-test may have a different cutoff Xmin value than a chi-square test. It is important to choose the appropriate test and corresponding cutoff Xmin value based on the research question and data being analyzed.

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