# Math Help - Geometric Distribution

1. ## Geometric Distribution

Question:

"The Geometric Distribution and Shorts in NiCad Batteries"

In their article "A Case Study of the Use of an Experimental Design
Snyder describe a series of experiments conducted in order to reduce
the proportion of cells being scrapped by a battery plant because of
internal shorts.

The experimental program was successful in reducing the proportion of
manufactured cells with internal shorts to around 0.03.

The following procedure is being developed in order to monitor the
process for increases in the true proportion of manufactured cells
with internal shorts.

Suppose that testing of batteries for internal shorts begins
on a production run in this plant for monitoring/control purposes,

let B = the number of batteries required until
the first internal short is discovered.

[a]
What is the distribution and expected value of B,
if the proportion of manufactured cells with internal shorts
remains at 0.03?

What is the distribution and expected value of B,
if the proportion of manufactured cells with internal shorts
increases to 0.10?

What happens, in general, to the expected value of B
as the proportion of manufactured cells with internal shorts
increases? Choose one: [Larger or Smaller]

[b]
Consider using B to test the following hypotheses:

H_0: % = 0.03 versus H_a: % > 0.03

where % = true proportion of manufactured cells with internal shorts.

B has been proposed as the TEST STATISTIC,
however three Rejection Regions are being considered.

A choice needs to be made concerning which of the three
Rejection Regions to implement in the procedure.

Rejection Region 1: Reject H_0 if B = 1.

Rejection Region 2: Reject H_0 if B <= 2.

Rejection Region 3: Reject H_0 if B <= 3.

[b.i]
Calculate the probability of a Type I Error for Rejection Region 1.

Calculate the probability of a Type I Error for Rejection Region 2.

Calculate the probability of a Type I Error for Rejection Region 3.

Based upon this information which Rejection Region do you recommend? WHY?

[b.ii]
Now suppose, in reality, the true proportion of manufactured cells
with internal shorts has increased to 0.10

Calculate the power for Rejection Region 1.

Calculate the power for Rejection Region 2.

Calculate the power for Rejection Region 3.

Based upon this information which Rejection Region do you recommend? WHY?

[b.iii]
For each Rejection Region determine an equation/formula/function that
relates power to %, true proportion of manufactured cells w/ internal shorts.

For Rejection Region 1: Power = ??function of %??

For Rejection Region 2: Power = ??function of %??

For Rejection Region 3: Power = ??function of %??

[b.iv]
Graph each of the above power functions on the same set of axes
use EXCEL, MATLAB, etc. to get a "nice" graph.

Place % on the x-axis and power on the y-axis.
Note: the range for both % and power is 0 to 1.

Now make a final recommendation concerning the Rejection Region
why you selected the Rejection Region you did.

[c]
Now suppose the experiment has been conducted and the observed value of B = 5.
Calculate the p-value.

(a) Given that, in batteries, an experimental process successfully reduced the proportion of manufactured cells with internal shorts to around 0.03. Let B = the number of batteries required until the first internal short is discovered. Given, the proportion of manufactured cells with internal shorts to around 0.03.

p = 0.03
q = 1 - p = 1 - 0.03 = 0.97

B ~ Geometric (p = 0.03)

P(X = x) = (0.03)*(0.97)^x for x = 1, 2,...

E(x) = q/p = 0.97/0.03 = 32.333

so, the expected value of B is 32

I'm wondering if my attempt at (a) is correct and if anyone has any helpful hints about how I should start part (b)

2. ## Re: Geometric Distribution

wouldn't $P(X=x) = (0.03)(0.97)^{x-1}$ ?

you had $x-1$ good batteries before you found the defective one at try $n$

3. ## Re: Geometric Distribution

oh, yes, you're right.

For the second part of part (a) would I use that same formula, except this time instead of using p = 0.03, I would use p = 0.10?

yes

5. ## Re: Geometric Distribution

okay, so to finish up part (a) this is what I did:

p = 0.10

q = 1 - 0.10 = 0.90

P(X = x) = (0.10)*(0.90)^(x - 1) for x = 1, 2, ...

E(x) = q/p = 0.90/0.10 = 9

So, in general, as the proportion of manufactured cells with internal shorts
increases, the expected value of B decreases.

Then for (bi)

P[Type I Error for Rejection Region 1] = P[ b = 1 | % = 0.03 ] = ??

this part I'm stuck at because I'm not sure if I would use an integral going from 0 to 0.03 of the geometric distribution or another method to calculate the probability. My teacher uses the computer software to figure out the probability, but doesn't show us how to figure it out by hand.

6. ## Re: Geometric Distribution

Would the above be correct?

Also, for (bi) would this be correct:

P[Type I Error for Rejection Region 1] = P[ b = 1 | % = 0.03 ] = the integral from 0 to 1 of (0.03)*(0.97)^(x-1) dx = 0.0304616

And then I'd use that integral again for the other rejection regions, except I would have the integral going from 0 to 2 for Rejection Region 2 and then have the integral going from 0 to 3 for Rejection Region 3

7. ## Re: Geometric Distribution

Originally Posted by AwesomeHedgehog
okay, so to finish up part (a) this is what I did:

p = 0.10

q = 1 - 0.10 = 0.90

P(X = x) = (0.10)*(0.90)^(x - 1) for x = 1, 2, ...

E(x) = q/p = 0.90/0.10 = 9
this isn't quite right. Since we never choose 0 batteries. B has a support of {1,2,3, ....}
For this support $E[B] = \dfrac 1 p$ not $\dfrac {1-p} p$ as you've used. This last is the formula for the mean when 0 is a possible value.

So, in general, as the proportion of manufactured cells with internal shorts
increases, the expected value of B decreases.
yes as it better had. If it's more likely the batteries are bad I'd expect to see a bad one faster.

Then for (bi)

P[Type I Error for Rejection Region 1] = P[ b = 1 | % = 0.03 ] = ??

this part I'm stuck at because I'm not sure if I would use an integral going from 0 to 0.03 of the geometric distribution or another method to calculate the probability. My teacher uses the computer software to figure out the probability, but doesn't show us how to figure it out by hand.
no.

Consider the experiment for Region 1. We grab a battery and test it. It either passes or not with probability p=0.03, thus

Pr[Type I error] = Pr[H0=True | rejected at B=1] = Pr[we get a bad battery first try | p=0.03] = 0.03

For region 2, we sample a battery, if it fails we reject, if it passes, sample again and pass if it passes else reject. In other words we fail on {fail} or {pass, fail}

The probability of failing is thus p+(1-p)p = 0.059. Similarly for Region 3. We fail on {fail}, {pass, fail}, {pass, pass, fail}

figure out Pr[Type I error] in region 3

8. ## Re: Geometric Distribution

Okay here's what I have:

The completed part of part (a):

When p = 0.03

q = 1 - p = 1 - 0.03 = 0.97

B ~ Geometric (p = 0.03)

P(X = x) = (0.03)*(0.97)^(x - 1) for x = 1, 2,...

E(x) = q/p = 0.97/0.03 = 32.333

so, the expected value of B is 32

Then when p = 0.10

q = 1 - 0.10 = 0.90

P(X = x) = (0.10)*(0.90)^(x - 1) for x = 1, 2, ...

E(x) = 1/p = 1/0.10 = 10

So, in general, as the proportion of manufactured cells with internal shorts increases, the expected value of B decreases.

For part (bi):

You said that the probability of a type I error for region one would be equal to 0.03. I used this formula and got the same answer:

P[Type I Error for Rejection Region 1] = P[ b = 1 | % = 0.03 ] = the integral from 0 to 1 of (0.03)*(0.97)^(x-1) dx = 0.0304616

I used the geometric distribution in the integral to show my work as for how I got 0.03.

But for the Rejection Region 2 and 3, the integral wouldn't be going from 0 to 1 anymore, that part would change.

So, P[Type I Error for Rejection Region 2] = P[ b <= 2 | % = 0.03 ] = the integral from 0 to 2 of (0.03)*(0.97)^(x-1) dx = 0.06

which is the same answer that you got, except I calculated it differently.

And then, P[Type I Error for Rejection Region 3] = P[ b <= 3 | % = 0.03 ] = the integral from 0 to 3 of (0.03)*(0.97)^(x-1) dx = 0.089

9. ## Re: Geometric Distribution

Originally Posted by AwesomeHedgehog
Okay here's what I have:

The completed part of part (a):

When p = 0.03

q = 1 - p = 1 - 0.03 = 0.97

B ~ Geometric (p = 0.03)

P(X = x) = (0.03)*(0.97)^(x - 1) for x = 1, 2,...

E(x) = q/p = 0.97/0.03 = 32.333

so, the expected value of B is 32

Then when p = 0.10

q = 1 - 0.10 = 0.90

P(X = x) = (0.10)*(0.90)^(x - 1) for x = 1, 2, ...

E(x) = 1/p = 1/0.10 = 10

So, in general, as the proportion of manufactured cells with internal shorts increases, the expected value of B decreases.

For part (bi):

You said that the probability of a type I error for region one would be equal to 0.03. I used this formula and got the same answer:

P[Type I Error for Rejection Region 1] = P[ b = 1 | % = 0.03 ] = the integral from 0 to 1 of (0.03)*(0.97)^(x-1) dx = 0.0304616

I used the geometric distribution in the integral to show my work as for how I got 0.03.

But for the Rejection Region 2 and 3, the integral wouldn't be going from 0 to 1 anymore, that part would change.

So, P[Type I Error for Rejection Region 2] = P[ b <= 2 | % = 0.03 ] = the integral from 0 to 2 of (0.03)*(0.97)^(x-1) dx = 0.06

which is the same answer that you got, except I calculated it differently.

And then, P[Type I Error for Rejection Region 3] = P[ b <= 3 | % = 0.03 ] = the integral from 0 to 3 of (0.03)*(0.97)^(x-1) dx = 0.089
you are messed up because you are including 0 in your support. You never choose 0 batteries. n = {1, 2, 3 ...}

10. ## Re: Geometric Distribution

Okay, so I tried it your way and did:

P[Type I Error for Rejection Region 1 ] = 0.03

P[Type I Error for Rejection Region 2] = p + (1 - p)*p = 0.0591 = 0.06

P[Type I Error for Rejection Region 3] = p + (1 - p)*p + (1 - p)*(1 - p)*p = 0.087

but what I do not understand is how you figured out how to use that to find the probability. I get what you're saying now with how my integrals should not include 0, but I do not get how you figured out how to do it this other way. I wish my teacher would have taught us how to do it by hand instead of using the computer to calculate the values.

11. ## Re: Geometric Distribution

Originally Posted by AwesomeHedgehog
Okay, so I tried it your way and did:

P[Type I Error for Rejection Region 1 ] = 0.03

P[Type I Error for Rejection Region 2] = p + (1 - p)*p = 0.0591 = 0.06

P[Type I Error for Rejection Region 3] = p + (1 - p)*p + (1 - p)*(1 - p)*p = 0.087

but what I do not understand is how you figured out how to use that to find the probability. I get what you're saying now with how my integrals should not include 0, but I do not get how you figured out how to do it this other way. I wish my teacher would have taught us how to do it by hand instead of using the computer to calculate the values.
I just envisioned what the guy/gal in the battery checking booth had to do.

12. ## Re: Geometric Distribution

For the Rejection Region that I'd recommend, would I choose Rejection Region 3 because it makes the rejection region better and it allows for variation to occur?

and for part (bii)

to solve for power, that whole thing just confuses me with the geometric distribution. Do you have any hints on how to start that?

13. ## Re: Geometric Distribution

For part (bii) I remembered that power = 1 - beta = P[Reject H_0 when it is false] = P[Type II Error]

Now, I just need to figure out what the formula is in order to calculate the power for each rejection region

14. ## Re: Geometric Distribution

Originally Posted by AwesomeHedgehog
For part (bii) I remembered that power = 1 - beta = P[Reject H_0 when it is false] = P[Type II Error]

Now, I just need to figure out what the formula is in order to calculate the power for each rejection region
For part (bii)

μ = 0.10

P[Type II Error for Rejection Region 1] = P[ b = 1 | μ = 0.10] = 1 - [the integral from 0 to 1 of (0.10)*(0.90)^(x - 1) dx ] = 0.8945

I'm guessing that that would be similar for the other Rejection Regions, but I'm not sure.

Also, (bii) relates to (biii) which really confuses me. Any help would be appreciated.