# Help with Sample Size Needed

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- This topic has 9 replies, 3 voices, and was last updated 4 years, 7 months ago by John.

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- May 26, 2015 at 6:20 pm #55037

JohnParticipant@john.lombardo**Include @john.lombardo in your post and this person will**

be notified via email.Hi Everyone,

I’m new to LSS and am currently taking a course.

One of the assignments seems very straight-forward, but I simply cannot wrap my head around it. Can anyone help?

There are 4 parts to this assignment, but I’ve only included the 1st part as well as my response for it.

Be kind because I am new to the stats world…

Question Part 1:

A part is not to exceed a failure rate of 3.4 failures in one million.

(a) Determine the sample size needed for a hypothesis test of the equality of this failure rate to a 3.4 parts per million failure rate. Use risks of 0.05 and an uncertainty of +/- 10% of the failure rate target.Response to Part 1:

(a) Sample size needed (n) for a failure rate criterion of 3.4 DPMO (3.4 / 1,000,000 = 0.0000034)

• Risk = a = 0.05 (or 95% confidence level)

Or Za = 1.645

• St.Dev. = +/- 10% = 1,000,000 * 0.1 = +/- 100,000

Or 1,000,000 +/- 100,000

900,000 <= p <= 1,100,000

• n = ((Ua * (pa * (1- pa))1/2) + (Ub * (pb * (1-pb)) 1/2))/(pb – pa)

pa = 3.4 / 1,100,000 = 3.0909090909090909090909090909091e-7

• pa = 0.0000003091

pb = 3.4 / 900,000 = 3.74777777777777777777777777777778e-6

• pb = 0.00000375

Ua = 1.645 (Table B, Appendix A, Single-Sided Normal Distribution)

Ub = 1.645 (Table B, Appendix A, Single-Sided Normal Distribution)

n = ((1.645 * (9.091 * (1 – 0.0000003091))1/2) + (1.645 * (1.111 * (1 – 0.00000375))1/2))/(0.00000375 – 0.0000003091)

n = ((1.645 * (9.091 * 0.9999996909)1/2) + (1.645 * (1.111 * 0.99999625)1/2))/(0.0000034409)

n = ((1.645 * (9.0909971899719)1/2) + (1.645 * (1.11099583375)1/2))/(0.0000034409)

n = ((1.645 * 3.015) + (1.645 * 1.054))/(0.0000034409)

n = 1.73383/0.0000034409

n = 503888.52 = 503,8890May 26, 2015 at 6:21 pm #198319

JohnParticipant@john.lombardo**Include @john.lombardo in your post and this person will**

be notified via email.Sorry, the formatting came out wrong…

Also, I have access to Minitab 16 but am new to that too…

0May 27, 2015 at 7:21 am #198324

Amit Kumar OjhaParticipant@AmitOjha**Include @AmitOjha in your post and this person will**

be notified via email.Hi John,

Please find below the solution for the given problem:

It is given thatTolerance = 10

Alpha = 0.05 and Hence Z = 1.96

Standard Deviation corresponding to 3.4 parts per million defect rate = 6

**Hence Sample Size = ((1.96*6)/10)^2 = 32.14****So you need a sample of size = 33**(Rounded Off)Good Luck !!!

0May 27, 2015 at 8:40 am #198326

JohnParticipant@john.lombardo**Include @john.lombardo in your post and this person will**

be notified via email.Thank you AmitOjha,

How did you determine Z?

I know 3.4 dpmo is the 6 sigma standard (really its equal to 4.5 sigma).

And that is used as the st.dev., but are you using reference tables or are these calculated?Also, is this related to a type of 1 proportion test

Regards,

John0May 27, 2015 at 5:59 pm #198329

JohnParticipant@john.lombardo**Include @john.lombardo in your post and this person will**

be notified via email.Ok, i realize my last question on Z is silly. I realize the z-table contains my answer.

But I am being pointed towards a 1 proportion type test to solve this for the assignment.

Can you help clarify?

0May 27, 2015 at 11:07 pm #198330

Amit Kumar OjhaParticipant@AmitOjha**Include @AmitOjha in your post and this person will**

be notified via email.Hi,

by 1 proportion test for determining sample size what exactly do you imply ? Sorry I dint get that :-(

0May 28, 2015 at 8:16 am #198335

NigelParticipant@Nigelbloomy**Include @Nigelbloomy in your post and this person will**

be notified via email.@John.Lombardo Do you mean a 1 tailed test vs a 2 tailed test? Amit probably knows better but I think that would mean you need to use a Z-score of 1.645 instead of 1.96.

This article explains the difference between the 2 tests:

http://www.ats.ucla.edu/stat/mult_pkg/faq/general/tail_tests.htm0May 28, 2015 at 2:20 pm #198336

JohnParticipant@john.lombardo**Include @john.lombardo in your post and this person will**

be notified via email.Hi,

So the final solution that I submitted (and was accepted) is below.

For the 1 proportion test, this is done through minitab as follows:

Go to: Stat –> Power and sample size –> 1 Proportion …

Then enter into the dialoge:

Comparison propotion = 0.00000374

Power value = 0.95

Hypothesized proportion: 0.00000306Or you can do it the manual way using the following formula:

n = (((Ua * (pa * (1- pa))^1/2) + (Ub * (pb * (1-pb))^1/2))/(pb – pa))^2

• pa = 0.0000034 * 1.1 = 0.00000374

• pb = 0.0000034 * 0.9 = 0.00000306

• Ua = 1.645 (Table B, Appendix A, Single-Sided Normal Distribution)

• Ub = 1.645 (Table B, Appendix A, Single-Sided Normal Distribution)

• n = (((1.645 * (0.00000374 * (1 – 0.00000374))1/2) + (1.645 * (0.00000306 * (1 – 0.00000306))1/2))/(0.00000306 – 0.00000374))^2

• n = 79,389,227In any case, the sample size is deemed unmanageable!

0May 28, 2015 at 11:10 pm #198339

Amit Kumar OjhaParticipant@AmitOjha**Include @AmitOjha in your post and this person will**

be notified via email.Hi John,

See as far as I know, there can be only one point which you need to take care about the formula for sample size calculation i.e. whether you are concerned about one tailed or two tailed test.

In this context Nigel is correct.However, the formula which you used to calculate the sample size is not correct.

Refer Levin and Rubin – Statistic for Management for sample size calculation. This is best book I have ever known for statistics for beginners.

0June 9, 2015 at 8:23 am #198391

JohnParticipant@john.lombardo**Include @john.lombardo in your post and this person will**

be notified via email.Thank you Amit, I will certainly take a look at it!

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