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Mathematics 9 Online
OpenStudy (vane11):

Stats Question A manufacturer receives parts from two suppliers. An SRS of 400 parts from supplier 1 finds 20 defective; an SRS of 100 parts from supplier 2 finds 10 defective. Let p1 and p2 be the proportion of all parts from suppliers 1 and 2, respectively, that are defective. Is there evidence of a difference in the proportion of defective parts produced by the two suppliers? To make this determination, you test the hypotheses H0 : p1 = p2 and Ha : p1 ≠ p2.The P-value of your test is A. 0.1164. B. 0.0602. C. 0.0301.

OpenStudy (vane11):

I got p1=.5 p2= .1 p=.0012 HO=30/500 I think... so now what?

jimthompson5910 (jim_thompson5910):

do they expect you use a calculator? or a table? for the p value

OpenStudy (vane11):

I can use a calculator and a table if necessary

jimthompson5910 (jim_thompson5910):

ok because I'm getting something close to the answer, but wonder if they're using a table (instead of a calculator)

OpenStudy (vane11):

Oh I've only been using a calculator, but if there Is a table needed, then I have 5 to choose from, do you know which one is needed?

jimthompson5910 (jim_thompson5910):

you would use the normal cdf table

jimthompson5910 (jim_thompson5910):

or standard normal cdf table

OpenStudy (vane11):

ok I got that one, do you need me to attach it? Im not sure how I would use it in this problem

jimthompson5910 (jim_thompson5910):

ok they are definitely used a table (I hate when they do that when you can easily use a calculator)

jimthompson5910 (jim_thompson5910):

They're using something similar to this http://homes.cs.washington.edu/~jrl/normal_cdf.pdf

jimthompson5910 (jim_thompson5910):

sure attach yours so we can compare

OpenStudy (vane11):

its the same one, just that mine's cut in half, one is the negative values and the other the positive values, so when does the table come into play?

jimthompson5910 (jim_thompson5910):

ok we'll get to that later (towards the end), so ignore the tables for now

OpenStudy (vane11):

alright

jimthompson5910 (jim_thompson5910):

p1 = x1/n1 p1 = 20/400 p1 = 0.05 --------------- p2 = x2/n2 p2 = 10/100 p2 = 0.10

jimthompson5910 (jim_thompson5910):

p = (x1 + x2)/(n1 + n2) p = (20+10)/(400+100) p = 0.06

jimthompson5910 (jim_thompson5910):

next we find the standard error SE = sqrt(p(1-p)(1/(n1)+1/(n2))) SE = sqrt((0.06)*(1-0.06)*(1/(400)+1/(100))) SE = 0.0265518360947035

jimthompson5910 (jim_thompson5910):

then we find the test statistic z=(p1-p2)/(SE) z=(0.05-0.1)/(0.0265518360947035) z = -1.88310894288677 z = -1.88

jimthompson5910 (jim_thompson5910):

we then use this test statistic to find the area to the left of it under the curve this is where the table or calculator will come in

OpenStudy (vane11):

ok ohhhh alright duh

jimthompson5910 (jim_thompson5910):

so using the table I posted, what is that area?

OpenStudy (vane11):

.0301

jimthompson5910 (jim_thompson5910):

now you double this to get 2*0.301 = 0.0602

jimthompson5910 (jim_thompson5910):

you double this area because this is a two tailed test

OpenStudy (vane11):

ok, answered my question before I could ask it

jimthompson5910 (jim_thompson5910):

they threw in the trap in C a lot of people would have stopped at 0.0301

OpenStudy (vane11):

I would have, thankyou!

jimthompson5910 (jim_thompson5910):

you're welcome

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