An inspector inspects a shipment of medications to determine the efficacy in terms of the proportion p in the shipment that failed to retain full potency after 60 days of production. Unless there is clear evidence that this proportion is less than 0.05, she will reject the shipment. To reach a decision, she will test the following hypotheses using the large-sample test for a population proportion: H0 : p = 0.05, Ha : p < 0.05 To do so, she selects an SRS of 200 pills. Suppose that eight of the pills have failed to retain their full potency. The p-value of the test is?
Huge.
these are the options
. 0.2352. B. 0.2582. C. 0.4704. D. 0.5164.
This is a binomial distribution test with mean np and variance npq. n=200 p=.05 and q=1-p. Standard Deviation =sqrt(variance).
so it would be d?
So mean = 200*5% = 10 variance = 95%*10=9.5 and Standard Deviation is sqrt(9.5) = 3.08. Her observation was 8 (2 below the mean of 10) which is 2/3.08 of a standard deviation.
im confused.
so the answer is 2/3.08=.65?
Whatever the z-score of -0.649 standard deviations is is your p value.
thank you!!! will post two more questions.
What did you get?
.382
That is too high.
i did invnorm(.649)
i dont know
I don't know what program that is but in Excel, =NORM.S.DIST(-0.649,1) = 0.2582
thank you so much! thats all i have for now
Yeah, invnorm is for when you know p and need to find the number of standard deviations, this is the opposite.
you really know a lot of about statistics! i have my final tomorrow.
Good luck
ahhhh...im nervous
Do you feel like you understand what I am doing with the np, npq stuff?
honestly, not really
So a binomial distribution occurs when you have a collection of items/observations/whatever where each thing is either yes or no. In this case each pill inspected was either good or bad.
n is the number of items you sample and p is the chance that any one item is bad.
oh, i see. its one way or another.
So this could work for shooting free throws in basketball, for example.
Or manufacturing iPods or whatever.
okay
On average the number of bad pills we find is going to be n*p.
so its the proportion?
If p is 5% and we test 200 pills then we expect that .05 * 200 = 10 pills will be bad, usually.
oh i see
But the binomial distribution can be approximated by a normal distribution if n and p are large enough (which in this case is an okay assumption). And the variance of a Binomial distribution = n*p*(1-p)
that makes sense
what is n* and p*?
* is just a multiplication sign.
n is still the number of tests and p is the expected failure rate. 1-p is the expected success rate.
oh right....that makes sense
are you a teacher?
So in this example the variance of the expected distribution is np(1-p) = 200*(0.05)*(0.95)
No, I'm an Actuary.
that explains it well!
And you understand the relationship between variance and standard deviation, right?
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