The problem: There are 2 large containers ( A & B ) both filled with white and red marbles. 10 marbles are randomly drawn from container A; 0 red and 10 white. 11 marbles are drawn randomly from container B; 3 red and 8 white. Based on these samples, can it be concluded that the population of red marbles in container B is significantly (p=0.95) different from that in container A? Also, if a “statistic” is utilized in the solution, show that it is valid at these sample sizes. Help, I have already submitted 3 solutions, all rejected …
No definite conclusions can be made. Are you asking us to help you win a contest?
Is this homework?
Sorry, couldn’t figure out how to get back here. Both somewhat correct. Probability class extra credit. Is that ok to post?
Univeristy is kind of a haze, but I think you might be missing a variable or two.
See if this calculator works.
Not as you’ve posted it. We have a standing policy against doing homework for people.
However, if you post what you’ve done so far, and where you’re stuck, you can usually get some hints.
Hoping I don’t screw this up too …
Thanks for the link, but as I understand it, to obtain a 0.95/0.95 confidence/significance level, as in the poll calculator you linked me to, the sample size would be set to obtain a 0.95 confidence level at a 0.95 significance. In this problem the sample size is 10 vs 11 …
This is only for extra credit …
Been awhile since stat I probibly did it wrong but
Test and Confidence Interval for Two Proportions
Success = 1
Variable X N Sample p
b 3 11 0.272727
a 0 10 0.000000
Estimate for p(b) - p(a): 0.272727
95% CI for p(b) - p(a): (0.00954012, 0.535914)
Test for p(b) - p(a) = 0 (vs not = 0): Z = 2.03 P-Value = 0.042
- NOTE * The normal approximation may be inaccurate for small samples.
Divide P by 2 and there is your p score. So yes I think its significant beyond 95%
On second though don’t divide by two.
Insufficient information. How many marbles are in the containers?
Why should that matter?
With the given as “large containers” that are “both filled”, is that necessary? I think you can safely assume an infinite amount for the purposes of the exercise.
Thanks. One approach, similar to yours, used binomial proportions. p(B) - p(A) = 0.2727 with s = 0.1529 resulted in Z = 1.784, i.e. not significant at p=0.95. This solution was not accepted since I could not prove or find a proof that the statistic follows a normal distribution with this number of samples. Using your result for Z = 2.03 and Student’s-t of 2.093 with n(A) + n(B) - 2 d.f., would also indicate that the sampling results are not significant at p=0.95. However, as you mentioned, need to prove that the statistic follows a normal distribution at 10 vs 11 samples.
The standard for significant results is p = 0.05, not 0.95. You got it reversed.
Roughly speaking, a test that results in p=0.05 would mean that under the null hypothesis (both buckets are pretty much the same), if we repeated this experiment many times, we would see results like this only 5% of the time. Since 5% is pretty low, we conclude that the null hypothesis is false, and we accept the alternative hypothesis, namely that the buckets are different.