Question

Consider this testing situation. A box contains 16 chips (with some mixture of red and black chips). Suppose we have the following hypotheses:

            H_O:     The box contains R=8 red and B=8 black chips.

            H_A:     The box contains some other mixture of red and black chips.

We randomly select 5 chips simultaneously from the box without replacement. Our Test Statistic is the Y = # of Black chips found in the sample.

Suppose we use the following decision rule: Reject H_O if Y= 0 or Y=5.

You might recall (wonderful if you do, no problem if you don’t), the variable Y has what is called the hypergeometric probability distribution given by:

p(y) = C(B,y) * C(R, 5 – y) p(y) = for y = 0,1,2,3,4,5 C(16,5)

1. Find alpha, the probability of a Type I error in this hypothesis test
2. Find the power of this test as a function of B, the true number of Black chips in the box, and graph your power curve against value of B for B = 0, 1, 2, 3, ...., 15, 16.

Answer 1

Solo Given, A box contains 16 chips. mixture of red and black chips. The following hypothesis is : Ho : The box contains R=8 and 13=8 black chips. HA: The box contains some other mixture of red & black chips Randomly select s clips from box without replacement. Test statistic is y-# of Black chips. we reject Ho: Of Yoo or 4-5 gregoored states tocs are of a type 1 error (a), find alpha, the probability hypothesis test. in this d = PHO is true (reject Ho) 2 = PHO is true creject Hol - p Creject tolto as true) a. (resect Hon Ho is true) - P(Ho is true) - 2* $ 1C (+) d = 0.8205

(b). Black chips in the box, and against value of BP graph your power wie The Power of test designated by B is B = P(reject Ho/ Ho ºs false) not 16-b b 16-b ts sctu cx, ct 3 ct 2 ct 2 cxc + b cx 16-be 16-b The Power of test designated by ß CX 4 ct sc The graph is an shown above.