## ProblemSuppose that the proportion, p, of defective items in a population is unknown, and that we want to test:_{0}: p = 0.2 H _{1}: p not equal 0.2
a) The value of the power function at the points p=0,0.1,0.2,...,0.9,1; and sketch the power function. b) The size of the test. ## Solution:Recall that the power is just the probability of rejecting the null as a function of the parameter. Let's call it w(p). Hence, |

> w := p -> 1-sum(binomial(20,k)*p^k*(1-p)^(20-k),k=0..6)+

> sum(binomial(20,k)*p^k*(1-p)^(20-k),k=0..1);

/ 6 \ |----- | | \ k (20 - k)| w := p -> 1 - | ) binomial(20, k) p (1 - p) | | / | |----- | \k = 0 / / 1 \ |----- | | \ k (20 - k)| + | ) binomial(20, k) p (1 - p) | | / | |----- | \k = 0 /> seq(evalf(w(j/10),3),j=1..9);

.394, .156, .400, .751, .942, .994, 1.000, 1.000, 1.000> evalf(w(0.01),3);

.983> evalf(w(0.001),3);

1.00> plot(w(p),p=.001..0.999);

as you can see the picture also solves the other question. The size is just w(0.2) = .156. |

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Carlos Rodriguez <carlos@math.albany.edu> Last modified: Tue Nov 17 11:36:07 EST 1998