Question

Two different companies have applied to provide cable television service in a certain region. Let p denote the proportion of all potential subscribers who favor the first company over the second. Consider testing H_o: p=0.5 versus H_a: p?0.5 based on a random sample of 25 individuals. Let the test statistic X be the number in the sample who favor the first company and x represent the observed value of X.

a) Describe type I and type II errors in the context of this problem situation.

b) Suppose that x=6. Which values of X are at least as contradictory to H_o as this one?

c) What is the probability distribution of the test statistic X when H_o is true? Use it to compute the P-value when x=6

d) If H_o is to be rejected when P-value?0.044, compute the probability of a type II error when p=0.4, again when p=0.3, and also when p=0.6 and p=0.7 [Hint: P-value > 0.044 is equivalent to what inequalities involving x]

e) Using the test procedure of (d), what would you conclude if 6 of the 25 queried favored company 1?

Answer 1

The null and alternative hypotheses are, Но: р-0.5 На р # 0.5 The sample size is n-25 The rejection region is x:xs7or x 218) corresponding acceptance region is (x:x28 orxs17) a) The type I error is rejecting null hypothesis when it is true. The type I error in the situation is to conclude that one company is favoured over other company when none of the companies is actually favoured The type II error is failing to reject null hypothesis when it is false. The type II error in the situation is to conclude that no company is favoured when actually one of the companies is favoured b) In a sample ofn-25 , suppose x-6, the values that are at least as contradictory to Ho are, 25-619

c) When Ho is true, the probability distribution of test statistic X follows binomial distribution with n-25 and p-0.5. X ~ Bin(n= 25, p = 0.5 The probability value when x-6 is, 25 25-6 0.5 (1-0.5 - 0.0053 d) The probability of type II error when p 0.4 is β(0.4)-P(Accept Ho l X ~ Bin(25.0.4)) Px28 or x s17x Bin(25.0.4) P(8 s x s17x Bin(25,0.4)) P(xs17 x ~Bin(25,0.4)-P(xs8x ~Bin(25,0.4) = 0.9988-0.2735 = 10.7253 The probability of type II error when p-0.3 is, β(0.3)-P(Accept Ho l X ~ Bin(25.0.3)) Px 28 orxs17xBin(25,0.3)) -P(8 sxs17xBin(25,0.3)) -P(xs 17 X ~ Bin(25,0.3))-P(xs8l X ~ Bin(25,0.3)) = 1.0000-0.6769 0.3231

The probability of type II error when p- 0.6 is, B(0.6)- P(Accept Ho X Bin(25,0.6)) = P(12 8 or x 17 1 X ~ Bin(25.0.6) -P(85 x 171 X ~ Bin(25.0.6) = P(X17 , x ~ Bin(25.0.6))-Plx S8, x ~ Bin(25.0.6 0.8464-0.0043 -0.8421 The probabılity of type ll error when p-0.7 is β(0.7) = P(Accept Ho , X ~ Bin( 25, 0.7)) )) =Plx28 or x 171 X ~ Bin(25.0.7 P(8 s17|x Bin(25,0.7)) P(x s17| x ~Bin(25,0.7))- P(xs8| X ~ Bin(25,0.7) 0.4882 -0.0001 10.4881 e) The value X=6 is within the rejection region.~{x: x 7 or12 18} . Therefore, Ho will be rejected in favour of H, and it is concluded that that company 2 is favoured