Question
QUESTION 18 A carpet company advertises that it will deliver your carpet within 15 days of purchase. A sample of 49 past customers is taken. The average delivery time in the sample was 16.6 days. Assume the population standard deviation is known to be 5.6 days. Answer the following questions: . What is the cut-offpoint at alpha = .05? 2. What is the cut-offpoint ifType I error is 1%? 3, Would you reject the Null at 5% level of significance? 4. Would you reject the Null at 1% level of significance? 5. If suppose the true delivery time for the population is 17 days. What is the Type II error at Alpha .01? 2Ss
Explanation / Answer
Given that, Standard deviation, =5.6 Sample Mean, X =16.6 Null, H0: =15 Alternate, H1: !=15 Level of significance, = 0.01 From Standard normal table, Z /2 =2.5758 Since our test is two-tailed Reject Ho, if Zo < -2.5758 OR if Zo > 2.5758 Reject Ho if (x-15)/5.6/(n) < -2.5758 OR if (x-15)/5.6/(n) > 2.5758 Reject Ho if x < 15-14.42448/(n) OR if x > 15-14.42448/(n) ----------------------------------------------------------------------------------------------------- Suppose the size of the sample is n = 49 then the critical region becomes, Reject Ho if x < 15-14.42448/(49) OR if x > 15+14.42448/(49) Reject Ho if x < 12.93936 OR if x > 17.06064 Implies, don't reject Ho if 12.93936 x 17.06064 Suppose the true mean is 17 Probability of Type II error, P(Type II error) = P(Don't Reject Ho | H1 is true ) = P(12.93936 x 17.06064 | 1 = 17) = P(12.93936-17/5.6/(49) x - / /n 17.06064-17/5.6/(49) = P(-5.0758 Z 0.0758 ) = P( Z 0.0758) - P( Z -5.0758) = 0.5302 - 0 [ Using Z Table ] = 0.5302 For n =49 the probability of Type II error is 0.5302