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Chapter 8 Section 1 Homework B 1. You want to conduct a test of significance for p the population proportion. The test you will run is H0: p = 0.4 , Ha: p > 0.4, n = 80, you decide that the critical value will be What is the probability of a type I error? p̂c = 0.51. 2. An airline’s public relations department says that the airline rarely loses passengers’ luggage. It further claims that on those occasions when luggage is lost, 90% is recovered and delivered to its owner within 24 hours. A consumer group wants to sample 122 air travelers from this company who lost their luggage. The consumer group will ran the following test: H0: p = 0.9 Ha: p < 0.9, n = 122. Suppose that the significance level is 5%, and let us say that the effect size is 0.05, that is, a change of 5% would be deemed important by the consumer group enough to warrant action. This means the alternative hypothesis is 0.85. a) Calculate the power of the test of hypothesis (significance test) for this possible alternative value of p. b) What is the probability of a type II error. 3. You wish to run a significance test (test of hypothesis) H0: p = 0.2 Ha: p > 0.2, n = 20 Because the criteria np > 10 and n(1 – p) > 10 is no longer met, then I can no longer use a normal approximation method used previously. Below is the table of possible p̂ values that can be attained from a sample of 20 along with the probability of each (the symbol 1E-08 means 1 x 10-8). Table for the situation where p = 0.2 and n = 20. Successes out of 20 0 1 2 3 4 5 6 7 8 9 p̂ 0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 Probability, P( p̂ = k) 0.0115 0.0576 0.1369 0.2054 0.2182 0.1746 0.1091 0.0545 0.0222 0.0074 Successes out of 20 10 11 p̂ 0.5 0.55 Probability, P( p̂ = k) 0.002 0.0005 12 13 14 15 16 17 18 19 20 0.6 0.65 0.7 0.75 0.8 0.85 0.9 0.95 1 9E-05 1E-05 2E-06 2E-07 1E-08 8E-10 3E-11 8E-13 1E-14 a) I decide to consider that there is enough evidence against the null if my sample proportion, p̂ , is greater than or equal to 0.4 (8 successes out of 20 attempts). What is the probability of a type I error. b) Suppose that in reality the value of the population proportion, p is 0.35, alternative hypothesized value. Then the table above is not reality, and reality is the table featured below. Keep in mind that you will still run the test given originally, meaning you will reject the null hypothesis if p̂ , is greater than or equal to 0.4 (8 successes out of 20 attempts). What is the power of the hypothesis test you will run, with respect to alternative proportion. Table for the situation where p = 0.2 and n = 20. Successes ouf of 20 p̂ Probability, P( p̂ = k) Successes ouf of 20 p̂ Probability, P( p̂ = k) 0 1 2 3 4 5 6 7 8 9 0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.0002 0.002 0.01 10 11 12 13 14 15 16 17 18 19 20 0.5 0.55 0.6 0.65 0.7 0.75 0.8 0.85 0.9 0.95 1 6E-06 5E-07 3E-08 8E-10 0.0323 0.0738 0.1272 0.1712 0.1844 0.1614 0.1158 0.0686 0.0336 0.0136 0.0045 0.0012 0.0003 4E-05 4. Is there such a thing as home field advante in football? In order to provide evidence that field advantage is real, a group will sample NFL games from 1990 to 2005. The assumption will be that there is no such thing as field advantage. This then will mean that half the time the home team will win and half the time the home team will lose. The proportion p, will represent the proportion of times the home team wins. The significance test run will be H0: p = 0.5 Ha: p > 0.5, n = 100, a) We will reject the null hypothesis if p̂ > 0.65. What is the probability of a type I error? What is the value 0.65 called? b) Suppose that an effect size of 0.15 is deemed to be important. What is the probability of a type II error? c) What is the power of this test? 5. Below are two discreet distributions for the random variable X, the number of free-throws my friend can make out of 10 attempts (Assumption on the use of normal calculations not appropriate for this scenario; does not meet np> 10 and n(1 – p) > 10. I will assume that my friends ability is modeled by the following distribution. Null hypothesis view Number of freethrows made, X 0 1 2 3 4 5 6 7 8 9 10 probability of making X freethrows 0.0010 0.0098 0.0439 0.1172 0.2051 0.2461 0.2051 0.1172 0.0439 0.0098 0.0010 The alternative view is given by Number of freethrows made, X 0 probability of making X freethrows 0.107 Alternative hypothesis view 1 2 3 4 5 0.268 0.302 0.201 0.088 0.026 6 7 8 9 10 0.006 8E-04 7E-05 4E-06 1E-07 Note the notation 8E-04 is equivalent to 8 x 10-4. If the number of free-throws made is 2 or less I will assume that the null hypothesis view is not correct, and choose the alternative view. a) What is the probability of a type I error? b) What is the probability of a type II error? c) What is the power?