HW#8_f08_sol - CE 3102 Fall 2008 Homework#8 Due 9:45 AM...

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Unformatted text preview: CE 3102 Fall 2008 Homework #8 Due: 9:45 AM, Thursday December 4, 2008 This is a team assignment. You may work together in teams of up to, but not exceeding, three people. Only one written submission is required from each team. However, every team member is responsible for knowing all of the material. 1. In propositional logic, two compound statements are equivalent if they have the same true/false pattern. Use truth to tables to prove or disprove: (a) A is equivalent to ~(~A) (b) ~(A&~B) is equivalent to (A->B) (c) (~A)orB is equivalent to (A->B) A B ~A ~B ~(~A) A->B ~A or B A&~B ~(A&~B) T T F F T T T F T F T T F F T T F T T F F T T F F T F F F T T F T T F T ~(~A) is just the opposite of the ~A column. ~A or B is true whenever either the ~A column is true of the B column is true A&~B is true only when both the A and ~B columns are true, and ~(A&~B) is its opposite. 2. California DMV records indicate that all vehicles undergoing emissions testing during the previous year, 70% passed on the first try. A random sample of 200 cars tested in Orange County during the current year yields that 156 passed on the initial test. (a) Does this suggest that the true proportion for Orange County during the current year differs from the statewide proportion? Test the relevant hypothesis using a significance level of α =0.05. Clearly state your null hypothesis and alternate hypothesis, and you conclusion. (b) Compute a 95% confidence interval for the proportion passing based on the sample. (a) Let p denote the true proportion of vehicles in Orange County that would pass the emissions test on the first try. (1) State hypotheses: H0: p=0.70 HA:p ≠ 0.70 (2) Determine test statistic and sampling distribution: Let X denote that number of vehicles in our sample passing the test on the first try, then if the null hypothesis is true, X would be a binomial random variable, with p=0.70 and n=200. Using the normal approximation to the binomial, the maximum likelihood estimate for p (x/n), is approximately normal, with mean equal to p and variance equal to p(1-p)/n, so we can use the standard normal (Z) distribution as our test statistic. That is, if the null hypothesis is true, the statistic n n X Z )) 7 (. 1 )( 7 (. 70 . /-- = is approximately distributed as a normal random variable, with mean equal to 0 and variance equal to 1.0....
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HW#8_f08_sol - CE 3102 Fall 2008 Homework#8 Due 9:45 AM...

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