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docker41 [41]
2 years ago
10

A major traffic problem in the Greater Cincinnati area involves traffic attempting to cross the Ohio River from Cincinnati to Ke

ntucky using Interstate 75. Let us assume that the probability of no traffic delay in one period, given no traffic delay in the preceding period, is 0.9 and that the probability of finding a traffic delay in one period, given a delay in the preceding period, is 0.6. Traffic is classified as having either a delay or a no-delay state, and the period considered is 30 minutes.Required:a. Assume that you are a motorist entering the traffic system and receive a radio report of a traffic delay. What is the probability that for the next 60 minutes (two time periods) the system will be in the delay state?b. What is the probability that in the long run the traffic will not be in the delay state?c. An important assumption of the Markov process model presented here has been the constant or stationary transition probabilities as the system operates in the future. Do you believe this assumption should be questioned for this traffic problem? Explain.
Mathematics
1 answer:
myrzilka [38]2 years ago
8 0

Answer:

a) 0.36

b) 0.3

c) Yes

Step-by-step explanation:

Given:

Probability of no traffic delay in one period, given no traffic delay in the preceding period = P(No_Delay) = 0.9

Probability of finding a traffic delay in one period, given a delay in the preceding period = P(Delay) = 0.6

Period considered = 30 minutes

a)

Let A be the probability that for the next 60 minutes (two time periods) the system will be in the delay state:

As the Probability of finding a traffic delay in one period, given a delay in the preceding period is 0.6 and one period is considered as 30 minutes.

So probability that for the next two time periods i.e. 30*2 = 60 minutes, the system in Delay is

P(A) = P(Delay) * P(Delay) =  0.6 * 0.6 = 0.36

b)

Let B be the probability that in the long run the traffic will not be in the delay state.

This statement means that the traffic will not be in Delay state but be in No_Delay state in long run.

Let C be the probability of one period in Delay state given that preceding period in No-delay state :

P(C) =  1 - P(No_Delay)

        = 1 - 0.9

P(C) = 0.1

Now using P(C) and P(Delay) we can compute P(B) as:

P(B) = 1 - (P(Delay) + P(C))

      = 1 - ( 0.6 + 0.10 )

      = 1 - 0.7

P(B) = 0.3

c)

Yes this assumption should be questioned for this traffic problem because it implies that  traffic will be in Delay state for the 30 minutes and just after 30 minutes, it will be in No_Delay state. However, traffic does not work like this in general and it makes this scenario unrealistic. Markov process model can be improved if probabilities are modeled as a function of time instead of being presented as constant (for 30 mins).

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Suppose you are an expert on the fashion industry and wish to gather information to compare the amount earned per month by model
Ann [662]

Answer:

(1) The degrees of freedom for unequal variance test is (14, 11).

(2) The decision rule for the 0.01 significance level is;

  • If the value of our test statistics is less than the critical values of F at 0.01 level of significance, then we have insufficient evidence to reject our null hypothesis.      
  • If the value of our test statistics is more than the critical values of F at 0.01 level of significance, then we have sufficient evidence to reject our null hypothesis.  

(3) The value of the test statistic is 0.3796.

Step-by-step explanation:

We are given that you are an expert on the fashion industry and wish to gather information to compare the amount earned per month by models featuring Liz Claiborne's attire with those of Calvin Klein.

The following is the amount ($000) earned per month by a sample of 15 Claiborne models;

$3.5, $5.1, $5.2, $3.6, $5.0, $3.4, $5.3, $6.5, $4.8, $6.3, $5.8, $4.5, $6.3, $4.9, $4.2 .

The following is the amount ($000) earned by a sample of 12 Klein models;

$4.1, $2.5, $1.2, $3.5, $5.1, $2.3, $6.1, $1.2, $1.5, $1.3, $1.8, $2.1.

(1) As we know that for the unequal variance test, we use F-test. The degrees of freedom for the F-test is given by;

\text{F}_(_n__1-1, n_2-1_)

Here, n_1 = sample of 15 Claiborne models

         n_2 = sample of 12 Klein models

So, the degrees of freedom = (n_1-1, n_2-1) = (15 - 1, 12 - 1) = (14, 11)

(2) The decision rule for 0.01 significance level is given by;

  • If the value of our test statistics is less than the critical values of F at 0.01 level of significance, then we have insufficient evidence to reject our null hypothesis.      
  • If the value of our test statistics is more than the critical values of F at 0.01 level of significance, then we have sufficient evidence to reject our null hypothesis.  

(3) The test statistics that will be used here is F-test which is given by;

                          T.S. = \frac{s_1^{2} }{s_2^{2} } \times \frac{\sigma_2^{2} }{\sigma_1^{2} }  ~ \text{F}_(_n__1-1, n_2-1_)

where, s_1^{2} = sample variance of the Claiborne models data = \frac{\sum (X_i-\bar X)^{2} }{n_1-1} = 1.007

s_2^{2} = sample variance of the Klein models data = \frac{\sum (X_i-\bar X)^{2} }{n_2-1} = 2.653    

So, the test statistics =  \frac{1.007}{2.653 } \times 1  ~ \text{F}_(_1_4,_1_1_)

                                   = 0.3796

Hence, the value of the test statistic is 0.3796.

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