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NikAS [45]
2 years ago
9

Consider two competing firms in a declining industry that cannot support both firms profitably. Each firm has three possible cho

ices, as it must decide whether or not to exit the industry immediately, at the end of this quarter, or at the end of the next quarter. If a firm chooses to exit then its payoff is 0 from that point onward. Each quarter that both firms operate yields each a loss equal to -1, and each quarter that a firm operates alone yields it a payoff of 2. For example, if firm 1 plans to exit at the end of this quarter while firm 2 plans to exit at the end of next quarter then the payoffs are 1,1) because both firms lose -1 in the first quarter and firm 2 gains 2 in the second. The payoff for each firm 1.
a. Write down this game in matrix form
b. Are there any strictly dominated strategies? Are there any weakly dominated strategies?
c. Find the pure-strategy Nash Equilibria.
d. Find the unique mixed-strategy Nash equilibrium.
Mathematics
1 answer:
yaroslaw [1]2 years ago
4 0

Answer:

a) attached below

b)  ( T,T )

c) The Pure-strategy Nash equilibria are : ( N,E ) and ( E,N )

d) The mixed-strategy Nash equilibrium for Firm 1 = ( 1/3 , 0, 2/3 )

while the mixed -strategy Nash equilibrium for Firm 2 = ( 1/3 , 0, 2/3 )

Step-by-step explanation:

A) write down the game in matrix form

let: E = exit at the industry immediately

     T = exit at the end of the quarter

     N = exit at the end of the next quarter

matrix is attached below

B) weakly dominated strategies is ( T,T )

C) Find the pure-strategy Nash equilibria

The Pure-strategy Nash equilibria are : ( N,E ) and ( E,N )

D ) Find the unique mixed-strategy Nash equilibrium

The mixed-strategy Nash equilibrium for Firm 1 = ( 1/3 , 0, 2/3 )

while the mixed -strategy Nash equilibrium for Firm 2 = ( 1/3 , 0, 2/3 ) since T is weakly dominated then the mixed strategy will be NE

Assume that P is the probability of firm 1 exiting immediately ( E )

and q is the probability of firm 1 staying till next term ( N ) ∴ q = 1 - P.

hence the expected utility of firm 2 choosing E = 0 while the expected utility of choosing N = 4p - 2q .

The expected utilities of E and N to firm 2 =

0 = 4p - 2q = 4p - 2 ( 1-p) = 6p -2 which means : p = 1/3 , q = 2/3

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Answer:

A) Option B is correct.

H₀: μ₁ = μ₂

Hₐ: μ₁ - μ₂ < 0

B) t = -2.502

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C) Option A is correct.

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D) Option A is correct.

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Step-by-step explanation:

The complete Question is presented in the two attached images to this answer.

A) To perform this test we first define the null and alternative hypothesis.

The null hypothesis plays the devil's advocate and usually takes the form of the opposite of the theory to be tested. It usually contains the signs =, ≤ and ≥ depending on the directions of the test.

While, the alternative hypothesis usually confirms the the theory being tested by the experimental setup. It usually contains the signs ≠, < and > depending on the directions of the test.

For this question, we want to test if the extra carbonation of cola results in a higher average compression strength. That is, that cola has a higher average compression strength than the strawberry drink.

Hence, the null hypothesis would be that there isn't significant evidence to suggest that the extra carbonation of cola results in a higher average compression strength, that is, cola has a higher average compression strength than the strawberry drink.

The alternative hypothesis is that there is significant evidence to suggest that the extra carbonation of cola results in a higher average compression strength, that is, cola has a higher average compression strength than the strawberry drink.

Mathematically, if the average compression strength of strawberry drink is μ₁, the average compression strength of cola is μ₂ and the difference in compression strengths is μ = μ₁ - μ₂

The null hypothesis is represented as

H₀: μ = 0 or μ₁ = μ₂

The alternative hypothesis is represented as

Hₐ: μ < 0 or μ₁ - μ₂ < 0

B) So, to perform this test, we need to compute the test statistic

Test statistic for 2 sample mean data is given as

Test statistic = (μ₁ - μ₂))/σ

σ = √[(s₂²/n₂) + (s₁²/n₁)]

μ₁ = average compression strength of strawberry drink = 537

n₁ = sample size of the sample of strawberry drink in cans surveyed = 10

s₁ = standard deviation of the compression strength of strawberry drink in cans surveyed= 22

μ₂ = average compression strength of cola = 559

n₂ = sample size of the sample of cola in cans surveyed = 10

s₂ = standard deviation of the compression strength of strawberry drink in cans surveyed = 17

σ = [(17²/10) + (22²/10)] = 77.5903160379 = 8.792

We will use the t-distribution as no information on population standard deviation is provided

t = (537 - 559) ÷ 8.792

= -2.502 = -2.50

checking the tables for the p-value of this t-statistic

Degree of freedom = df = n₁ + n₂ - 2 = 10 + 10- 2 = 18

Significance level = 0.05

The hypothesis test uses a one-tailed condition because we're testing in only one direction (whether compression strength of cola in can is greater).

p-value (for t = -2.50, at 0.05 significance level, df = 18, with a one tailed condition) = 0.011154 = 0.0112 to 4 d.p.

C) The interpretation of p-values is that

When the (p-value > significance level), we fail to reject the null hypothesis and when the (p-value < significance level), we reject the null hypothesis and accept the alternative hypothesis.

So, for this question, significance level = 0.05

p-value = 0.0112

0.0112 < 0.05

Hence,

p-value < significance level

This means that we reject the null hypothesis accept the alternative hypothesis & say that the extra carbonation of cola results in a higher average compression strength, that is, cola has a higher average compression strength than the strawberry drink.

D) The necessary conditions required before a t-test is deemed valid include.

- The samples used must be a random sample of the population distribution with each variable in the sample independent of other one.

- The distribution of the population where the samples were extracted from must be normal or approximately normal to ensure some degree of normality for the samples.

Hence, the necessary assumption for this t-test among the options is that the distributions of compression strengths are approximately normal.

Hope this Helps!!!

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