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MAVERICK [17]
4 years ago
9

Suppose that a monopoly has a profit function π(q) = pq - q2 where p denotes price and q denotes quantity. The demand function i

s faces is q(p) = 10 - p/2
Find:

the inverse demand function
the profit when the firm produces 5 units of output
the level of output that maximizes the firm’s profit
Mathematics
1 answer:
horrorfan [7]4 years ago
4 0

a. R\p = (10 - q)*2

The inverse demand function is just the inverse function of the demand function. In other words, we just have to isolate p in the demand function:

p = (10 - q)*2

b. R\25

The price for 5 units of output is given by the inverse demand function:

p = (10 - 5)*2 = 10

We replace p in the profit function:

π(q) = 10 * 5 - 5² = 25

c. 3

For this one, we replace the inverse demand function in the profit function and derivate for q, then equate to 0 and solve:

π(q) = ((10 - q)*2)*q - q² = 20q - 2q² - q² = 20q - 3q²

dπ/dq = 20 - 6*q

20 - 6q = 0

q = 20/6 = 3.33333

Now, a decimal level of output makes no sense. So, now we try the nearest integers 3 and 4, and find the respectives profits. The output that has that maximum profit will be the one that maximizes the profit. Keep in mind, that this will only be true in this particular case because the profit function has the form of a quadratic equation:

π(3) = 20 * 3 - 3*(3)² = 33

π(4) = 20 * 4 - 3*(4)² = 32

The answer is 3.

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A fabric manufacturer believes that the proportion of orders for raw material arriving late isp= 0.6. If a random sample of 10 o
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Answer:

a) the probability of committing a type I error if the true proportion is p = 0.6 is 0.0548

b)

- the probability of committing a type II error for the alternative hypotheses p = 0.3 is 0.3504

- the probability of committing a type II error for the alternative hypotheses p = 0.4 is 0.6177

- the probability of committing a type II error for the alternative hypotheses p = 0.5 is 0.8281

Step-by-step explanation:

Given the data in the question;

proportion p = 0.6

sample size n = 10

binomial distribution

let x rep number of orders for raw materials arriving late in the sample.

(a) probability of committing a type I error if the true proportion is  p = 0.6;

∝ = P( type I error )

= P( reject null hypothesis when p = 0.6 )

= ³∑_{x=0 b( x, n, p )

= ³∑_{x=0 b( x, 10, 0.6 )

= ³∑_{x=0 \left[\begin{array}{ccc}10\\x\\\end{array}\right](0.6)^x( 1 - 0.6 )^{10-x

∝ = 0.0548

Therefore, the probability of committing a type I error if the true proportion is p = 0.6 is 0.0548

b)

the probability of committing a type II error for the alternative hypotheses p = 0.3

β = P( type II error )

= P( accept the null hypothesis when p = 0.3 )

= ¹⁰∑_{x=4 b( x, n, p )

= ¹⁰∑_{x=4 b( x, 10, 0.3 )

= ¹⁰∑_{x=4 \left[\begin{array}{ccc}10\\x\\\end{array}\right](0.3)^x( 1 - 0.3 )^{10-x

= 1 - ³∑_{x=0 \left[\begin{array}{ccc}10\\x\\\end{array}\right](0.3)^x( 1 - 0.3 )^{10-x

= 1 - 0.6496

= 0.3504

Therefore, the probability of committing a type II error for the alternative hypotheses p = 0.3 is 0.3504

the probability of committing a type II error for the alternative hypotheses p = 0.4

β = P( type II error )

= P( accept the null hypothesis when p = 0.4 )

= ¹⁰∑_{x=4 b( x, n, p )

= ¹⁰∑_{x=4 b( x, 10, 0.4 )

= ¹⁰∑_{x=4 \left[\begin{array}{ccc}10\\x\\\end{array}\right](0.4)^x( 1 - 0.4 )^{10-x

= 1 - ³∑_{x=0 \left[\begin{array}{ccc}10\\x\\\end{array}\right](0.4)^x( 1 - 0.4 )^{10-x

= 1 - 0.3823

= 0.6177

Therefore, the probability of committing a type II error for the alternative hypotheses p = 0.4 is 0.6177

the probability of committing a type II error for the alternative hypotheses p = 0.5

β = P( type II error )

= P( accept the null hypothesis when p = 0.5 )

= ¹⁰∑_{x=4 b( x, n, p )

= ¹⁰∑_{x=4 b( x, 10, 0.5 )

= ¹⁰∑_{x=4 \left[\begin{array}{ccc}10\\x\\\end{array}\right](0.5)^x( 1 - 0.5 )^{10-x

= 1 - ³∑_{x=0 \left[\begin{array}{ccc}10\\x\\\end{array}\right](0.5)^x( 1 - 0.5 )^{10-x

= 1 - 0.1719

= 0.8281

Therefore, the probability of committing a type II error for the alternative hypotheses p = 0.5 is 0.8281

3 0
3 years ago
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