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Firdavs [7]
3 years ago
13

The reading on a voltage meter connected to a test circuit is uniformly distributed over the interval (θ, θ + 1), where θ is the

true but unknown voltage of the circuit. Suppose that Y1,Y2,...,Yn denotearandomsampleofsuchreadings. a Show that Y is a biased estimator of θ and compute the bias. b Find a function of Y that is an unbiased estimator of θ.
Mathematics
1 answer:
rewona [7]3 years ago
3 0

Answer:

Step-by-step explanation:

The complete question is

The reading on a voltage meter connected to a test circuit is uniformly distributed over the interval (θ, θ + 1), where θ is the true but unknown voltage of the circuit. Suppose that Y1,Y2,...,Yn denotearandomsampleofsuchreadings.Let Y be the sample mean. a) Show that Y is a biased estimator of θ and compute the bias. b Find a function of Y that is an unbiased estimator of θ.

Recall that an unbiased estimator Y of a parameter \theta is  a function of a random sample for which we have that

E[Y] = \theta. When this is not the case, the quantity E[Y]-\theta is called the biased of the estimator.

Recall that for each i, Y_i is uniformly distributed on the interval (\theta,\theta+1), then E[Y_i] = \frac{\theta + \theta +1 }{2} = \theta + \frac{1}{2}.

Then, using the linear property of the expeted value, we have that

E[Y] = E[\frac{1}{n}\sum_{i=1}^{n} Y_i] = \frac{1}{n}\sum_{i=1}^{n} E[Y_i] = \frac{n (\theta+0.5)}{n} = \theta + 0.5

So, Y is a biased estimator of [tex]\theta [/tex} and the bias is 0.5.

b) We can easily obtain an unbiased estimator of theta by simply substracting the bias to the biased estimator, that is Y-0.5 is an unbiased estimator of the parameter theta.

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Allen�s hummingbird (Selasphorus sasin ) has been studied by zoologist Bill Alther A small group of 15 Allen� s hummingbirds has
Bogdan [553]

Answer:

1) 80% CI: [3.04; 3.26]gr

d= 0.11

2) n= 28 hummingbirds

Step-by-step explanation:

Hello!

The study variable of this experiment is:

X: the weight of a hummingbird. (gr)

And it has a normal distribution, symbolically: X~N(μ;σ²)

And (I hope I got it correctly) its population standard deviation is σ= 0.33

There was a sample of n= 15 hummingbirds taken, its sample mean X[bar]= 3.15 gr

1) You need to construct an 80% Confidence Interval for the population mean of the hummingbird's weight.

Since the study variable has a normal distribution, you can use either the standard normal distribution or the Student's t distribution. Both are useful to estimate the population mean. Since the population standard variance is known, the best choice is the Standard normal.

Z= <u> X[bar] - μ </u>~ N(0;1)

       σ/√n

The formula for the interval is:

X[bar] ± Z_{1- \alpha /2} * (σ/√n)

Z_{1- \alpha /2}= Z_{0.90} = 1.28

3.15 ± 1.28 * (0.33/√15)

[3.04; 3.26]gr

The margin of error (d) of a confidence interval is hal its amplitude (a)

a= Upper bond - Lower bond

d= (Upper bond - Lower bond)/2

d= \frac{(3.26-3.04)}{2} = 0.11

2) You need to calculate a sample size for a 80% Confidence interval for the average weight of the hummingbirds with a margin of error of d= 0.08

As I said before, the margin of error is half the amplitude of the interval, the formula you use to estimate the population mean has the following structure:

"point estamator" ± "margin of error"

Then the margin of error is:

d= Z_{1- \alpha /2} * (σ/√n)

Now what you have to do is rewrite the formula based on the sample size

d= Z_{1- \alpha /2} * (σ/√n)

\frac{d}{Z_{1- \alpha /2}}= σ/√n

√n * \frac{d}{Z_{1- \alpha /2}}= σ

√n = σ * \frac{Z_1- \alpha /2}{d}

n = (σ * \frac{Z_1- \alpha /2}{d})²

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n= 27.8784 ≅ 28 hummingbirds.

I hope it helps!

4 0
3 years ago
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timurjin [86]

Answer:

A = (2x² + 4x + 1) (8 − x)

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Step-by-step explanation:

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A = (2x² + 4x + 1) (8 − x)

If you wish, you can simplify with distribution.

A = 8 (2x² + 4x + 1) − x (2x² + 4x + 1)

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Let z represent the hypotenuse.

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z = 34 miles

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68dz/dt = 1024 + 3600

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dz/dt = 68 mph

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