2 years ago

Parametric test based on standard error

1 answer:

8 0

Answer:

The parametric hypothesis tests that we introduced in Chapter 5 were all based upon the means of the sample and population.

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Irina18 [472]

I hope this helps you

6 0

3 years ago

Sladkaya [172]

The awnser for question 5 is 4/6

7 0

3 years ago

zhannawk [14.2K]

Assuming you are referring to simple interest not compound that that it is annual you would gain $100 putting you at $600

7 0

3 years ago

tangare [24]

Yes, it’s a function because there is no repeating x-values,

7 0

3 years ago

Luba_88 [7]

Let's take the first derivative:

$f'(x) = (-5e^{-x/3})' = -5 \times (e^{-x/3})' = -5e^{-x/3} \times \left(-\dfrac{1}{3}\right) = \dfrac{5}{3}e^{-x/3}.$

Notice that we can write this as:

$f'(x) = -\dfrac{f(x)}{3}.$

By taking the derivative of both sides $n$ times, we get:

$f^{(n+1)}(x) = -\dfrac{f^{(n)}}{3}.$

This means that each time you take a derivative, a factor of $-\dfrac{1}{3}$ will appear. So we conclude that:

$f^{(n)}(x) = \left(-\dfrac{1}{3}\right)^nf(x).$

Taking $n=6$ and $x=1$ , we get:

$f^{(6)}(1) = \left(-\dfrac{1}{3}\right)^6 f(1) = \dfrac{-5e^{-1/3}}{729} = -\dfrac{5}{729}e^{-1/3}.$

So we finally get:

$\boxed{-\dfrac{5}{729}e^{-1/3}}.$

6 0

3 years ago

Parametric test based on standard error ​