Simple random sampling is not the most statistically efficient method, because __________.

1 answer:

8 0

Simple random sampling is not the most statistically efficient method, because 
it may not get a good representation of subgroups in a population.
Given that it is random, your results will not be representative of the whole population. Perhaps you want to get a sample of a specific group - then, this method wouldn't really be useful as it is random, meaning that anybody could be taken into consideration.

You might be interested in

Evaluate the following functions for the given value. 19. f(x) = -x3 + x2; -2

suter [353]

Answer:

Step-by-step explanation:

f(x) = -x^3 + x^2

Let x = -2

Evaluate the expression

f(-2) = -(-2)^3 + (-2)^2

= - (-2)*(-2) * (-2) + ( -2) * (-2)

= -(-8) + ( 4)

= 8 + 4

= 12

3 0

2 years ago

As part of a screening process, computer chips must be operated in an oven at 145 °C. Ten minutes after starting, the temperatur

Novosadov [1.4K]

Answer:

Step-by-step explanation:

I solved this using initial conditions and calculus, so I hope that's what you are doing in math. It's actually NOT calculus, just a concept that is taught in calculus.

The initial condition formula we need is

$y=Ce^{kt}$

Filling in our formula with the 2 conditions we are given:

$65=Ce^{10k}$ and $85=Ce^{15k}$

With those 2 equations, we have 2 unknowns, the C (initial value) and the k (the constant). We know that the initial value (or starting temp) for both conditions is the same, so we solve for C in one equation, sub it into the other equation and solve for k. If

$65=Ce^{10k}$ then

$\frac{65}{e^{10k}}=C$ which, by exponential rules is the same as