Respuesta :
Answer:
D. The center remains constant, and the area in the tails of the distribution decreases.
Step-by-step explanation:
Hello!
Be it two independent random variables, X~N(μ;σ²) and U~Xₙ², the variable t is determined by the quotient between a random variable N(0;1) and the square root of a Chi-Square variable divided by its degrees of freedom:
[tex]t= \frac{(X-Mu)/Sigma}{\sqrt{U/n} }[/tex]
As a consequence of this, the structure of the distribution depends on the parameter n (degrees of freedom), it is centered in zero and has a bell-shape similar to the normal distribution.
It has a mean and variance:
E(t)= 0 for n > 1
V(t)= [tex]\frac{n}{n - 2}[/tex] n > 2
As you can see the variance of the distribution is directly affected by its degrees of freedom, which means that when the degrees of freedom change, the variance of the distribution change and so does its shape.
When ↑n ⇒ ↑V(t) ⇒ The area under the tails increases.
When ↓n ⇒ ↓V(t) ⇒ The area under the tails decreases.
In this example, the degrees of freedom of the distribution decreased from 40 to 20, then the variance of the distribution decreases and it "flattens", i.e. the area under the tails gets lowered.
The E(t) isn't affected by the modification of n, so the distribution remains centered in zero.
I hope this helps!
The center remains constant, and the area in the tails of the distribution decreases.
Given that,
A researcher studying the sleep habits of teens will select a random sample of n teens from the population to survey.
The researcher will construct a t -interval to estimate the mean number of hours of sleep that teens in the population get each night.
We have to determine,
Which of the following is true about the t -distribution as the value of n decreases from 40 to 20.
According to the question,
Two independent random variables, (μ;σ²) and U~Xₙ²,
The variable t is determined by the quotient between a random variable N(0;1) and the square root of a Chi-Square variable divided by its degrees of freedom:
[tex]t = \frac{\dfrac{x-\mu}{\sigma} }{\dfrac{\sqrt{\mu} }{n} }[/tex]
The structure of the distribution depends on the parameter n (degrees of freedom), it is centered in zero and has a bell-shape similar to the normal distribution.
It has a mean and variance:
E(t) = 0 for n > 1
V(t) = [tex]\frac{n}{n-2}[/tex] n > 2
The variance of the distribution is directly affected by its degrees of freedom, which means that when the degrees of freedom change, the variance of the distribution change and so does its shape.
Whenever the value of n increase V(t) also increase so that the area under the tails increases.
Whenever the value of decreases n the value of V(t) also decreases so that the area under the tails decreases.
The degrees of freedom of the distribution decreased from 40 to 20,
Then,
The variance of the distribution decreases and it "flattens", i.e. the area under the tails gets lowered.
The E(t) isn't affected by the modification of n, so the distribution remains centered in zero.
Hence, The center remains constant, and the area in the tails of the distribution decreases.
To know about Area under curve click the link given below.https://brainly.com/question/16525438
