Respuesta :
Answer:
B. Increasing the sample size by a factor of 4 decreases the standard error by a factor of 2. Changing the standard error doubles the magnitude of the standardized Z-value.
Step-by-step explanation:
Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".
The central limit theorem states that "if we have a population with mean μ and standard deviation σ and take sufficiently large random samples from the population with replacement, then the distribution of the sample means will be approximately normally distributed. This will hold true regardless of whether the source population is normal or skewed, provided the sample size is sufficiently large".
Let X the random variable of interest. We know from the problem that the distribution for the random variable X is given by:
[tex]X\sim N(\mu ,\sigma )[/tex]
We take a sample of n . That represent the sample size.
From the central limit theorem we know that the distribution for the sample mean [tex]\bar X[/tex] is also normal and is given by:
[tex]\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})[/tex]
The standard error is given by this formula:
[tex]Se=\frac{\sigma}{\sqrt{n}}[/tex]
If we increase the sample size by a factor of 4 the standard error needs to decrease since the value of n is on the denominator for the standard error.
A. Increasing the sample size by a factor of 4 increases the standard error by a factor of 2. Changing the standard error doubles the magnitude of the standardized Z-value.
False, if we increase the sample size the standard error not increase, decrease.
B. Increasing the sample size by a factor of 4 decreases the standard error by a factor of 2. Changing the standard error doubles the magnitude of the standardized Z-value.
True, when we increase the sample size by a factor of two we have this:
[tex]Me_f = \frac{\sigma}{\sqrt{4n}}=\frac{1}{2} \frac{\sigma}{\sqrt{n}}=\frac{1}{2} Me_i[/tex]
And makes sense this part "Changing the standard error doubles the magnitude of the standardized Z-value", beacuse the z value is given by:
[tex]z=\frac{\sqrt{n}(\bar X -\mu)}{\sigma}[/tex]
[tex]z_f=\frac{\sqrt{4n}(\bar X -\mu)}{\sigma}=2\frac{\sqrt{n}(\bar X -\mu)}{\sigma}[/tex]
C. Increasing the sample size by a factor of 4 decreases the standard error by a factor of 2. Changing the standard error decreases the standardized Z-value to half of its original value.
False, the first part of the statement is true but the second part no "Changing the standard error decreases the standardized Z-value to half of its original value", the standard error is a doubles from the original since we increase the sample size by a factor of 2 and [tex]\sqrt 4=2[/tex].
D. Increasing the sample size by a factor of 4 increases the standard error by a factor of 2. Changing the standard error decreases the standardized Z-value to half of its original value.
False, if we increase the sample size the standard error not increase, decrease.
