Respuesta :
Answer:
The standard deviation of the sampling distribution would change by a factor of [tex]\frac{1}{\sqrt{2}}[/tex]
Step-by-step explanation:
We use the central limit theorem to solve this question.
The Central Limit Theorem estabilishes that, for a random variable X, with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], a large sample size can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]s = \frac{\sigma}{\sqrt{n}}[/tex]
In this problem, we have that:
The standard deviation of the population is [tex]\sigma = 0.1[/tex]
Sample size n
[tex]s_{n} = \frac{0.1}{\sqrt{n}}[/tex]
Sample size 2n
[tex]s_{2n} = \frac{0.1}{\sqrt{2n}}[/tex]
What factor would the standard deviation of the sampling distribution of change?
[tex]F = \frac{s_{2n}}{s_{n}} = \frac{\frac{0.1}{\sqrt{2n}}}{\frac{0.1}{\sqrt{n}}} = \frac{\sqrt{n}}{\sqrt{2n}} = \frac{1}{\sqrt{2}}[/tex]
So the standard deviation of the sampling distribution would change by a factor of [tex]\frac{1}{\sqrt{2}}[/tex]
