Respuesta :
Answer:
a) The mean is 0.15 and the standard error is 0.056.
b) 1. Yes
Step-by-step explanation:
The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]s = \frac{\sigma}{\sqrt{n}}[/tex].
For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.
For proportions p, in samples of size n, the mean is [tex]\mu = p[/tex] and the standard error is [tex]s = \sqrt{\frac{p(1-p)}{n}}[/tex]. The Central Limit Theorem applies is np > 5 and np(1-p)>5.
In this question:
[tex]n = 40, p = 0.15[/tex]
So
(a) Find the mean and the standard error of the distribution of sample proportions.
[tex]\mu = 0.15, s = \sqrt{\frac{0.15*0.85}{40}} = 0.056[/tex]
So the mean is 0.15 and the standard error is 0.056.
(b) Is the sample size large enough for the Central Limit Theorem to apply?
np = 40*0.15 = 6 > 5
np(1-p) = 40*0.15*0.85 = 5.1>5
So yes
The standard error of the distribution of sample proportions is 0.056 and mean is 0.15.
Yes, the sample size is enough for the Central Limit Theorem to apply.
(a). Given that, size of sample, [tex]n=40[/tex]
Proportion, [tex]p=0.15[/tex]
In the distribution of sample proportions, mean [tex]\mu=p[/tex]
and, standard error = [tex]\sqrt{\frac{p(1-p)}{n} }[/tex]
So, mean [tex]\mu=0.15[/tex]
Standard error =[tex]\sqrt{\frac{0.15(1-0.15)}{40} }=0.056[/tex]
(b). The Central Limit Theorem applies if np > 5 .
[tex]np=40*0.15=6>5[/tex]
Thus, the Central Limit Theorem is applied.
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