Respuesta :
Answer:
a) p-hat (sampling distribution of sample proportions)
b) Symmetric
c) σ=0.058
d) Standard error
e) If we increase the sample size from 40 to 90 students, the standard error becomes two thirds of the previous standard error (se=0.667).
Step-by-step explanation:
a) This distribution is called the sampling distribution of sample proportions (p-hat).
b) The shape of this distribution is expected to somewhat normal, symmetrical and centered around 16%.
This happens because the expected sample proportion is 0.16. Some samples will have a proportion over 0.16 and others below, but the most of them will be around the population mean. In other words, the sample proportions is a non-biased estimator of the population proportion.
c) The variability of this distribution, represented by the standard error, is:
[tex]\sigma=\sqrt{p(1-p)/n}=\sqrt{0.16*0.84/40}=0.058[/tex]
d) The formal name is Standard error.
e) If we divided the variability of the distribution with sample size n=90 to the variability of the distribution with sample size n=40, we have:
[tex]\frac{\sigma_{90}}{\sigma_{40}}=\frac{\sqrt{p(1-p)/n_{90}} }{\sqrt{p(1-p)/n_{40}}}}= \sqrt{\frac{1/n_{90}}{1/n_{40}}}=\sqrt{\frac{1/90}{1/40}}=\sqrt{0.444}= 0.667[/tex]
If we increase the sample size from 40 to 90 students, the standard error becomes two thirds of the previous standard error (se=0.667).
Using the Central Limit Theorem, we get that:
a) Sampling distribution of sample proportions of size 40.
b) Symmetric.
c) 0.0580.
d) Standard error.
e) It would decrease.
The Central Limit Theorem establishes that, for a normally distributed random variable X, with mean and standard deviation , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean and standard erro .
For a skewed variable, we need n of 30 and greater.
For a proportion p in a sample of size n, we have that: and standard deviation .
In this problem:
- Proportion of 16%, thus [tex]p = 0.16[/tex].
- Sample of 40 students, thus [tex]n = 40[/tex]
Item a:
We are working proportions, and sample of 40, thus:
Sampling distribution of sample proportions of size 40.
Item b:
- Sample size of 40, thus, by the Central Limit Theorem, approximately normal, which is symmetric.
Item c:
This is the standard error, thus:
[tex]s = \sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.16(0.84)}{40}} = 0.0580[/tex]
It is of 0.0580.
Item d:
- The formal name is standard error.
Item e:
The formula is:
[tex]s = \frac{\sigma}{\sqrt{n}}[/tex]
- Since n is in the denominator, we can see that the standard error and the sample size are inversely proportional, and thus, increasing the sample size to 90, the variability would decrease.
A similar problem is given at https://brainly.com/question/14099217
