Question

Repeated student samples. Of all freshman at a large college, 16% made the dean’s list in the current year. As part of a class project, students randomly sample 40 students and check if those students made the list. They repeat this 1,000 times and build a distribution of sample proportions.
(a) What is this distribution called?

(b) Would you expect the shape of this distribution to be symmetric, right skewed, or left skewed? Explain your reasoning.

(c) Calculate the variability of this distribution.

(d) What is the formal name of the value you computed in (c)?

(e) Suppose the students decide to sample again, this time collecting 90 students per sample, and they again collect 1,000 samples. They build a new distribution of sample proportions. How will the variability of this new distribution compare to the variability of the distribution when each sample contained 40 observations?

Answer 1

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:

$\sigma=\sqrt{p(1-p)/n}=\sqrt{0.16*0.84/40}=0.058$

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:

$\frac{\sigma_{90}}{\sigma_{40}}=\frac{\sqrt{p(1-p)/n_{90}} }{\sqrt{p(1-p)/n_{40}}}}= \sqrt{\frac{1/n_{90}}{1/n_{40}}}=\sqrt{\frac{1/90}{1/40}}=\sqrt{0.444}= 0.667$

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).