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 p

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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 p

roject, 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?

Mathematics

1 answer:

AnnZ [12.1K] · Answer 1 · 2026-02-17T22:20:40+03:00

Answer:

a) p-hat (sampling distribution of sample proportions)

b) Symmetrical

c) σ=0.058

d) Standard error

e) By augmenting the sample size from 40 to 90 students, the standard error adjusts to two-thirds of the prior standard error (se=0.667).

Step-by-step explanation:

a) This distribution is defined as the sampling distribution of sample proportions (p-hat).

b) The expected form of this distribution is somewhat normal, symmetrical, and centered around 16%.

This occurs since the anticipated sample proportion is 0.16. Certain samples may exceed 0.16 while others may fall short, but most will cluster around the population mean. In essence, the sample proportions serve as an unbiased estimator of the overall proportion.

c) The distribution’s variability, denoted by the standard error, is:

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

d) The standard term is Standard error.

e) If we compare the variability between a sample size of 90 and a sample size of 40, we find:

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

When the sample size increases from 40 to 90 students, the standard error is reduced to two-thirds of the earlier standard error (se=0.667).