36-303 Sampling, Surveys & Society Homework 01 Solutions

February 10, 2010

1 Question 1

See Table 1.

(b) I couldn’t find any information about computer assistance, although I think they probably used computer assistance in randomizing geographic points and telephone numbers. Also, it is simply stated that one student from each household was selected by a systematic procedure to provide an approximate balance of respondents by sex, readers would like to know more details about the systematic procedure to decide on the validity of the survey. It is generally common to get nonresponse in telephone survey, I couldn’t find information about how was this problem dealt with in the survey so there might be some bias in the findings. I am not sure if the stratification method used in this survey was enough to draw conclusions about complex issues such as diversity of America’s youth.

(c)

\frac{2 ∗ \sqrt{0.5 ∗ 0.5}}{\sqrt{1005}} = 3.15%

2 Question 2

See Table 2.

(b) The information we actually have about the design and application of this survey is minimal. We do not know basic information such as the sampling frame and sampling design.

(c)

\frac{2 ∗ \sqrt{0.5 ∗ 0.5}}{\sqrt{302}} = 5.75%

Table 1: Survey: The State of Our Nation’s Youth

Sponsor The Horatio Alger Association of Distinguished Ameri-

can

Collector  collected by Peter D. Hart Research Associates

Purpose to give voice to the issues, opinions and thoughts of our

nation’s young people.

Year started  2005

Target Population students age 13 to 19 who identified themselves as ninth-

through-twelfth graders

Sampling Frame compiled list provided by american student list, the wellrespected national list management firm, which specializes in maintaining lists of k-12 students

Sample Design sample was created by selecting 505 geographic points randomly and proportionate to the population of each region and, within each region, by size of place. All telephone numbers have an equal chance to be included. One student from each selected household was included, selected by a systematic procedure to provide a balance of respondents by sex

Use of Interviewer  interviewer administered
Mode of Administration  telephone interview
Computer Assistance  not clear
Reporting Unit  student
Time dimension  conducted from May 9 to 12, 2005
Frequency  every year
Interviews per Round of Survey one
Levels of observation  student

Web Link  http://www.horatioalger.com/pdfs/state05.pdf/

Table 2: Survey: Raising the Bar: Employers’ views on college learning in the wake of the economic downturn

Sponsor  The Association of American Colleges and Universities
Collector  hart Research Associates

Purpose (1)find out if the majority of employers are expect to increase their hiring in the next year. If so, what kind of education background are most favorable to them for hiring (2) find out what kind of skills for new graduates do employers favor for hiring (3)find out if the employers are satisfied with college today in educating their employees to face challenges. If not, what needs to be improved in their opinion

Year started  2009

Target Population employers whose organizations have at least 25 employees and report that 25% or more of their new hires hold either an associates degree from a two-year college or a bachelor’s degree from a four-year college

Sampling Frame  not clear
Sample Design  not clear
Sample Size  302 employers
Use of Interviewer  yes, by associates at Hart research
Mode of Administration  not clear
Computer Assistance  not clear

Reporting Unit selected executives at private sector and non-profit organizations, including owners, CEOs, presidents, C-suite level executives, and vice presidents

Time dimension  conducted from October 27 to November 17,2009
Frequency  single application
Interviews per Round of Survey one
Levels of observation  employer, company

Web Link  http://www.aacu.org/leap/documents/2009_EmployerSurvey.pdf

3 Question 3

3.1 (a)

E[X_{i}] = E[X_{1}] (because the X_{i}’s are iid)
= 1 · p + 0 · (1 − p)
= p
V [X_{i}] = E[X_{i}^{2}] − E[X_{i}]^{2}
= E[X_{1}^{2}] − E[X_{1}]^{2}
= 1^{2} · p + 0^{2} · p − p^{2}
= p − p^{2} = p(1 − p)

3.2 (b)

E [∑_{i=1}^{n} X_{i}] = ∑_{i=1}^{n} E[X_{i}]
= ∑_{i=1}^{n} E[X_{1}]
= np
V [∑_{i=1}^{n} X_{i}] = ∑_{i=1}^{n} V [X_{i}]
= ∑_{i=1}^{n} V [X_{1}]
= np(1 − p)

Note that the identity V [∑_{i=1}^{n} X_{i}] = ∑_{i=1}^{n} V [X_{i}] is true because we know that the random variables X_{1}, X_{2}, ..., X_{n} are independent. This is not true in general.

3.3 (c)

Using the results from the previous sections,

E \left[{ \atop p̂}\right] = E [\frac{Y}{n} ]
= \frac{1}{n}E [Y]
= \frac{1}{n} · np
= p

The above property is called ”unbiasedness”. In this case we say that p̂ is an unbiased estimator of p.

Again using the results from the previous sections,

V \left[{ \atop p̂}\right] = V [\frac{Y}{n} ]
= \frac{1}{n^{2}} V [Y]
= \frac{1}{n^{2}} · np(1 − p)
= \frac{p(1 − p)}{n}

4 Question 4

4.1 (a)

P [X = 1] = P [X = 1, Y = 4] + P [X = 1, Y = 3] = \frac{1}{2}

P [X = 2] = 1 − P [X = 1] = \frac{1}{2}

P [Y = 3] = P [X = 1, Y = 3] + P [X = 2, Y = 3] = \frac{1}{2}

P [Y = 4] = 1 − P [Y = 3] = \frac{1}{2}

Now, to prove that X and Y are not independent, we just need to show a counter example:

(P [X = 1, Y = 4] = \frac{1}{8}) ≠ (P [X = 1] ∗ P [Y = 4] = \frac{1}{4})

4.2 (b)

assuming X and Y are independent, then

P [X = x|Y = y] = \frac{P [X = x, Y = y]}{P [Y = y]} = \frac{P [X = x] ∗ P [Y = y]}{P [Y = y]} = P [X = x]