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ECON10062 Introductory Statistics 2010-11 2nd Sem es ter Course Content Coursew ork 1 Review Test Submission: Coursework1

Review Test Submission: Coursework1

User Shu Lim

Submitted 04/03/11 20:24

Name Coursework1

Status Completed

Score 4 out of 10 points

Time

Elapsed

34 minutes and 10 seconds out of 2 hours.

Instructions Read the instructions file that was publ ished in the relevant Blackboard Folder.

Follow the ins tructions given in that file BEFORE you start this tes t. You will be

presented with 10 ques tions which are randomly drawn from a larger pool of

questions. Each question has only one correct answer and is worth one point. This

assess ment will contribute 5% of your overall grade.

ECON10062

Introductory

Statistics 2010-11

2nd Semester

Noticeboard

Course Content

Additional Media

Communication

Course Information

Assessment

My Grades

Learning Resources

eLearning Support

Staff DetailsQuestion 1

This ques tion refers to the infstudy and the cjspolb variables of the British Crime

Survey dataset on Blackboard. What is the observed joint probability of picking a

respondent that is not full-time s tudent and not confident (this includes 'not very'

and 'not at all') that the police is effective at catching criminals?

Selected

Answer:

0.47232

Correct

Answer:

0.00855

Answer

Feedback:

wrong. This is P(not confident) = P(not very confident) + P(not

at all confident).

Response

Feedback:

revise the s tructure of jo int frequency and probability tables. Also

remember that you can add the probabili ties of two (or more)mutually exclusive categories .

Question 2

You have the following sample observations for the variable X: 45,78,56,74,63

and 45. Which of the following are correct describtive s tatistics?

Selected Answer: Mean = 60.1667, variance = 198.9667, skewness =

0.1418

Correct Answer: Mean = 60.1667, variance = 198.9667, skewness =

0.1418

Answer

Feedback:

correct

Response Feedback: well done

Question 3

0 out of 1 points

1 out of 1 points

0 out of 1 points

04/03/2011 Review Test Submission: Coursework1

online.manchester.ac.uk/…/review.jsp… 1/4



This question refers to variable (fairatt5) in the British Crime Survey dataset on

Blackboard, ignoring all respondents that do not answer the ques tion. Which of

the following statements is correct?

Selected

Answer:

About 40.8% of respondents believe that the British Criminal

Justice System is too soft on those accused of crime.

Correct

Answer:

More than 80% of respondents believe that the British Criminal

Justice System is too soft on those accused of crime.

Answer

Feedback:

Incorrect, 40.8% is the proportion of respondents that s trongly

agree 'that the CJS is soft on accused offenders'. However there is

another 39.9% who tend to agree with this statement. You shouldadd these two categories.

Response

Feedback:

These are ordinal data. Once you established the frequency for the

respons es 1 to 4, and then calculated the corresponding

probabilities you should easily answer this question.

Question 4

You have the following sample observations for the variable X: 45,78,56,74,63

and 45. What are the standardis ed Z value for the 3rd and 4th of these

observations?

Selected Answer: Z(X3) = -0.2954, Z(X4) = 0.9807

Correct Answer: Z(X3) = -0.2954, Z(X4) = 0.9807

Answer Feedback: correct


Question 5

Which of the following s tatements regarding measures of central tendency is

correct?

Selected

Answer:

When a distribution is symmetric the mean and median

will be very similar.

Correct

Answer:

When a distribution is symmetric the mean and median

will be very similar.

Answer

Feedback:

correct

Response Feedback: good work

Question 6

What is the correct measure of central tendency for the variable fairatt5 of the

British Crime Survey? Ignore all respondents that did not answer.

Selected

Answer:

Either Median = 2 or Mean = 1.8217

Correct

Answer:

Either Median = 2 or Mode = 1

Answer

Feedback:

Incorrect, these are categorical (ordinal) data. The mean cannot be

interpreted. The Median was correctly calculated

Response

Feedback:

revise how to calculate measures of central tendency and ens ure

you know when the mean can be calculated

Question 7

1 out of 1 points

1 out of 1 points

0 out of 1 points

0 out of 1 points





You have the following joint frequency table for the two variables pubeve (PV,

asks for frequency of visits to a pub, 1=no visits, 2=less than once a week,

3=once to twice a week, 4=at least 3 times a week) and nchil (NC, asks for

number of children under 16 in household):

pubeve

nchil 1 2 3 4

none 4374 2037 1491 591 8493

one 603 501 218 88 1410

at least 2 749 593 267 57 1666

5726 3131 1976 736 11569

What is the probability distribution for number of children (NC) conditional on the

respondent going to the pub at least three times a week (PV = 4)?

Selected

Answer:

P(NC=0|PV=4) = 0.0511, P(NC=1|PV=4) = 0.0076,

P(NC=2|PV=4) = 0.0049

Correct

Answer:

P(NC=0|PV=4) = 0.8030, P(NC=1|PV=4) = 0.1196,

P(NC=2|PV=4) = 0.0774

Answer

Feedback:

wrong, you report the joint probabilities, e.g. P(PV=1 AND NC=2).

This cannot be a probability distribution as the probabilities do not

sum to one.

Response

Feedback:

revise how to derive conditional probabilities from a

frequency table.

Question 8

This ques tion refers to the infstudy and the cjspolb variables of the British Crime

Survey dataset on Blackboard. What is the probability of a respondent being fairly

confident in the pol ice (being effective at catching criminals) conditional on the

respondent not being a full-time s tudent?

Selected

Answer:

0.4754

Correct

Answer:

0.4844

Answer Feedback:

wrong, you multiplied P(fairly confident) with P(not student). That isnot the observed conditional probability.

Response

Feedback:

revise the structure of joint frequency and probability tables and

the definition of conditional probabili ties.

Question 9

Which of the following s tatements regarding his torical probabilities is correct?

Selected

Answer:

Historical probabilities can be a guide for the probability of

future events if they have been derived from a large s ample that is

representative for the circumstances considered.

Correct

Answer:

Historical probabilities can be a guide for the probability of

future events if they have been derived from a large s ample that isrepresentative for the circumstances considered.

Answer

Feedback:

correct


Question 10

0 out of 1 points

1 out of 1 points

0 out of 1 points





Friday, 4 March 2011 20:25:23 o'clock GMT

You have the following joint frequency table for the two variables pubeve (asks for

frequency of visits to a pub, 1=no visits, 2=less than once a week, 3=once to

twice a week, 4=at least 3 times a week) and nchil (asks for number of children

under 16 in hous ehold):

pubeve

nchil 1 2 3 4

none 4374 2037 1491 591 8493

one 603 501 218 88 1410

at least 2 749 593 267 57 1666

5726 3131 1976 736 11569

What is the probability distribution for num ber of pub visits (PV) conditional on

there being at least two children in the household (NC = 2)?

Selected

Answer:

P(PV=1|NC=2) = 0.0647, P(PV=2|NC=2) = 0.0513,

P(PV=3|NC=2) = 0.0231, P(PV=4|NC=2) = 0.0049

Correct

Answer:

P(PV=1|NC=2) = 0.4496, P(PV=2|NC=2) = 0.3559,

P(PV=3|NC=2) = 0.1603, P(PV=4|NC=2) = 0.0342

Answer

Feedback:

wrong, you report the joint probabilities, e.g. P(PV=1 AND NC=2).

This cannot be a probability distribution as the probabilities do not

sum to one.

Response

Feedback:

revise how to derive conditional probabilities from a

frequency table.

OK



review test submission_ coursework1

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