College of Doctoral Studies
RES-845: Module 3 Problem Set
Problem 1:
In this module, you are shown how to estimate the probability of getting a certain z-score value
equal to or higher than the one that is observed (i.e., more extreme in the tail), as well as the
proportion of all z-values that would NOT be in the tail of the distribution of all possible z-scores
in a normally shaped distribution. To do this, you compute the z-score value and then look up the
probabilities/proportions that match that z-score in Table B.1 in the back of your textbook.
(See pp. 178-179 and pp. 699-702 in textbook.)
Raw X
Value
Mean
SD
z-score
Prop./Prob. Prop./Prob.
in tail
in body
A.
38.25
30
5
B.
39.80
30
5
C.
17.00
14
1.5
D.
11.5
14
1.5
Summary
Which of the above raw score/z-values (a, b, c, and/or d) would be extreme
enough to occur 5% or less of the time (i.e., p < .05) within its distribution of
scores?
Problem 2:
An actual outcome can be compared with the probability of getting that outcome by chance
alone. This is the basis of inferential statistics. In inferential statistics, we are comparing what we
really observe with what would be expected by chance alone. That which would be expected by
chance alone would be the null hypothesis (that is, nothing is going on here but chance alone).
If we were to throw a coin, there would be a 50% chance it would come up heads, and a
50% chance it would come up tails by chance alone. By extension, if we threw the coin
20 times, we’d expect 50% (p = .5 or pn = 20*.5 = 10) of the tosses to come up heads,
and 50% (q = .5 or qn = 20*.5 = 10) to come up tails by chance alone if this is a fair coin.
a. You aren’t sure if your friend is using a fair coin when he offers to toss the coin to
decide who will win $100. You ask him to let you toss the coin 25 times to test it
out before you decide whether you will take the bet, using this coin. You toss the
coin 25 times and it comes up heads 19 times. Is this a fair coin (the null
hypothesis)? What is the probability of getting 19 heads in 25 tosses by chance
alone? You have decided that if the outcome of 19/25 tosses as heads would occur
less than 5% of the time by chance alone, you will reject the idea that this is a fair
coin. (see pp. 184-196 and 699-702 in textbook).
b. Now, suppose the outcome of your trial tosses was 15 heads in 25 tosses. What is
the probability of 15 heads in 25 tosses? Would you decide this is a fair coin,
using the 5% criterion as in question a (see pp. 184-196 and 699-702 in
textbook)?
Problem 3:
A teacher, Mrs. Jones, tests her 8th grade class on a standardized math test. Her class of 20
students (n) gets a mean score (M) of 80 on the test. She wants to know how her class did in
comparison with the population of all 8th grade classes that have taken this test. She goes to a
national database and finds out that the national average () of scores for the population of all 8th
graders who took this test is 78, with a population standard deviation ( of 3 points.
a. Based on the population mean and standard deviation, what is the expected mean
and standard deviation (standard error) for the distribution of sample means
based on the sample size of 20 students in a class? (See pp. 201-211)
b. If this distribution of the sample means is normal, what would be the z-score
equal to a mean test score of 80 that Mrs. Jones’ class received? (See pp. 211215, Table B.1 in textbook)
c. When you look up the z-score you computed in part b, what is the probability of
obtaining a sample mean greater than M = 80 for a sample of 20 in this
population? (See Table B.1 in textbook)
d. Mrs. Jones wants to know if her class did significantly better than the average 8th
grade class on this test. (See Chapters 7 and 8)
• What is the null hypothesis?
• What is the alternative hypothesis?
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•
Is the mean score obtained in Mrs. Jones’ class (sample) significantly
different from the population mean, using the criterion that her class’s
score would have to fall in the part of the distribution of all scores in the
population that is above the mean and has frequencies of occurrence of 5%
or less of all scores in the population (i.e., her class’s mean score would
have a probability of occurring by chance alone of p < .05)?
© 2012. Grand Canyon University. All Rights Reserved.
College of Doctoral Studies
RES-845: Module 3 Problem Set Solutions
Problem 1:
In this module, you are shown how to estimate the probability of getting a certain z-score value
equal to or higher than the one that is observed (i.e., more extreme in the tail), as well as the
proportion of all z-values that would NOT be in the tail of the distribution of all possible z-scores
in a normally shaped distribution. To do this, you compute the z-score value and then look up the
probabilities/proportions that match that z-score in Table B.1 in the back of your textbook.
(See pp. 178-179 and pp. 699-702 in textbook.)
Raw X
Value
Mean
SD
z-score
Prop./Prob. Prop./Prob.
in tail
in body
A.
38.25
30
5
B.
39.80
30
5
C.
17.00
14
1.5
D.
11.5
14
1.5
Summary
Which of the above raw score/z-values (a, b, c, and/or d) would be extreme
enough to occur 5% or less of the time (i.e., p < .05) within its distribution of
scores?
Solution:
Formula for 𝑧 =
𝑋−𝑀
𝑆𝐷
After computing the z-value, go to Table B.1 (pp. 699-702). Find the matching z-value in the
first column (A). Following across row where this z-value is, find the matching proportion (%)
of all scores that would fall in the tail of the distribution in column C. Find the matching
proportion (%) of all scores in the distribution that would fall in the in the body in Column B.
Raw X Value
Mean
SD
z-score
Prop./Prob. in tail Prop./Prob. in body.
a.
38.25
30
5
_1.65_
_.0495____
_.9505___
b.
39.80
30
5
_1.96_
_.025____
_.975___
c.
17.00
14
1.5
2.00_
_.0228___
_9772___
d.
11.50
14
1.5
- 1.67
_.0475___
_.9525___
In summary, which of the above raw score/z-values (a, b, c, and/or d) would be extreme enough
to occur 5% or less of the time (i.e., p < .05) within its distribution of scores?
All of them_(.0495, .025, .0228, and .0475 are all less than the value, .05)_ The extremes
are in the tail.
© 2012. Grand Canyon University. All Rights Reserved.
Problem 2:
An actual outcome can be compared with the probability of getting that outcome by chance
alone. This is the basis of inferential statistics. In inferential statistics, we are comparing what we
really observe with what would be expected by chance alone. That which would be expected by
chance alone would be the null hypothesis (that is, nothing is going on here but chance alone).
If we were to throw a coin, there would be a 50% chance it would come up heads, and a
50% chance it would come up tails by chance alone. By extension, if we threw the coin
20 times, we’d expect 50% (p = .5 or pn = 20*.5 = 10) of the tosses to come up heads,
and 50% (q = .5 or qn = 20*.5 = 10) to come up tails by chance alone if this is a fair coin.
a. You aren’t sure if your friend is using a fair coin when he offers to toss the coin to
decide who will win $100. You ask him to let you toss the coin 25 times to test it
out before you decide whether you will take the bet, using this coin. You toss the
coin 25 times and it comes up heads 19 times. Is this a fair coin (the null
hypothesis)? What is the probability of getting 19 heads in 25 tosses by chance
alone? You have decided that if the outcome of 19/25 tosses as heads would occur
less than 5% of the time by chance alone, you will reject the idea that this is a fair
coin. (see pp. 184-196 and 699-702 in textbook).
b. Now, suppose the outcome of your trial tosses was 15 heads in 25 tosses. What is
the probability of 15 heads in 25 tosses? Would you decide this is a fair coin,
using the 5% criterion as in question a (see pp. 184-196 and 699-702 in
textbook)?
Solutions:
a. The probability of getting 19 heads in 25 tosses of the coin is: _____.
pn = .5(25) = 12.5 and qn = .5(25) = 12.5 (Each is greater than 10, so the
approximation to the binomial distribution can be used.)
Mean () = pn = 12.5
SD () = √𝑛𝑝𝑞 = √25 ∗ .5 ∗ .5 = √6.25 = 2.5
X – pn
19 – 12.5
Z = √𝑛𝑝𝑞 = 2.5
= 2.6
© 2012. Grand Canyon University. All Rights Reserved.
Go to Table B.1 and find z = 2.6. What is the value for prop. in tail: .0047. This is
your probability: p = .0047. (Remember, to change a number to a percentage,
move the decimal point 2 places to the right.) So, this probability that this
outcome would happen by chance alone would be 00.47%, which is a lot less
than 5% (actually less than 1% as well). Reject the null hypothesis that says
this is a fair coin.
b. The probability of getting 15 heads in 25 tosses of the coin is: _____.
pn = .5(25) = 12.5 and qn = .5(25) = 12.5 (Each is greater than 10, so the
approximation to the binomial distribution can be used.)
Mean () = pn = 12.5
SD () = √𝑛𝑝𝑞 = √25 ∗ .5 ∗ .5 = √6.25 = 2.5
X – pn
15 – 12.5
Z = √𝑛𝑝𝑞 = 2.5
= 1.0
Go to Table B.1 and find z = 1.0. What is the value for prop. in tail: .1587? This
is your probability: p = .1587. (Remember, to change a number to a
percentage, move the decimal point 2 places to the right.) So, this would
mean that by chance you’d expect the outcome to occur 15.87% of the time,
which is more than 5%. Thus, you would not reject the null hypothesis that
this is a fair coin. Thus, you’d believe that the outcome was by chance alone
and go on to use this coin if you accepted the bet.
In the second example, 15 of 25 tosses coming up heads could be expected to
happen .1587 (or 15.87% ) of the time. Thus, this is not so rare (and is much more
© 2012. Grand Canyon University. All Rights Reserved.
than only 5% of the time that I said would be my cut off for deciding that this is
not a fair coin.
Problem 3:
A teacher, Mrs. Jones, tests her 8th grade class on a standardized math test. Her class of 20
students (n) gets a mean score (M) of 80 on the test. She wants to know how her class did in
comparison with the population of all 8th grade classes that have taken this test. She goes to a
national database and finds out that the national average () of scores for the population of all 8th
graders who took this test is 78, with a population standard deviation ( of 3 points.
a. Based on the population mean and standard deviation, what is the expected mean
and standard deviation (standard error) for the distribution of sample means
based on the sample size of 20 students in a class? (See pp. 201-211)
b. If this distribution of the sample means is normal, what would be the z-score
equal to a mean test score of 80 that Mrs. Jones’ class received? (See pp. 211215, Table B.1 in textbook)
c. When you look up the z-score you computed in part b, what is the probability of
obtaining a sample mean greater than M = 80 for a sample of 20 in this
population? (See Table B.1 in textbook)
d. Mrs. Jones wants to know if her class did significantly better than the average 8th
grade class on this test. (See Chapters 7 and 8)
• What is the null hypothesis?
•
What is the alternative hypothesis?
•
Is the mean score obtained in Mrs. Jones’ class (sample) significantly
different from the population mean, using the criterion that her class’s
score would have to fall in the part of the distribution of all scores in the
population that is above the mean and has frequencies of occurrence of 5%
© 2012. Grand Canyon University. All Rights Reserved.
or less of all scores in the population (i.e., her class’s mean score would
have a probability of occurring by chance alone of p < .05)?
Solutions:
a. The distribution of sample means would have a mean of 78 (same as the
population mean) and a
standard error ofM = pop. √𝑠𝑎𝑚𝑝𝑙𝑒 𝑠𝑖𝑧𝑒 = √20
b.
𝑧=
𝑋−𝜇
𝜎𝑀
=
80−78
.67
=
2
.67
= 2.98
c. Proportion in tail = .0014, p = .0014, only .14% of all means in the distribution of
means would be equal to or greater than a score of 80.
d. Null hypothesis: The mean score of Mrs. Jones’ class is not different from the
mean score of all 8th graders who took this test.
Alternative hypothesis: Mrs. Jones’ class mean differed significantly (was
significantly higher) from the mean score of all 8th graders who took this test.
Significantly different? Yes, the actual probability of a class of 20 receiving a
mean score of 80 has a probability of occurring less than p < .05; actual p = .0014.
© 2012. Grand Canyon University. All Rights Reserved.
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