Franklin, thank you for your intuitive answers to our discussion questions this week:
D3.4.1 (a) What is the mean visualization test score?
After recalculating my entire worksheet, I agree that the output on 4.1b the answer for the mean
visualization test score is 5.2433. The mean visualization test is located on page 70 in our text
(Morgan, Leech, Gloeckner, & Barrett, 2020).
D3.4.1 (b) What is the skewness statistic for math achievement test? What does this tell us?
I agree that the output of 4.1b, which are descriptives for variables initially labeled as Scale. The
skewness statistic for the math achievement test is .044. Skewness is when the tail of a frequency
distribution is longer than the other, and if the mean and median are very different, the curve is
skewed (Morgan, Leech, Gloeckner, & Barrett, 2020). The skewness of a perfectly normal curve has
a skewness of zero. The skewness of .044 tells us that the variable is not very skewed since on
the normal curve the skewness is zero. The .044 is approximately normally distributed.
D3.4.1 (c) What is the minimum score for mosaic pattern test? How can that be?
I disagree with your answer for D3.4.1(c) in the output 4.1b, which are descriptives for variables
initially labeled as Scale. The minimum core on the mosaic pattern test is -4.0. Well, if this were
an error, the best place to find an error, is looking into the codebook or going back over the data
set that was used. In looking into the codebook, you can see that the visualization scores range
from -4.0 to 16. The visualization score of -4.0 would tell us that at least one person scored the
lowest possible score, which is probably negative due to guessing wrong on the questions
(Morgan, Leech, Gloeckner, & Barrett, 2020).
D3.4.2 (a) For which variables that we called scale, is the skewness statistic more than 1.00
or less than –1.00?
I agree that the output in 4.1b, the variable that has a skewness of more than 1.00 or less than -
1.00 is known as the “competence scale” and is negatively skewed data.
D3.4.2 (b) Why is the answer important?
I couldn’t find the answer to this question for why this is set of data is important but in regards to
the variable “competence scale” is negatively skewed which mean there is no symmetricity and
the values of the mean, median and mode will not be equal. When the data is skewed this means
that the curve is not normally distributed (Morgan, Leech, Gloeckner, & Barrett, 2020).
D3.4.2 (c) Does this agree with the boxplot for Output 4.2? Explain.
Yes, I agree that Output 4.2 does agree with the boxplot also. The skewed data shows a lopsided
box plot, where the median cuts the box into two equal pieces. If the longer part of the box is to
the right or above the median, the data is said to be skewed right. If the longer part is to the left
or below the median, as our case, the data is skewed left or negative.
D3.4.3 (a) How many participants have missing data?
I agree that the Output 4.2b, Boxplots of Competence and Motivation Scales in the processing
summary shows that both the competence scale and motivation scale are both missing 4 cases.
D3.4.3 (b) What percent of students have a valid (non-missing) motivation scale or
competence scale score?
I agree with your answer for Output 4.2b, Boxplots of Competence and Motivation Scales. The
percent of students that have a valid (non-missing) motivation scale or competence scale score is
94% or a little greater.
D3. 4.3 (c) Can you tell from Outputs 4.1b and 4.2b how many are missing both motivation
scale and competence scale scores? Explain.
I agree with both answers for Output 4.1b and 4.2b. In 4.1b the competence scale has 73 and the
motivation scale has 73 and the population is 75. In this output both are missing two scores. In
the output 4.2b, the competence scale is 71 and the motivation scale 71. Therefore, no one is
missing both motivation and competence scores, because two are missing each of the scores, and
four are missing at least one of the scores (Morgan, Leech, Gloeckner, & Barrett, 2020).
D3.4.4 (a) Can you interpret the means? Explain.
I agree with your answers for the mean Output in 4.4, Descriptives for Dichotomous Variables.
When dealing with dichotomous variables, you can use the mean to understand what percentage
of participants fall into each of the two groups. For example, the mean of academic track is .55,
which indicates that 55% of the participants were coded as 1 (regular track); thus 45% were
coded 0 (fast track) . Because the mean is greater than .50, there are more students in the regular
track than in the fast track. If the mean is close to 1 or 0 (see algebra 1 in h.s. and calculus in
h.s.), then splitting the data on that dichotomous variable might not be useful because there will
be many participants in one group and very few participants in the other (Morgan, Leech,
Gloeckner, & Barrett, 2020).
D3.4.4 (b) How many participants are there altogether?
I also agree that Output 4.4, Descriptives for Dichotomous Variables participants is 75 which is
the population or N.
D3.4.4 (c) How many have complete data (nothing missing)?
I agree that in Output 4.4, we can see that all the variables have a population of 75 which means
all of these have complete data (nothing missing).
D3.4.4 (d) What percent are in the fast track?
I disagree with your answer of 75% on the fast track. After calculating that the first variable,
academic track we know that the mean is .55 which states that 55% of the participants were
coded as 1 (regular track) and the other 45% would be coded to 0 (fast track).
The breakdown is as follows:
Algebra 1 in h.s. we know that the mean is .79 which states that 79% of the participants were
coded as 1 (regular track) and the other 21% would be coded to 0 (fast track). Algebra 2 in h.s.
we know that the mean is .47 which states that 47% of the participants were coded as 1 (regular
track) and the other 53% would be coded to 0 (fast track). Geometry in h.s. we know that the
mean is .48 which states that 48% of the participants were coded as 1 (regular track) and the
other 52% would be coded to 0 (fast track). Trigonometry in h.s. we know that the mean is .27
which states that 27% of the participants were coded as 1 (regular track) and the other 73%
would be coded to 0 (fast track). Calculus in h.s. we know that the mean is .27 which states that
27% of the participants were coded as 1 (regular track) and the other 73% would be coded to 0
(fast track). Math grade. we know that the mean is .41 which states that 41% of the participants
were coded as 1 (regular track) and the other 59% would be coded to 0 (fast track).
D3.4.4. (e) What percent took algebra 1 in h.s.?
I disagree with your answer of 75% getting the percent that took algebra 1 in h.s. First, we must
take the mean and multiply it by the population which is .79 x 75 and divide this by the number
of variables 7 and multiply by the population 75 and then multiply by 100% to get the
percentage. The total percentage that took algebra 1 in h.s. would be 11%.
D3.4.5. (a) 9.6% of what group are Asian-Americans?
I agree that the percentage of Asian Americans were 9.6%. In looking at the chart, you can see
that it says “valid” and “missing”. The 9.6% is the percentage of subjects in the study who made
a valid answer to this question and listed themselves as Asian-Americans (Morgan, Leech,
Gloeckner, & Barrett, 2020).
D3.4.5 (b) What percent of students have visualization 2 scores of 6?
I disagree with your visualization of students’ scores. When going back to the visualization data
scores using output 4.5, there is no missing data in this variable, so the valid percent and the
percent are both the same. The percent is 5.3% (Morgan, Leech, Gloeckner, & Barrett, 2020).
D3.4.5 (c) What percent had such scores of 6 or less?
I disagree with your percentage of scores of 6 or less. When using output 4.5, you can find this
by using the cumulative percent column which is 70.7%.
Franklin,
I was guilty of not taking my time and answering these discussion questions producing wrong
answers. For you especially, I would break down the questions and answer the individual
problems first. Combining these questions is confusing the readers only because your answers
are not within the question. I am guilty of time management, and we should try harder in the
upcoming weeks. God Bless and thank you for your time.
References
References
Morgan, G., Leech, N., Gloeckner, G., & Barrett, K. (2020). IBM SPSS for Introductory
Statistics: Use and Interpretation, Sixth Edition (6th Ed.). Routledge.
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