Give an example of a qualitative variable and an example of a quantitative variable (discrete or continuous.) Explain the common
graphical displays designed for use with
(a) qualitative data and
(b) quantitative data.
Note: Please answer all questions above by Wednesday and respond to at least three peers by Saturday night. Please, give page numbers of your textbook or some web links to support your answer.
(a) qualitative data and
(b) quantitative data.
Note: Please answer all questions above by Wednesday and respond to at least three peers by Saturday night. Please, give page numbers of your textbook or some web links to support your answer.
2 answers:
Answer:
a) Pie and Bar Chart b) Dot Plot, Bar Graph, Box Plot among others.
Step-by-step explanation:
a) Qualitative data examples.
There are some good examples of qualitative data, or variables like gender, level of study, marital status, etc.
These qualitative variables can be better displayed graphically with the use of several graphics. But the options are more restricted since there is not much information to be taken out of these when the graphic is solely dedicated to qualitative variables.
(Check below)
b) Quantitative data.
On the other hand, quantitative variables are the countable ones. The Quantitative Data can be either Discrete (for Integer numbers) or Continuous for any number ∈ R.
There is a wider set of options for displaying quantitative data, for quantitative variables, because we can extract more information from them. Since all we want is to display the position of that variable, then we can use Dot plots, Bar Graph, etc. Trace curves, etc.
(Check it below)
Examples:
- Means of 6 groups (Dot plots)
- Ages (Bar Graphs).
- Volume of Largest Dams in South America
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Answer:
a) Reject H₀
b) [0.31; 3.35]
Step-by-step explanation:
Hello!
a) The objective of this example is to compare if the population means of the production rate of the assembly lines A, B and C. To do so the data of the production of each line were recorded and an ANOVA was run using it.
The study variable is:
Y: Production rate of an assembly line.
Assuming that the study variable has a normal distribution for each population, the observations are independent and the population variances are equal, you can apply a parametric ANOVA with the hypothesis:
H₀ μ₁= μ₂= μ₃
H₁: At least one of the population means is different from the others
Where:
Population 1: line A
Population 2: line B
Population 3: line C
α: 0.01
This test is always one-tailed to the right. The statistic is the Snedecor's F, constructed as the MSTr divided by the MSEr if the value of the statistic is big, this means that there is a greater variance due to the treatments than to the error, this means that the population means are different. If the value of F is small, it means that the differences between populations are not significant ( may differ due to error and not treatment).
The critical region is:
If F ≥ 3.36, the decision is to reject the null hypothesis.
Looking at the given data:
= 11.32653
With this value the decision is to reject the null hypothesis.
Using the p-value method:
p-value: 0.001005
α: 0.01
The p-value is less than the significance level, the decision is to reject the null hypothesis.
At a level of 5%, there is significant evidence to say that at least one of the population means of the production ratio of the assembly lines A, B and C is different than the others.
b) In this item, you have to stop paying attention to the production ratio of the assembly line A to compare the population means of the production ratio of lines B and C.
(I'll use the same subscripts to be congruent with part a.)
The parameter to estimate is μ₂ - μ₃
The populations are the same as before, so you can still assume that the study variables have a normal distribution and their population variances are unknown but equal. The statistic to use under these conditions, since the sample sizes are 6 for both assembly lines, is a pooled-t for two independent variables with unknown but equal population variances.
t= (X[bar]₂ - X[bar]₃) - ( μ₂ - μ₃) ~t
Sa√(1/n₂+1/n₃)
The formula for the interval is:
(X[bar]₂ - X[bar]₃) ± *
Sa= 0.827 ≅ 0.83
X[bar]₂ = 43.33
X[bar]₃ = 41.5
(43.33-41.5) ± 3.169 *
1.83 ± 3.169 * 0.479
[0.31; 3.35]
With a confidence level of 99% you'd expect that the difference of the population means of the production rate of the assemly lines B and C.
I hope it helps!
