STAT 245 Lecture Notes - Lecture 8: Unimodality

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pinkgoldfish18

23 Sep 2017

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The mean is most affected by skewness and outliers, i. e. the mean is pulled in the direction of the tail or the direction of the skewness. The median somewhat affected by skewness and outliers. The mode is unaffected by skewness and outliers. Given a data set, we usually want to know its mean and a measure of the spread. Data is not so spread out, it is quite concentrated near the mean, i. e. variability is small. Data is quite spread out, it is not so concentrated near the mean, i. e. variability is large. There are a few ways to define the spread, The standard deviation (by far the most popular way to define data spread). All three data sets have the same mean, i. e. for {50,50, 50} , the mean is. But their ranges are different, i. e. for {50,50, 50} , the range is. The bigger the range, the more spread out the data is.

Related Questions

Consider the following Minitab display of two data sets.

Variable	N	Mean	SE Mean	StDev	Minimum	Q1	Median	Q3	Maximum
C1	20	20.00	1.62	7.26	7.00	15.00	20.00	25.00	31.00
C2	20	20.00	1.30	5.79	7.00	20.00	22.00	22.00	31.00

(1) What are the respective means? The respective ranges?

mean

of c1: range of c1:

mean of c2:

range of c2:

(2) Which data set seems more symmetric? Why?

a. C2 seems more symmetric because the median is greater than the mean.

b. C1 seems more symmetric because the standard deviation is greater than the standard deviation for C2.

c. C1 seems more symmetric because the mean is equal to the median.

d. C2 seems more symmetric because the interquartile range is smaller than the interquartile range for C1.

(3) Compare the interquartile ranges of the two sets. How do the middle halves of the data sets compare?

a. The C1 distribution has a larger interquartile range that is symmetric around the median.

b. The C2 distribution has a smaller interquartile range that is symmetric around the median.

c. The C1 distribution has a smaller interquartile range that is symmetric around the median.

d. The C2 distribution has a larger interquartile range that is symmetric around the median.

cyanwildebeest393

In data mining, the test set is the data set used to ..........

(1) Build the data mining models

(2) estimate the accuracy of candidate models on unseen data

(3) estimate the accuracy of the final model on unseen data

(4) show actual versus predicted class values.

cerisefox680

QUESTION 1What is the purpose of post hoc tests? Under what circumstances are post hoc tests not needed or appropriate?

a. Under what circumstances is ANOVA a more appropriate statistical technique to use than a t test for independent samples? Please provide an example(s) to support your answer. (2 marks)

b. (2 marks)

c. In testing the difference between group means, how are between group and within group variability related to each other? Provide an example to support your answer. (2 marks)

QUESTION 2

a. If a two-factor ANOVA results in a significant main effect for factor A and a significant AxB interaction, explain why you should be cautious in interpreting the effect of factor A. (1 mark)

b. For the following research situations what would be your main effects and interaction effect. If there was a significant interaction effect what would that mean? (please be as clear as possible in your explanation):

A researcher hypothesizes that the effect of alcohol consumption on one’s motor skills depends on the person’s weight (underweight, normal, overweight), but that this may differ for men versus women. (2 marks)
A teacher is interested in seeing how attending class and reading the assigned chapters is related to her students’ performance on tests. She assesses each student in terms of how often they attend class (rarely, sometimes, regularly) and how often they read the assigned chapters (rarely, sometimes, regularly). (2 marks)

QUESTION 3

Ethel is interested in two methods of note-taking strategies and the effect of these strategies on the overall GPAs of college freshmen. She believes that men would benefit most from Method 1, while women would benefit most from Method 2. After obtaining 30 men and 30 women volunteers in freshmen orientation, she randomly assigns 10 men and 10 women to Method 1, 10 men and 10 women to Method 2, and 10 men and 10 women to a control condition. During the first month of the spring semester individuals in the two note-taking method groups receive daily instruction on the particular note-taking method to which they were assigned. The control group receives no note-taking instruction. Fall and spring GPAs for all participants are recorded and a score that is the difference between the spring and fall semester GPAs is calculated.

Below are the results from running a two-way ANOVA. Please answer the following:

What is the dependent variable? (0.5 marks)
What are the independent variables (factors)? (0.5 marks)
What is the research question(s) being addressed? (1 mark)
What does the descriptive statistics (graphical) tell us about the effects under study?

(1 mark)
What assumptions are being tested in the R output? Are these assumptions met? Please

provide statistical evidence. (2 marks)
What effects are being tested in the R output? (1 mark)
What does output from the ANOVA table tell us? (1 mark)
What type of follow-up analyses was conducted and why? Please interpret. (2 marks)

: Men
: Method1

      median
coef.var

0.300

0.469

: Women
: Method1

      median
coef.var

0.150

0.496

: Men
: Method2

      median
coef.var

0.275

0.027

: Women
: Method2

      median
coef.var

0.600

0.629

: Men
: Control

      median
coef.var

0.175

mean 0.335

mean 0.170

mean 0.305

mean 0.640

mean 0.165

mean 0.105

skew.2SE
   0.408

SE.mean CI.mean.0.95

var 0.052

var 0.033

var 0.037

var 0.032

var 0.022

var 0.021

normtest.W
     0.936

std.dev
  0.229

0.682
    skewness

1.076
    skewness

0.630
    skewness

0.072

SE.mean CI.mean.0.95

normtest.p 0.830 -----------------------------------------------------------------------------

skew.2SE
   0.342

kurtosis
  -0.645

kurt.2SE
  -0.242

normtest.W
     0.964

0.058

SE.mean CI.mean.0.95

std.dev
  0.183

normtest.p 0.096 -----------------------------------------------------------------------------

skew.2SE
   0.361

kurtosis
  -1.363

kurt.2SE
  -0.511

normtest.W
     0.869

0.061

SE.mean CI.mean.0.95

std.dev
  0.192

normtest.p 0.857 -----------------------------------------------------------------------------

0.278
    skewness

0.904
    skewness

skew.2SE
  -0.124

kurtosis
  -1.096

kurt.2SE
  -0.411

normtest.W
     0.979

-0.171

: Women
: Control

      median
coef.var

0.100

1.392
    skewness

0.560

0.046

kurtosis
  -0.759

0.105

kurt.2SE
  -0.284

   std.dev
     0.146

normtest.p
     0.512

skew.2SE
   0.019

kurtosis
  -1.443

kurt.2SE
  -0.541

normtest.W
     0.967

skew.2SE
   0.458

kurtosis
  -0.770

kurt.2SE
  -0.289

normtest.W
     0.906

0.056

SE.mean CI.mean.0.95

std.dev
  0.178

normtest.p 0.254 -----------------------------------------------------------------------------

                                                                      std.dev
                                                                        0.149

normtest.p 0.958 -----------------------------------------------------------------------------

0.047

SE.mean CI.mean.0.95

0.164

0.131

0.137

0.127

0.107

> leveneTest(gpa$gpaimpr~gpa$gender, data=gpa,center=mean)

Levene's Test for Homogeneity of Variance (center = mean) Df F value Pr(>F)

group 1 7.09 0.01* 58

---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1 > leveneTest(gpa$gpaimpr~method, data=gpa,center=mean) Levene's Test for Homogeneity of Variance (center = mean)

      Df F value Pr(>F)
group  2    1.94   0.15

> #running Levene's test for assumption of HOV (INTERACTION EFFECT)
> leveneTest(gpa$gpaimpr, interaction(gpa$gender,gpa$method),center=mean) Levene's Test for Homogeneity of Variance (center = mean)

      Df F value Pr(>F)
group  5    0.57   0.72

Df Sum Sq Mean Sq F value Pr(>F) gender 1 0.020 0.020 0.61 0.43753

method 2 1.174 0.587 17.81 1.1e-06 *** gender:method 2 0.695 0.348 10.54 0.00014 *** Residuals 54 1.780 0.033
---

Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

gender        0.0055
method        0.3200
gender:method 0.1894

0.011
0.397
0.281

eta.sq eta.sq.part

#selecting only Method1, run ANOVA, then follow-up tests

  Tukey multiple comparisons of means
    95% family-wise confidence level

Fit: aov(formula = gpaimpr ~ gender, data = gpaMethod1)
$`gender`
Women-Men -0.165 -0.3594851  0.02948506    0.0915581

diff        lwr        upr     p adj

#selecting only Method2, run ANOVA, then follow-up tests

  Tukey multiple comparisons of means
    95% family-wise confidence level

Fit: aov(formula = gpaimpr ~ gender, data = gpaMethod2)
$`gender`
Women-Men 0.335 0.161153 0.508847   0.000754

diff      lwr      upr    p adj

aquamarineox434

STAT 245 Lecture Notes - Lecture 8: Unimodality

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