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According to the national health survey, heights of adults may have a normal model with mean heights of 69.1" for men and 64.0" for women. Based on this information, on average how much taller are men then women?

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Karol E.
Karol E.Lv8
3 Apr 2021
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Related textbook solutions

Introductory Statistics
OER Edition, 2013
Openstax
ISBN: 9781938168208

Related questions

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):

  1. 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)

  2. 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)

1

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:

  1. What is the dependent variable? (0.5 marks)

  2. What are the independent variables (factors)? (0.5 marks)

  3. What is the research question(s) being addressed? (1 mark)

  4. What does the descriptive statistics (graphical) tell us about the effects under study?

    (1 mark)

  5. What assumptions are being tested in the R output? Are these assumptions met? Please

    provide statistical evidence. (2 marks)

  6. What effects are being tested in the R output? (1 mark)

  7. What does output from the ANOVA table tell us? (1 mark)

  8. What type of follow-up analyses was conducted and why? Please interpret. (2 marks)

2

: 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

3

> 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

57

> #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

54

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

4

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

5

Voldyne 5000 chart

The table at the bottom of this page accompanies the Voldyne 5000 Volumetric Exerciser. The table gives the typical lung capacity (in millilitres) for women for various ages and heights. Based on the chart, it is possible to conclude that the partial derivative of the lung cancer capacity with respect to age and with respect to height has constant values. What are those values?

 

  Height(in.) 58'' 60'' 62'' 64'' 66'' 68'' 70'' 72'' 74''
  20 1900 2100 2300 2500 2700 2900 3100 3300 3500
A 25 1850 2050 2250 2450 2650 2850 3050 3250 3450
G 30 1800 2000 2200 2400 2600 2800 3000 3200 3400
E 35 1750 1950 2150 2350 2550 2750 2950 3150 3350
  40 1700 1900 2100 2300 2500 2700 2900 3100 3300
I 45 1650 1850 2050 2250 2450 2650 2850 3050 3250
N 50 1600 1800 2000 2200 2400 2600 2800 3000 3200
  55 1550 1750 1950 2150 2350 2550 2750 2950 3150
Y 60 1500 1700 1900 2100 2300 2500 2700 2900 3100
E 65 1450 1650 1850 2050 2250 2450 2650 2850 3050
A 70 1400 1600 1800 2000 2200 2400 2600 2800 3000
R 75 1350 1550 1750 1950 2150 2350 2550 2750 2950
S 80 1300 1500 1700 1900 2100 2300 2500 2700 2900

1) Use descriptive statistics to investigate if there is an association between having had cholesterol and having had diabetes for the subjects in the sample. Then use an inference method to make a conclusion about the association between having had high cholesterol and having had diabetes in the population of all U.S. adults. 

2) Use descriptive analysis to compare the distributions of age between subjects having high cholesterol and those who have never had it. Then use an inference method to compare the mean age of U.S. adults having had high cholesterol to the mean age of U.S. adults who have never had high cholesterol.

 

This has to be done with RStudio with the file provided.

Thanks in advance!

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