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A quality control system selects a sample of 10 items from each batch of products for testing. If 2 or more of the sample are defective the whole batch is rejected. If the probability of an item being defective is 0.05, what is the probability: i. of having 2 defects in the sample? ii. that the batch is rejected? iii. mean number of defects in a batch? iv. standard deviation of number of defects in a batch?

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Introductory Statistics
OER Edition, 2013
Openstax
ISBN: 9781938168208

Related questions

Rules of dice game:

• All players start with 10 points

• A dice is rolled, and if the number on the dice is 1, 2, 3, 4 or 5, this number is added to each player's score. If the number is 6, the score off all players is reduced to zero

• Before the dice is rolled each player is given the choice of continuing or taking their current score as their final score.

• This process is repeated until a 6 is rolled and the game ends. All players still playing at this point receive zero as their final score.

Suppose that you choose to play the first dice roll.

(i) Compute the expected value and standard deviation of your score after the first dice roll.

(ii) On average, do you do worse or better than someone who choose not to play and take 10 as their final score?

Now suppose that a 6 was not rolled and you chooses to continue playing

(iii) On average, what score would you have before the dice is rolled?

(iv) Compute the expected value for your score after the second dice roll.

(v) On average, do you do worse or better than someone who choose to stop playing after 1 dice roll?

(vi) Is there an optimal number of dice rolls you should continue playing for? Explain your reasoning.

(vii) If the rules are changed such that when the number is 2, 3, 4, 5 or 6, this number is added to your score. While if the number is 1, your score is reduced to zero. Discuss how this affects the optimal number of dice rolls you should continue playing for.

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

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