STAT 8010 Lecture Notes - Lecture 8: Confidence Interval, Standard Deviation, Sampling Distribution

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18 Aug 2016

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Describe the sampling distribution for the random variable (n-1)s2/ 2. The distribution is on the nonnegative side of the real line. The mean of the distribution is equal to the degrees of freedom. It is the sampling distribution for a function of s2. For 8 degrees of freedom, find the chi-square value such that 5% of the area is below it. The rejection region is found using a chi-square distribution with df=(n-1). Small-sample confidence intervals for a single variance n. Example: a medical supplies manufacturer claims that its new thermometers are so precise that the standard deviation in its measurements is smaller than . 25 f. To test this claim, a hospital took 10 measurements in an incubator. Below are the results of the experiment measured in degrees. Conduct a hypothesis test to determine if the manufacturer"s claim is true. {notes: 1) sampling distribution of s2 is right-skewed (chi-squared, 2); 2) chi-square practice on page 59; 3) formulas p. 60}

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2. (12 marks) In a study of 85 individuals, food questionnaires were used to measure the change in polyunsaturated fatty acids (PUFA) intake from the beginning to the end of the study. In addition researchers measured a genetic marker in the TNF-αgene; for this analysis, we will take the two values of this TNF-αvariable to be A- or A+. The researchers also collected information on gender, age and BMI. The primary question of interest is whether TNF-αmodiﬁes the eﬀect of PUFA on HDL.

(a) (1 mark) Let TNFa be a variable with value 1 for A+ and 0 for A- TNF-αstatus and

let gender be 1 for males and 0 for females. State a linear model for mean change in HDL that includes the predictors PUFA, TNFa, PUFA×TNFa interaction, and eﬀects for gender, age and BMI.

(b) (1 mark) A summary of the ﬁtted model is:

Estimate Std. Error t value Pr(>|t|)

(Intercept) 0.037531 0.196980 0.191 0.8494

pufa 0.019143 0.009245 2.071 0.0417

TNFa 0.074433 0.034128 2.181 0.0322

gender -0.008988 0.031793 -0.283 0.7781

age -0.003895 0.002156 -1.806 0.0747

bmi 0.006777 0.004332 1.565 0.1217

pufa:TNFa -0.024575 0.017844 -1.377 0.1724

Write a short sentence to describe why we would even consider removing gender from the model (i.e., why we would not just be happy with the above model from

the outset).

(c) (2 marks) Below is a summary of the model ﬁt without gender. Using our 10% rule-of-thumb, is there any evidence that gender confounds the PUFA×TNFa eﬀect?

If so why, if not why not?

Estimate Std. Error t value Pr(>|t|)

(Intercept) 0.024467 0.190364 0.129 0.8981

pufa 0.018841 0.009130 2.064 0.0423

TNFa 0.074850 0.033897 2.208 0.0301

age -0.003783 0.002107 -1.795 0.0765

bmi 0.006835 0.004301 1.589 0.1161

pufa:TNFa -0.024496 0.017737 -1.381 0.1712

(d) (2 marks) The researchers decided to use the model with PUFA, TNFa, PUFA×TNFa, age and BMI for inference of PUFA×TNFa interaction. What are the degrees of

freedom for the t-distribution whose critical value we need to construct conﬁdence intervals for the interaction? What upper tail probability should you use for a 95%

conﬁdence interval?

(e) (1 mark) The appropriate critical value is 1.96. Construct a 95% conﬁdence interval for the interaction.

(f) (2 marks) State the null hypothesis and a two-sided alternative hypothesis for testing whether TNFa modiﬁes the eﬀect of PUFA on HDL.

(g) (1 mark) From the computer output, what is the p-value for this test?

(h) (2 marks) For ﬁxed age and BMI, what is the estimated eﬀect of a one-unit change in PUFA for a subject who is A- at TNF-α? For ﬁxed age and BMI, what is the estimated eﬀect of a one-unit change in PUFA for a subject who is A+ at TNF-α.

crimsonyellowjacket200

Hypothesis testing with ANOVA opinions about whether caffeine enhances test performance differ. You design a study to test the impact of drinks with different caffeine contents on students' test-taking abilities. You choose 21 students at random from your introductory psychology course to participate in your study. You randomly assign each student to one of three drinks, each with a different caffeine concentration, such that there are seven students assigned to each drink. You then give each of them a plain capsule containing the precise quantity of caffeine that would be consumed in their designated drink and have them take an arithmetic test 15 minutes later.

The students receive the following arithmetic test scores:

	Cola	Black Tea	Coffee
Caffeine Content (mg/oz)	2.9	5.9	13.4
	85	85	92	∑X² = 147,641
	86	89	87	G = 1,755
	82	82	80	N = 21
	75	75	89	k = 3
	66	88	96
	78	76	83
	87	82	92
	T₁ = 559	T₂ = 559	T₃ = 559
	SS₁ = 338.86	SS₂ = 338.86	SS₃ = 338.86
	n₁ = 7	n₂ = 7	n₃ = 7
	M₁ = 79.8571	M₂ = 79.8571	M₃ = 79.8571

1.) You plan to use an ANOVA to test the impact of drinks with different caffeine contents on students' test-taking abilities. What is the null hypothesis?

(a) The population mean test score for the cola population is different from the population mean test score for the black tea population.

(b) The population mean test scores for all three treatments are equal.

(d) The population mean test scores for all three treatments are not all equal.

2.) Calculate the degrees of freedom and the variances for the following ANOVA table:

Source	SS	df	MS
Between	-	-	-
Within	702.28	-	-
Total	973.14	-

The formula for the F-ratio is:

3.) Using words, the formula of the F-ratio can be written as:

4.) Using the data from the ANOVA table given, the F-ratio can be written as:

5.) Calculate for F-ratio:

6.) At the level of significance, what is your conclusion?

(a)You can reject the null hypothesis; you do not have enough evidence to say that caffeine affects test performance.

(b)You cannot reject the null hypothesis; caffeine does appear to affect test performance.

(c)You cannot reject the null hypothesis; you do not have enough evidence to say that caffeine affects test performance.

(d)You can reject the null hypothesis; caffeine does appear to affect test performance.

ceruleanfly137

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

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STAT 8010 Lecture Notes - Lecture 8: Confidence Interval, Standard Deviation, Sampling Distribution

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