PBHL 5401 Lecture Notes - Lecture 21: F-Distribution, Null Hypothesis, Confidence Interval

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mauvehamster904

28 Mar 2016

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Week 10: hypothesis testing 3 or more means: We use anova to compare populations of subgroups. We compare 3 sample means to see if there is a difference among them. We are not asking if each mean is exactly equal. We are asking if each mean likely came from the larger overall population. =0. 5 so we test for 95% confidence interval. Variance formula or sum of squares: s2= ( x x )2 n 1 xbar is the average in the group. X is the value from one number in the group. (subtract the x from xbar) then square it ^2. You will do this for every number in the group individually then add them that"s the. Then the n-1 is sample size minus 1. So do that last then divide the total by the n-1. Step 1 sum of squares total / df total = mean squares total. Step 2 sum of squares between/ df between = mean squares between.

Related Questions

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

aquamarineox434

Introduction: A Chi-square test is used to compare observed data with expected data according to a hypothesis. For instance, if you were crossbreeding 2 heterozygous pea plants, you would expect to see a 3:1 phenotypic ratio in the offspring. In this case, if you were to breed 400 pea plants, you would expect to see 300 plants showing the dominant trait and 100 showing the recessive trait. But what happens if you observe only 260 plants with the dominant trait and 140 plants with the recessive trait? Does this mean something is wrong with Mendelian genetics or is this difference in expected results just due to chance (random sampling error)? These are the questions that can be answered using Chi-square statistics. The results of this statistical test is used to either reject or accept (fail to reject) the null hypothesis. The null hypothesis states there is no significant difference between the observed results and the expected results. This means that if the null hypothesis is accepted, the difference in observed and expected results was just a matter of chance and so the observed results basically "fit" with what was expected. Degrees of freedom (df) = number of independent outcomes (Y) being compared less 1 df = Y-1 At the 95% confidence interval we are 95% confident that there is a significant difference between the observed and expected results, therefore rejecting the null hypothesis. Probability Value - Is the decimal value determined from the X2 table and is the probability of accepting the null hypothesis. A 0.05 probability value equates to a 95% confidence interval.

The Chi-squared test formula is: Example: If we cross two pea plants that are heterozygous yellow pods, we would expect a 3:1 phenotypic ratio. So let's say we actually did the cross and got 280 plants with green pods and 120 plants with yellow pods. Question: Is this a 3:1 phenotypic ratio? This is the value of Chi-squared Test. We have a total of 400 plants and we expect a 300 green:100 yellow phenotypic ratio If the calculated Chi-squared value is less than the critical value listed in the Chi-squared table, then we accept the null hypothesis. This means that there is no significant difference between the observed and the expected values. Our degrees of freedom (df) = 2 outcomes - 1, or df = 1. Now we go the X2 table below and using the df = 1 and probability value of 0.05, our critical value is 3.84. Since our calculated X2 value is 5.33, and is larger than the critical value, we reject the null hypothesis and can say (at 95% confidence) that there is a significant difference between our observed and expected values.

The parent generation is yellowed podded and green podded pea plants. You cross a yellow podded pea plant with a green podded pea plant and you get 100% yellow podded plants in the F1 Generation (Phenotypic ratio 4 : 0, yellow to green). What will be the expected phenotypic ratio when you allow the F1 generation to reproduce?

Fill out the Punnett square.

If we actually did the cross and got 1150 yellow and 350 green. Would this be a consistent with what was expected?

Learning Outcomes Questions

1. Why would you run a Chi-squared test?

To determine if our data is consistent with expected results.
		a To determine if our data is consistent with expected results. b To determine if our data exactly matches the expected results.
		c To determine the expected results.
	d	To compare the phenotypic ratios to the genotypic ratios.

2. Determine the degrees of Freedom of the phenotypic ratio for this genetic cross.

a. 1

b. 2

c. 3

d. 4

e. 5

3. Using the data given, what is the result of your Chi-squared analysis? x2= ___.

	a.	2.22
	b	2.71
	c	4.36
	d	187.78
	e	448.27

4. Using the results of your Chi-squared analysis, do we fail to reject or reject the null hypothesis?

a.		Fail to reject the null
b.		Reject the null
c.		It cannot be determined from the data given

silversalmon27

PBHL 5401 Lecture Notes - Lecture 21: F-Distribution, Null Hypothesis, Confidence Interval

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