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Psyc 233         Exam 4           Online              Part I

Statistics

Psyc 233         Exam 4           Online

Part I. Computation  Show your work for computations.  No Work = No Credit!

1. A researcher is interested in testing whether men with beards have a different number of children than the unshaven general population.  The general population has on average µ = 1.3 children.  A sample of 25 bearded men was found to have an average number of children of = 1.8, with a standard deviation of 1.2.  Test whether bearded men have a different number of children.  Set alpha to α = .05

STEP 1: State your hypotheses in words (include and label both H0 and H1).    (2 points)

H0: The average number of children in the population of men with beards is equal to the average number of children in the general population

H1: The average number of children in the population of men with beards is not equal to the average number of children in the general population

State your hypotheses in symbols (include and label both H0 and H1). (2 points)

H0: µ = 1.3

H1: µ ≠ 1.3

STEP 2:  Set up the criteria for making a decision.  That is, find the critical value.

(2 points)

t (2-sided, 0.05, 24) = ±2.064

STEP 3:  Summarize the data into the appropriate test-statistic.  That is, compute the t statistic.                                                                                                           (3 points)

t = (- µ )/ Sx

Sx = 1.2/ 5 = 0.24

t = (1.8 – 1.3)/ 0.24

t = 2.083

STEP 4:  Evaluate the Null Hypothesis (Reject or Fail to reject?)               (1 point)

Reject the null hypothesis

STEP 5:  What is your conclusion?                                                               (1 point)

Conclude that the average number of children in the population of men with beards is significantly different from the average number of children in the general population

2.  A researcher wants to examine the effects of caffeine on task performance.  Participants are randomly assigned to one of three groups based on quantity of caffeine they are given just before the task.  Below are the number of correct responses on the task after caffeine therapy.  Is there a difference in the scores between the groups?  Set alpha to .05

300mg                         1000mg                       2000mg

n=4                              n=4                              n=4

STEP 1:  State the null and research hypotheses.                                                                                                                                                     (2 points)

H0: µ300 = µ1000 = µ2000

H1: At least one of the means is different from the others

STEP 2:  Set up the criteria for making a decision.  That is, find the critical value.

df between = k – 1

df between = 2

df within = n -k

df within = 9

F0.05 (2, 9) = 4.26

STEP 3:  Compute the appropriate test-statistic. Show your work!             (12 points)

(I have already computed SSBetween for you).

SSbetween = 142/ 4 + 322/ 4 + 452/ 4 - 912/ 12

SSbetween = 121.167

SSwithin = [58 – 142 /4] + [294 - 322 /4] + [600 - 452/ 4]

SSwithin = 9 + 38 + 93.75

SSwithin = 140.75

Source                           SS                     df                      MS                    F

Between (group)         121.15             __2__              _60.575____            ___3.874__

Within (error)             __140.75___   _9___              __15.638___

Total                           ___261.90__   _10___

STEP 4:  Evaluate the null hypothesis (based on your answers to the above steps).

(1 point)

We fail to reject the null hypothesis as the F statistic is smaller than the F critical

STEP 5:  Based on your evaluation of the null hypothesis, what is your conclusion?

(1point)

None of the three group pairings had significantly different performance in the allocated task

2B.  Based on your conclusion in the ANOVA problem you just worked, should you conduct a Tukey’s Post-hoc test? Highlight or underline one                        (1 point)

Yes                                          No

3.  A researcher conducts an ANOVA to see if 3 different levels of gas octane affect the number of miles per gallon a car achieves.  The omnibus ANOVA was significant with MSerror = 30.39.  The mean gas millage for each octane level is provided below.  Conduct Tukey’s post-test to determine which groups differed from one another by:  1. stating all mean differences,  2. computing the HSD, and 3. stating which groups differ based on your test.  Set alpha at .05.                                                                            (5 points)

X1(High Octane):

Mean = 32mpg; n = 21

X2(Moderate Octane):

Mean = 28mpg; n = 21

X3(Low Octane):

Mean = 35mpg; n=21

Step 1: Mean Differences

Pair 1: High Octane and Moderate Octane = 32 – 28 = 4

Pair 2: Low Octane and High octane =35 – 32 = 3

Pair 3: Low Octane and Moderate Octane = 35 – 28 = 7

Step 2: HSD

q (3, 60) = 3.40

HSD = 3.40 √ 30.39/ 21

HSD = 1.11

Step 3: Decision

4 > 1.11; High Octane and Moderate Octane differ

3> 1.11; Low Octane and High Octane differ

7 > 1.11; Low Octane and Moderate Octane differ

Part II Multiple Choice (Type the letter of the BEST answer in the blank before each question number).                                                                                     (1 point each)

__C___ 1.  If the ANOVA test statistic (F) is significant then it mean at a minimum:

a.  each mean must be different from every other mean

b.  there must be several differences among the means

c.  there must be at least one difference among the means

d.  all of the means must be equal

___C__ 2.  In an independent measures t-test, one group has 15 participants and the other has 26 participants.  What is the degrees of freedom for this test?

a.  41

b.  2

c.  39

d.  40

___A__ 3. A researcher varies the study time for a vocabulary test for 3 groups of students. To see how much learning take place, she then gives them a vocabulary test on the information.  An ANOVA shows that those with more study time recalled more words.  Eta-square for the test is found to be η2 = .51.  Which interpretation of Eta-Square is correct?

1. 51% of the variability in words recalled is due to amount of study.
2. 51% of the variability in study time is due to the number of words recalled.
3. There is a 51% chance that the number of words recalled is due to amount of study.
4. None of the above

_B____ 4.  When the null hypothesis is true, then F = MSbetween(group) / MSwithin(error) will be equal to:

a. 0

b. 1

c. greater than 1

d. not enough information given

Use the following information to answer the next few questions…..

A researcher interested in the effects of music on anxiety levels decided to have one group of subjects (n = 10) listen to 30 minutes of jazz, while the other group (n = 10) listened to 30 minutes of ‘white’ (background) noise on a sound machine.  She then gave both sets of subjects a standard anxiety rating scale, and found the Jazz group recalled , with a variance of 8.2  The white-noise groups recalled , with a variance of 6.2.  Did the jazz music significantly affect levels of anxiety? α = .05

_B____5. Which is the appropriate statistical test?

1. One-sample t-test
2. Two-sample t-test
3. ANOVA
4. Tukey’s post-hoc test

_B____6. Which is the appropriate research hypothesis?

1.  µjazz = 6.2
2.  µjazz  ≠ µwn
3.  µjazz = µwn
4.  µjazz  ≠ 6.2

__D___7.  Which is the appropriate null hypothesis?

1.  The population listening to jazz music has the same anxiety level as the general population.
2. The population listening to jazz music has a different level of anxiety than the general population.
3.  The population listening to jazz music has a different level of anxiety than those listening to white noise.
4.  The population listening to jazz music has the same level of anxiety as those listening to white noise.

___A__8.  Which is the appropriate critical value?

1. 2.101
2. 2.262
3. 2.086
4. 2.228

­­

__B___ 9.  If an experiment involves 5 groups with 5 participants in each group, then __

is the critical  F  at the =.01  level of significance.

a.  2.87

b.  4.43

c.  4.10

d.  none of the above

___C__10.  Tukey’s HSD test is used when:

a.  We want to find out how much of the variability in our data is due to the effect

b.  We want to avoid calculating an ANOVA

c. We want to find out which of the treatment means in our ANOVA were significantly different

d. None of the above

___D__ 11.  What is the critical value of a two-tailed one sample t-test, a = .01, n = 20?

1. 2.539                                       c. 2.845
2. 2.086                                      d. 2.861

___B__ 12.  An experimenter conducts an ANOVA test comparing 3 different groups with 21 participants in each condition.  What q-value (studentized range statistic) would be needed to conduct a Tukey's post-hoc test if Alpha = .01?

a. 3.40

b. 4.28

c. 3.58

d. 4.64

__A___ 13.  In ANOVA, the total variability can also be thought of as

a.  between variability + within variability

b.  group variability

c.  error variability

d.  between variability

___A__ 14.  When conducting an ANOVA, you decide to reject the null hypothesis.  Which of the following must be true?

a.  between variability > within variability

b.  between variability = within variability

c.  between variability < within variability

d.  between variability > total variability

__B___ 15.  For which of the following is it possible to get a negative value?

a. F

b. t

c. η2

d. power

Use the following information to answer the next two questions:

A researcher conducted a one-sample t-test which compared the average heart rate in the general population (µ = 85) to a sample of 25 professional athletes (= 82, S =1.5), and rejected the null hypothesis that there was no difference in heart rate for the two populations.

___B__16.  What is the correct tcritical value to use in computing the 95% confidence interval using the data above?

1. 1.77
2. 2.064
3. 2.492
4. 2.797

__B___ 17. After computing the confidence interval, the researcher found values of 81.5 and 82.5.  What is the correct interpretation for the confidence interval?

P(81.5 < μathlete > 82.5) = .95

1. We are 95% sure that the general population has a mean heart rate between 81.5 and 82.5
2. We are 95% sure that the population of athletes has a mean heart rate between 81.5 and 82.5
3. We are 95% confident that athletes have the same heart rate as the general population.
4. None of the above