Answers to the Practice Statistics Final

All Answers are in Yellow or Red

Part 1. Multiple Choice Problems

Study note: In this section of the final you will be required to answer 10-20 multiple choice questions. You should be able to answer them without looking in your notes. If you find that you need your notes, then you need more study time. Warning: If you have to use your notes on the final for these problems (which you are free to do), then you are likely to run out of time.

1.1) Which of the following statements about the normal curve are true?
a.) the distribution mean is equal to the mode for the distribution
b.) the size of an area under the normal curve is also a probability statement about the events in that area.
c.) converting the raw scores of any empirical normal distribution to z scores will give the
distribution a mean equal to 0 and a standard deviation equal to 1.00.
d.) all of these
e.) none of these

1.2.) Which of the following should be used to calculate the correlation between GPA and I.Q.?
a. Pearson's r b. coefficient of determination c. t test d. F e. r²

1.3.) You have an experimental design with 2 independent variables (consisting of 3 different groups of subjects each) and 1 dependent variable. The most appropriate inferential statistic would be?
a. Pearson's r b. t-test for independent samples c. one-way ANOVA d. Multifactorial ANOVA e. mixed design (ANOVA)

1.4.) An r2 of 0.79 calculated for any 2 variables (x and y).
a. means that 79% of the variablity in x and y are due to unidentified variables.
b. means that there is a 79% chance that these variables are related to each other.
c. means that the corresponding r indicated a strong relationship between variables x and y.
d. all of these

1.5.) You have an experimental design with one independent variable (consisting of 2 different groups of subjects) and one dependent variable. The most appropriate inferential statistic would be?
a. Pearson's r b. t-test for independent samples c. one-way ANOVA d. two-way ANOVA e. mixed design ANOVA

1.6.) You have just completed a One-way ANOVA which resulted in the following: F(2,35) = 16.09, p < .05. In addition, prior to the start of the study, you have determined that you are interested in testing all possible pair-wise comparisons. Which statistic should you calculate next?
a. Tukey's HSD b. t test c. Dunnett's t d. ANOVA e. plot the interaction effect f.) none, stop and write the conclusion

1.7. If a distribution is positively skewed, which descriptive statistic would be most appropriate?
a. mode b. F-ratio c. mean d. r e. t-test

1.8. If the mean, median and mode of a distribution are all equal, then this distribution would be?
a. positively skewed b. negatively skewed c. normal d. biimodial

1.9) You have just completed a One-way ANOVA which resulted in the following: F(4,132) = 1.09, p > .05. In addition, prior to the start of the study, you have determined that you are interested in comparing each experimental group with only the control group. Which statistic should you calculate next?
a. Tukey's HSD b. t test c. Dunnett's t d. Chi square e. plot the interaction effect f.) none, stop and write the conclusion

Part 2. Short Concept Problems

Instructions: If you make your calculations on a separate sheet of paper, please attach your work if you expect partial credit. Place all answers in the blanks.

2.1) At the begining of the 2000 NCAA mens basketball tournament Wisconsin was one of 64 teams invited to compete. By March 26, 2000 there were four teams left Wisconsin, Michigan State, North Carolina and Florida. Answer the following questions.

a.) At the start of the tournament what was the theoretical probability that Wisconsin would win their first game 1 in 2 or p = 0.50. Remember in simple probability p=number of desired outcomes/number of possible outcomes. Wisconsin could win one way and there were two possible outcomes (win or loss).

b.) On March 26, what was the theoretical probability that Wisconsin would win the championship 1 in 4 or p = 0.25. Wisconsin could win one way and there were a total four possible outcomes winners. Of course, Michigan State won %*%#!!!!???!?@#.

c.) By the way Citellus fans, Wisconsin won the Rose bowl this year. At the beginning of the season there were 11 teams in the Big Ten and 10 in the PAC 10. What were the theoretical odds (at the start of the season) that Wisconsin would win the Rose bowl? This is an example of conditional probability. That means two independent conditions must be met. In conditional probability p= pr{event 1} x pr{event 2}. The pr {Winning the big 10} was 1 in 11 or 0.09 and the pr{Winning the Rose bowl} was 1 in 2 or 0.5 so 0.5 x 0.09 = p=0.045.

d.) Oh and they won last year too! What were the theoretical odds of repeat titles? Same rationale as in c so 0.045 x 0.045 = pr{back-back titles}=0.002

2.2.) You have a 16-arm radial maze designed to test learning in rats. In this version of the test each arm is baited with a food reward. The animal is trained to find the rewards.

a.) What is the probability of a correct response on the animal's first choice at the start of the test session. This is another simple probability problem. So there are 16 correct choices and 16 possible responses so pr{correct response}= 16 in 16 or 1.00

b.) What is the probability of a correct response on the animal's fourth choice (assume that on each of the first three choices the animal found and ate a food reward)This is another simple probability problem. The rat ate 3 so there are only 13 food pellets or correct choices left and 16 possible responses so pr{correct response}= 13 in 16 or 0.8125.

c.) What is the probability that an animal will enter 16 arms without making any mistakes 1.00 Is this a likely event yes no (circle one)
This is a conditional probability problem. You are asking the animal to make a correct response each time assuming that the animal made a correct response on the previous choice.

So on choice 1 pr{correct}= 16 in 16 or 1.00
On choice 2 the animal has eaten one pellet on choice 1now pr{correct}= 15 in 16 or .9375 (we'll truncate to be conservative so use .93)
On choice 3 the animal has eaten two pellets on choices 1 & 2 now pr{correct}= 14 in 16 or .8775 (we'll truncate to be conservative so use .87)
On choice 4 the animal has eaten three pellets on choices 1, 2 & 3 so pr{correct}= 13 in 16 or .8125 (we'll truncate to be conservative so use .81)
On choice 5 the animal has eaten three pellets on choices 1, 2, 3 & 4 so pr{correct}= 12 in 16 or .75

and so on until you get to choice 16

On choice 16 the animal has eaten three pellets on choices 1 thru 15 so pr{correct}= 1 in 16 or .0625(we'll truncate to be conservative so use .06)

Then.....

Since each event depends on the others we multiple each event together. Thus,

pr{all arms correct} = 1.00 x .93 x .87 x .81 x .75 x .68 x .62 x .56 x .50 x .43 x .37 x.31 x .25 x .18 x .12 x .06
pr{all arms correct} = 0.0000009

2.3.) A researcher has an experimental design with 1 independent variable (consisting of 5 levels or groups) and 1 dependent variable. They have decided to complete all of the possible independent samples t-tests to analyze their data.

The researcher set alpha = 0.05.

What will be the experiment-wise error rate for these analyses: p= 0.50 ......... Each comparison will have an error rate of 0.05. There are 10 possible comparisons for this study. So the overall error rate is 0.05+0.05+0.05+0.05+0.05+0.05+0.05+0.05+0.05+0.05=0.50

If they use Tukey's HSD test instead, the experiment-wise error rate for these

analyses will be p= 0.05. Tukey's is designed to control for the inflation in the experiment-wise error rate that occurs when completing multiple comparisons.

2.4) Prior to the start of this study you determined that you were interested in comparing each experimental group with only the control group. You have just completed a One-way ANOVA with four levels of the independent variable and one level of a single dependent variable. There were 60 subjects in the experiment. You found that F= 4.61 .

First find the degrees of freedom...... Between groups = K-1 or 4-1 = 3, the Within groups = Ntot-K = 60-4 = 56.
Then look up the critical value in Table F from the back of your book. The value was 2.78 for an alpha of .05.
Since your F exceeds the expected value found in the table we reject the null hypothesis and write the equation as shown below.

Complete the F equation for these data: F(3 ,56 ) = 4.61, p <.05.

Which statistic should you calculate next?

Because you had set up the study to make your comparisons only to the control group, Dunnett's test is the best choice for post-hoc comparisons.

a. Tukey's HSD b. t test c. Dunnett's t d. F e. plot the interaction effect f.) none, stop and write the conclusion

2.5) Interpret the following Z score for Bill. Bill's performance on last year's history final was equal to a z score of 2.5. What does this score tell you about Bill's knowledge of history?

Answer: The z score tells you nothing about Bill's knowledge of history. What it does tell you is that Bill scored 2.5 standard deviations above the average score on this exam (or better than 99.38% of the rest of the class). Remember that the z score is a useful way to compare a subject against a group. If the group is standardized (e.g., an IQ test), then you can make predictions about behavior. But in this case you are comparing a single student against others in a nonstandardized group.

Study note: In this next section of the final you will be required to completely (but appropriately) analyze a subset of problems like those presented below.

Part 3. Analyzing Experiments & Writing Conclusions

Instructions: Complete all of these problems. Attach all work and graphs. Be sure to write your technical ID number on all work. Assume that all assumptions have been met, the data is normal and use an alpha of .05. Place all answers in the blanks.

Problem 3.1): There have been a large number of studies reporting improved memory in aging animals treated with the drug nimodipine. This led researchers in Italy to test the hypothesis that nimodipine would also improve memory in Alzheimer's patients.

Ten Alzheimer's patients were randomly selected from a normally distributed group of male Alzheimer's patients currently living in a local nursing home. Subjects were first monitored for six months without any medications. The score on the Folstein mini mental status exam was recorded for each patient. During the next six months, subjects were given an oral dose of nimodipine three times a day. The score on the Folstein mini mental status exam was again recorded for each patient at the end of this period. Analyze this study. Note: the Folstein evaluates memory and cognitive function. A perfect score is a 30. A score of 25 or lower can indicate Alzheimer's disease. An Alzheimer patient usually drops two points per year on this test as the disease progresses.

The data are as follows:

subject 1 scored 22 berfore treatment and 21 after 6 months of treatment
subject 2 scored 23 berfore treatment and 20 after 6 months of treatment
subject 3 scored 20 berfore treatment and 21 after 6 months of treatment
subject 4 scored 19 berfore treatment and 18 after 6 months of treatment
subject 5 scored 18 berfore treatment and 20 after 6 months of treatment
subject 6 scored 19 berfore treatment and 20 after 6 months of treatment
subject 7 scored 22 berfore treatment and 21 after 6 months of treatment
subject 8 scored 20 berfore treatment and 21 after 6 months of treatment
subject 9 scored 21 berfore treatment and 21 after 6 months of treatment
subject 10 scored 20 berfore treatment and 21 after 6 months of treatment

Put your answers in the blanks below:

Using the statistical format write the null hypotheses: Ho:m1=m2

Using the statistical format write the alternative hypotheses: H1:m1<m2

Before treatment Mean = 20.4 standard deviation = 0.93

After treatment Mean = 20.4 standard deviation = 1.58

Inferential Statistic = t(9)= 0, p > .05 critical value (from table) = 1.833

or

Pearson's r= Not appropriate r2 = Not appropriate

Write an appropriate conclusion: Nimodipine did not effect scores on the Folstein mini mental status exam t(9)= 0, p > .05.

Attach an appropriate graph.

Answer the following:

The scale of measurement for the dependent variable was? ratio ordinal interval nominal (circle one)
The independent variable was? Phase of drug treatment
The results from this study will generalize to what population(s)? Maile alzheimer's patients in nursing homes.

Problem 3.2): There have been a large number of studies reporting improved memory in aging animals treated with the drug nimodipine. This led researchers in Italy to test the hypothesis that nimodipine would also improve memory in Alzheimer's patients.

Ten Alzheimer's patients were randomly selected from a normally distributed group of male Alzheimer's patients currently living in a local nursing home. Subjects were randomly assigned to one of two groups. Group 1 received Nimodipine and Group 2 received placebo. The score on the Folstein mini mental status exam was recorded for each patient at the end of the six months of treatment. Analyze this study. Note: the Folstein evaluates memory and cognitive function. A perfect score is a 30. A score of 25 or lower can indicate Alzheimer's disease. An Alzheimer patient usually drops two points per year on this test as the disease progresses.

The data are as follows:

subject 1 was in Group 1 and scored 25 after 6 months of treatment
subject 2 was in Group 2 and scored 22 after 6 months of treatment
subject 3 was in Group 1 and scored 23 after 6 months of treatment
subject 4 was in Group 2 and scored 21 after 6 months of treatment
subject 5 was in Group 1 and scored 24 after 6 months of treatment
subject 6 was in Group 2 and scored 18 after 6 months of treatment
subject 7 was in Group 1 and scored 25 after 6 months of treatment
subject 8 was in Group 1 and scored 26 after 6 months of treatment
subject 9 was in Group 2 and scored 17 after 6 months of treatment
subject 10 was in Group 2 and scored 19 after 6 months of treatment

Put your answers in the blanks below:

Using the statistical format write the null hypotheses: Ho:m1=m2
Using the statistical format write the alternative hypotheses: H1:m1>m2

Nimodipine Mean = 24.60 standard deviation = 1.14

Placebo Mean = 19.40 standard deviation = 2.074

Inferential Statistic = t(8) = 4.914, p < .05 critical value (from table) = 1.86

Show work me the work for 3.2

Write an appropriate conclusion: Nimodipine treated male alzheimer's patients showed significantly higher t(8) = 4.914, p < .05 cognitive function as measured by the Folstein mental status exam (see Figure 1) compared to controls.

Attach an appropriate graph.

Answer the following:

The scale of measurement for the dependent variable was? ratio ordinal interval nominal (circle one)
The independent variable was? Type of drug obtained
The results from this study will generalize to what population(s)? Maile alzheimers patients in nursing homes.

Problem 3.3): There have been a large number of studies reporting impaired memory in elderly humans suffering from depression. This observation led researchers to test the hypothesis that depression may contribute to the memory impairment reported in Alzheimer's patients.

Ten Alzheimer's patients were randomly selected from a normally distributed group of female Alzheimer's patients currently living in a local nursing home. Subjects were given the Folstein mini mental exam and the Beck depression test.

Analyze this study. Note: the Folstein evaluates memory and cognitive function. A perfect score is a 30. A score of 25 or lower can indicate Alzheimer's disease. An Alzheimer patient usually drops 2 points per year on this test as the disease progresses. The Beck measures depression. Assume a score of 30 is maximum and indicates a high degree of depression and a score of 0 would indicate no depression.

The data are as follows:

subject 1 scored 22 on the Folstein and 23 on the Beck
subject 2 scored 23 on the Folstein and 21 on the Beck
subject 3 scored 25 on the Folstein and 30 on the Beck
subject 4 scored 24 on the Folstein and 22 on the Beck
subject 5 scored 22 on the Folstein and 21 on the Beck
subject 6 scored 21 on the Folstein and 25 on the Beck
subject 7 scored 20 on the Folstein and 2 on the Beck
subject 8 scored 22 on the Folstein and 0 on the Beck
subject 9 scored 23 on the Folstein and 21 on the Beck
subject 10 scored 17 on the Folstein and 0 on the Beck

Put your answers in the blanks below:

Using the statistical format write the null hypotheses: Ho: r = 0
Using the statistical format write the alternative hypotheses: H1: r <> 0
S

Not appropriate Mean = Not appropriate standard deviation = Not appropriate

Not appropriate Mean = Not appropriate standard deviation = Not appropriate

Inferential Statistic = Not appropriate critical value for r (from table) = .5494

Pearson's r= r(8) =0.72 , p < .05 r2 = 0.5184

calculations for r

Write an appropriate conclusion: There was a significant relationship between the degree of
depression and the amount of cognitive impairment r(8) =0.72 , p < .05.

Attach an appropriate graph. The appropriate graph would be a scatterplot (as shown below).

Answer the following:

The scale of measurement for the dependent variable was? both ratio ordinal interval nominal (circle 1)
The independent variable was? none, correlational designs compare two dependent variables
The results from this study will generalize to what population(s)? female alzheimer's patients living in nursing homes.

3.4 A researcher is interested in determining the usefulness of the "nicotine patch" in helping people quit smoking. Twenty-four smokers were randomly selected from a population of smokers. Subjects were then randomly assigned to one of three training programs. The number of cigarettes smoked per day were recorded for each subject. Subjects in program one were not required to quit smoking (mean= 15.5, s= 1.05). Subjects in program two were required to quit smoking without formal assistance (mean= 12.5, s= 1.55). Subjects in program three were also required to quit, but were given the "nicotine patch" during the duration of the study (mean= 9.5, s= 0.44). The analyses up to the ANOVA table (presented below) have been completed for you. Finish the analyses (including Tukey's HSD if needed), and include an appropriate conclusion (and construct the appropriate graph(s)). Use an alpha = 0.05 for all tests. Assume all hypotheses are one-tailed.

Answers to the Anova Table
Between Groups:
df = 2 (K-1, where K is the number of levels of the independent variable)
MS = 77.04 (154.09/2)
F= 2.22 (77.04/34.65)

Within Groups:
df = 21 (Ntot-K), where Ntot is the total number of subjects in the study and K is the number of levels of the independent variable)
MS = 34.65 (727.85/21)

The critical value for the effect is 3.47. Since the calculated F = 2.22 and did not exceed
the expected value we fail to reject the null and report the result as indicated below..

Conclusion: The nicotine patch did not help people quit smoking (F(2,21) = 2.22, p > .05 (see Figure 3.4).

3.5 A researcher is interested in determining the usefulness of the "nicotine patch" in helping people quit smoking. Twenty-four smokers were randomly selected from a population of smokers. Subjects were then randomly assigned to one of two types of training programs (no assistance or behavioral therapy) and one of two types of drugs (placebo patch or nicotine patch). The number of cigarettes smoked per day were recorded for each subject. The analyses up to the ANOVA table (presented below) have been completed for you. Finish the analyses (including the most appropriate post-hoc test - if needed), and construct the appropriate graph(s). Use an alpha = 0.05 for all tests. Assume all hypotheses are one-tailed.

********Summary Statistics*********

Answers to the Anova Table
Between Groups Drug:
df = 1 (A-1, where A is the number of levels of the independent variable drug)
MS = 154.09 (154.09/1)
F= 4.23 (154.09/36.39)

Between Groups Therapy:
df = 1 (B-1, where B is the number of levels of the independent variable therapy)
MS = 254.09 (254.09/1)
F= 6.98 (254.09/36.39)

Drug x Therapy interaction effect:
df = 1 (A- 1 * B-1)
MS = 387.78 (387.78/1)
F= 10.65 (387.78/36.39)

Within Groups:
df = 20 (Ntot-(A*B)), where Ntot is the total number of subjects in the study and A and B are the number of levels of the independent variables)
MS = 36.39 (727.85/20)

The critical value for the interaction effect is 4.35. Since the calculated F = 10.65 and exceeds
the expected value we reject the null and consider the alternative hypothesis. Because the interaction
is significant we will ignore the main effects of drug and therapy and write the conclusion as indicated below..

Conclusion: There was a significant interaction between the
type of drug used to quit smoking and the type of therapy
received (F(1,20) = 10.65, p < .05 (see Figure 3.5). Analysis of
the interaction shows that the nicotine patch was only effective if combined
with psychotherapy.

Part 4. Design and Analysis Matching Problems

Instructions: Use the choices at the bottom of the page. Put the letter of the answer in the blanks listed by each question. An answer may be used more than once.

4.1.) A researcher wanted to determine the effects of caffeine on learning. The researcher obtained a random sample of 100 college students and randomly assigned them to one of two equal sized treatment groups (placebo or caffeine). Each subject was asked to drink a solution containing the appropriate drug. Two hours later each subject was given a list of 25 words and asked to remember as many of the words as possible. Twenty-four hours later the subjects were asked to write down as many of the words as possible. The number of correctly recalled words were recorded for each subject.

4.1-a. Type of research design is a. experimental

4.1-b. Of those listed below, the most appropriate analyses for this design would be c. independent samples t-test, means, a bar graph, and a conclusion.

4.2.) A researcher wanted to determine the effects of overcrowding on prison violence. Our researcher obtained a random sample of 500 prisoners from six prisons in 3 midwestern states. For each prisoner the researcher recorded the number of prisoners confined in his/her cell and obtained the number of reprimands for violent acts from each prisoner's detention file.

4.2-a. Type of research design: c. correlational

4.4-b. Of those listed below, the most appropriate analyses for this design would be? d. Pearson's r, t-test for r, scatterplot, and a conclusion.