PSY520 Topic 5 Exercises 2019
Complete the following exercises located at the end of each chapter and put them into a Word document to be submitted as directed by the instructor.
Show all relevant work; use the equation editor in Microsoft Word when necessary.
1. Chapter 13, numbers 13.6, 13.8, 13.9, and 13.10
2. Chapter 14, numbers 14.11, 14.12, and 14.14
3. Chapter 15, numbers 15.7, 15.8, 15.10 and 15.14
PSY 520 â€“ Module 5
Submit your answers in the boxes provided. No credit will be given for responses not found in the correct answer area.
Chapter 13:
13.6Itâ€™s well established, weâ€™ll assume, that lab rats require an average of 32 trials in a complex water maze before reaching a learning criterion of three consecutive errorless trials. To determine whether a mildly adverse stimulus has any effect on performance, a sample of seven lab rats were given a mild electrical shock just before each trial.
Question: | Steps: | Calculations or Logic: | Answer: |
Given that X5 34.89 and s5 3.02, test the null hypothesis with t, using the .05 level of significance. | What is the research hypothesis? | ||
What is the null hypothesis? | |||
Is this a one-tailed or two-tailed test? | |||
What are the degrees of freedom? | |||
What is the t critical for .05 significance? | |||
What is the calculated t? | |||
Do you accept or reject the null hypothesis? | |||
Construct a 95 percent confidence interval for the true number of trials required to learn the water maze. | |||
Interpret this confidence interval. |
13.8Assume that on average, healthy young adults dream 90 minutes each night, as inferred from a number of measures, including rapid eye movement (REM) sleep. An investigator wishes to determine whether drinking coffee just before going to sleep affects the amount of dream time. After drinking a standard amount of coffee, dream time is monitored for each of 28 healthy young adults in a random sample. Results show a sample mean, X,of 88 minutes and a sample standard deviation, s, of 9 minutes.
Question: | Steps: | Calculations or Logic: | Answer: |
Use tto test the null hypothesis at the .05 level of significance. | What is the research hypothesis? | ||
What is the null hypothesis? | |||
Is this a one-tailed or two-tailed test? | |||
What are the degrees of freedom? | |||
What is the t critical for .05 significance? | |||
What is the calculated t? | |||
Do you accept or reject the null hypothesis? | |||
If appropriate (because the null hypothesis has been rejected), construct a 95 percent confidence interval and interpret this interval. |
13.9In the gas mileage test described in this chapter, would you prefer a smaller or a larger sample size if you were the car manufacturer? Why? a vigorous prosecutor for the federal regulatory agency? Why
Question: | Smaller or Larger? | Why? |
In the gas mileage test described in this chapter, would you prefer a smaller or a larger sample size if you were the car manufacturer? | ||
In the gas mileage test described in this chapter, would you prefer a smaller or a larger sample size if you were a vigorous prosecutor for the federal regulatory agency? |
13.10Even though the population standard deviation is unknown, an investigator uses zrather than the more appropriate tto test a hypothesis at the .05 level of significance.
Question: | Larger or smaller? |
Is the true level of significance larger or smaller than .05 | |
Is the true critical value larger or smaller than that for the critical z? |
Chapter 14:
14.11To test compliance with authority, a classical experiment in social psychology requires subjects to administer increasingly painful electric shocks to seemingly helpless victims who agonize in an adjacent room. Each subject earns a score between 0 and 30, depending on the point at which the subject refuses to comply with authorityâ€”an investigator, dressed in a white lab coat, who orders the administration of increasingly intense shocks. A score of 0 signifies the subjectâ€™s unwillingness to comply at the very outset, and a score of 30 signifies the subjectâ€™s willingness to comply completely with the experimenterâ€™s orders.
Ignore the very real ethical issues raised by this type of experiment and assume that you want to study the effect of a â€œcommittee atmosphereâ€ on compliance with authority. In one condition, shocks are administered only after an affirmative decision by the committee, consisting of one real subject and two associates of the investigator, who act as subjects but in fact merely go along with the decision of the real subject. In the other condition, shocks are administered only after an affirmative decision by a solitary real subject.
A total of 12 subjects are randomly assigned, in equal numbers, to the committee condition (X1) and to the solitary condition (X2). A compliance score is obtained for each subject. Use tto test the null hypothesis at the .05 level of significance.
COMMITTEE | SOLITARY |
2 | 3 |
5 | 8 |
20 | 7 |
15 | 10 |
4 | 14 |
10 | 0 |
Question: | Steps: | Calculations or Logic: | Answer: |
Use tto test the null hypothesis at the .05 level of significance. | What is the research hypothesis? | ||
What is the null hypothesis? | |||
Is this a one-tailed or two-tailed test? | |||
What are the degrees of freedom? | |||
What is the t critical for .05 significance? | |||
What is the calculated t? | |||
Do you accept or reject the null hypothesis? |
14.12To determine whether training in a series of workshops on creative thinking increases IQ scores, a total of 70 students are randomly divided into treatment and control groups of 35 each. After two months of training, the sample mean IQ (â€“X1) for the treatment group equals 110, and the sample mean IQ (â€“X2) for the control group equals 108. The estimated standard error equals 1.80.
Question: | Steps: | Calculations or Logic: | Answer: |
Using t, test the null hypothesis at the .01 level of significance. | What is the research hypothesis? | ||
What is the null hypothesis? | |||
Is this a one-tailed or two-tailed test? | |||
What are the degrees of freedom? | |||
What is the t critical for .01 significance? | |||
What is the calculated t? | |||
Do you accept or reject the null hypothesis? | |||
If appropriate (because the null hypothesis has been rejected), estimate the standardized effect size, construct a 99 percent confidence interval for the true population mean difference, and interpret these estimates. |
14.14.An investigator wishes to determine whether alcohol consumption causes deterioration in the performance of automobile drivers. Before the driving test, subjects drink a glass of orange juice, which, in the case of the treatment group, is laced with two ounces of vodka. Performance is measured by the number of errors made on a driving simulator. A total of 120 volunteer subjects are randomly assigned, in equal numbers, to the two groups. For subjects in the treatment group, the mean number of errors (â€“X1) equals 26.4, and for subjects in the control group, the mean number of errors (â€“X2) equals 18.6. The estimated standard error equals 2.4.
Question: | Steps: | Calculations or Logic: | Answer: |
Use tto test the null hypothesis at the .05 level of significance. | What is the research hypothesis? | ||
What is the null hypothesis? | |||
Is this a one-tailed or two-tailed test? | |||
What are the degrees of freedom? | |||
What is the t critical for .05 significance? | |||
What is the calculated t? | |||
Do you accept or reject the null hypothesis? | |||
Specify the p-value for this test result. | |||
If appropriate, construct a 95 percent confidence interval for the true population mean difference and interpret this interval. | |||
If the test result is statistically significant, use Cohenâ€™s dto estimate the effect size, given that the standard deviation, s p, equals 13.15. | |||
State how these test results might be reported in the literature, given s1 5 13.99 and s2 5 12.15. |
Chapter 15:
15.7An educational psychologist wants to check the claim that regular physical exercise improves academic achievement. To control for academic aptitude, pairs of college students with similar GPAs are randomly assigned to either a treatment group that attends daily exercise classes or a control group. At the end of the experiment, the following GPAs are reported for the seven pairs of participants:
GPAs | ||
PAIR NUMBER |
PHYSICAL EXERCISE(X1) |
NO PHYSICAL EXERCISE X2 |
1 | 4.00 | 3.75 |
2 | 2.67 | 2.74 |
3 | 3.65 | 3.42 |
4 | 2.11 | 1.67 |
5 | 3.21 | 3.00 |
6 | 3.60 | 3.25 |
7 | 2.80 | 2.65 |
Question: | Calculations or Logic: | Answer: | |
Using t, test the null hypothesis at the .01 level of significance. | Step 1 | What is the research problem? | |
Step 2 | What is the null hypothesis? | ||
What is the alternative hypothesis? | |||
Step 3 | What is the decision rule? | ||
Step 4 | What is the critical t? | ||
What is the value of t? (you will need to calculate this) | |||
Step 5 | What is the decision? (retain or reject the null hypothesis at the specified level of significance; note the relationship between the observed and critical t scores) | ||
Step 6 | What is your interpretation of the decision in relation to the original research problem? | ||
Specify the p-value for this test result. | |||
If appropriate (because the test result is statistically significant), use Cohenâ€™s dto estimate the effect size | |||
How might this test result be reported in the literature? |
15.8A school psychologist wishes to determine whether a new antismoking film actually reduces the daily consumption of cigarettes by teenage smokers. The mean daily cigarette consumption is calculated for each of eight teenage smokers during the month beforeand the month afterthe fi lm presentation, with the following results: (Note:When deciding on the form of the alternative hypothesis, H1 , remember that a positive difference score ( D5 X1 2 X2 ) reflects a declinein cigarette consumption.)
MEAN DAILY CIGARETTE CONSUMPTION | ||
SMOKER NUMBER | BEFORE FILM (X?) | AFTER FILM (X)? |
1 | 28 | 26 |
2 | 29 | 27 |
3 | 31 | 32 |
4 | 44 | 44 |
5 | 35 | 35 |
6 | 20 | 16 |
7 | 50 | 47 |
8 | 25 | 23 |
Question: | Calculations or Logic: | Answer: | |
Using t, test the null hypothesis at the .05 level of significance. | Step 1 | What is the research problem? | |
Step 2 | What is the null hypothesis? | ||
What is the alternative hypothesis? | |||
Step 3 | What is the decision rule? | ||
Step 4 | What is the critical t? | ||
What is the value of t? (you will need to calculate this) | |||
Step 5 | What is the decision? (retain or reject the null hypothesis at the specified level of significance; note the relationship between the observed and critical t scores) | ||
Step 6 | What is your interpretation of the decision in relation to the original research problem? | ||
Specify the p-value for this test result. | |||
If appropriate (because the null hypothesis was rejected), construct a 95 percent confidence interval for the true population mean for all difference scores. | |||
If appropriate, use Cohenâ€™s dto obtain a standardized estimate of the effect size. | |||
Interpret the effect size. | |||
What might be done to improve the design of this experiment? |
15.10In a classic study, which predates the existence of the EPO drug, Melvin Williams of Old Dominion University actually injected extra oxygen-bearing red cells into the subjectsâ€™ bloodstream just prior to a treadmill test. Twelve long-distance runners were tested in 5-mile runs on treadmills. Essentially, two running times were obtained for each athlete, once in the treatment or blood-doped condition after the injection of two pints of blood and once in the placebo control or non-blood-doped condition after the injection of a comparable amount of a harmless red saline solution. The presentation of the treatment and control conditions was counterbalanced, with half of the subjects unknowingly receiving the treatment first, then the control, and the other half receiving the conditions in reverse order.
Since the difference scores, as reported in the New York Times,on May 4, 1980, are calculated by subtracting blood-doped running times from control running times, a positive mean difference signifies that the treatment has a facilitative effect, that is, the athletesâ€™ running times are shorter when blood doped. The 12 athletes had a mean difference running time, D, of 51.33 seconds with a standard deviation, s D, of 66.33 seconds.
Question: | Calculations or Logic: | Answer: | |
Test the null hypothesis at the .05 level of significance. | Step 1 | What is the research problem? | |
Step 2 | What is the null hypothesis? | ||
What is the alternative hypothesis? | |||
Step 3 | What is the decision rule? | ||
Step 4 | What is the critical t? | ||
What is the value of t? (you will need to calculate this) | |||
Step 5 | What is the decision? (retain or reject the null hypothesis at the specified level of significance; note the relationship between the observed and critical t scores) | ||
Step 6 | What is your interpretation of the decision in relation to the original research problem? | ||
Specify the p-value for this test result. | |||
Would you have arrived at the same decision about the null hypothesis if the difference scores had been reversed by subtracting the control times from the blood-doped times? | |||
If appropriate, construct and interpret a 95 percent confidence interval for the true effect of blood doping. | |||
Calculate Cohenâ€™s dfor these results. | |||
Interpret the effect size. | |||
How might this result be reported in the literature? | |||
Why is it important to counterbalance the presentation of blood-doped and control conditions? | |||
Comment on the wisdom of testing each subject twiceâ€”once under the blood-doped condition and once under the control conditionâ€”during a single 24-hour period. (Williams actually used much longer intervals in his study.) |
15.14In Table 7.4 on page 173, all ten top hitters in the major league baseball in 2011 had lower batting averages in 2012, supporting regression toward the mean. Treating averages as whole numbers (without decimal points) and subtracting their batting averages for 2012 from those for 2011 (so that positive difference scores support regression toward the mean), we have the following ten difference scores: 14, 39, 61, 60, 13, 21, 50, 93, 16, 61.
Question: | Calculations or Logic: | Answer: | |
Test the null hypothesis (that the hypothetical population mean difference equals zero for all sets of top ten hitters over the years) at the .05 level of significance. | Step 1 | What is the research problem? | |
Step 2 | What is the null hypothesis? | ||
What is the alternative hypothesis? | |||
Step 3 | What is the decision rule? | ||
Step 4 | What is the critical t? | ||
What is the value of t? (you will need to calculate this) | |||
Step 5 | What is the decision? (retain or reject the null hypothesis at the specified level of significance; note the relationship between the observed and critical t scores) | ||
Step 6 | What is your interpretation of the decision in relation to the original research problem? | ||
Specify the p-value for this test result. | |||
Construct a 95% confidence interval. | |||
Calculate Cohenâ€™s d. | |||
How might these findings be reported? |