PSY 429 Experimental Psychology

Class Exercise #3

Hypothesis Testing with Single Sample

Answers

1. Suppose an investigator is interested in whether the mean score on an aptitude test for students who attend rural elementary schools is different from the typical score of elementary school students in general. The mean score for elementary school students in general is known to be 50. A random sample of 25 rural elementary school students is obtained, the aptitude test administered, and the mean score for the sample is found to be 56.00. The population standard deviation, s, is known to equal 10.00.

Test the null hypothesis: Ho : sample mean = population mean

 s_M = s / square root of n

s_M = 10.00/5

s_M = 2.00

z = M - m / s_M

z = 56.00 - 50.00/2.00

z = 3.00

Since 3.00 is greater than the critical value of z of +1.96, we reject the null hypothesis. We draw the conclusion that the aptitude score for children who attend rural elementary schools is significantly from the typical aptitude score of students in general.

2. An investigator is interested in whether the mean score on a test of short-term memory for Alzheimer's patients admitted to a private hospital is different from the mean score of Alzheimer's patients in general. The investigator believes that, in fact, the private hospital is too liberal in its criteria for diagnosing Alzheimer's Disease. Thus, he believes that he will find that the mean score for patients at the private hospital will be significantly higher than the mean score found in the general population of Alzheimer's patients (The higher the score, the stronger is short-term memory.) The mean score of typical patients diagnosed with Alzheimer's Disease is known to be 37. A random sample of 21 Alzheimer's patients at the private hospital is obtained and given the short-term memory test. The mean score for this sample is found to be 45. The standard deviation (s) of scores on this test, in general, is 6.15.

Test the experimenter's hypothesis with an alpha level set at .05.

 Since the investigator believes that he will find that the mean score for patients at a private hospital will be higher than that found in the general population of Alzheimer's patients, the hypothesis tested will be a directional one, using a one-tailed test.

s_M = s / square root of n

s_M = 6.15/square root of 21

s_M = 1.34

z = M - m / s_M

z = 45.00 - 37.00/1.34

z = 5.97

Since the observed or calculated z of 5.97 exceeds the critical value of z of 1.65, the investigator's hypothesis is supported. The mean score of the private hospital significantly exceeds the population mean.

 

3. Use the same experiment as described above, but with the following changes: The experimenter just wants to know if there will be a difference between his sample and the population mean.

Conduct this test with an alpha level of .01.

 Since the experimenter just wants to know if there is a difference, the experimental hypothesis will be tested using a non-directional or two-tailed test.

The only difference in the test is the critical value of z. For a two-tailed test it is plus or minus 2.58. Since the observed z of 5.97 still exceeds the critical value of z of 2.58, the investigator concludes that his experimental hypothesis is supported and we fail to reject the experimental hypothesis: there is a significant difference between the mean of the private hospital and the population mean.

4. If the mean score of the population is 500 and the standard deviation of the population is 100 while the mean of a sample is 650, test the hypothesis that there is no difference between the sample mean and the population mean if the sample size is 25. Use an alpha level of .05

 Since the hypothesis is that there is no difference, the null hypothesis using a two-tailed test will be used.

  s_M = s / square root of n

s_M = 100.00/5

s_M = 20.00

z = M - m / s_M

z = 650.00 - 500.00/20.00

z = 7.50

Since the observed z of 7.50 exceeds the critical value of z of plus or minus 1.96, the experimenter rejects the null hypothesis. A significant difference is found between the sample mean and the population mean.

5. If the mean score of the population is 150 and the standard deviation of the population is 15 while the mean of a sample is 160, test the hypothesis that the sample mean will be significantly larger than the population mean if the sample size is 15. Use an alpha level of .05.

  Since the hypothesis is that the sample mean will be larger than the population mean, the experimental hypothesis using a one-tailed test will be used.

  s_M = s / square root of n

s_M = 15.00/3.873

s_M = 3.873

z = M - m / s_M

z = 160.00 - 150.00/3.873

z = 2.58

Since the observed z of 2.58 exceeds the critical value of z of +1.65, the hypothesis is supported and we fail to reject the experimental hypothesis.

6. Test the same hypothesis and use the same data as in question #4 above except change the sample size to 45. What accounts for the different result?

  s_M = s / square root of n

s_M = 15.00/6.708

s_M = 2.236

z = M - m / s_M

z = 160.00 - 150.00/2.236

z = 4.47

Since the observed z of +4.472 exceeds the critical value of z of +1.65, the experimental hypothesis is supported and we fail to reject the experimental hypothesis.

With the increased sample size, the standard error of the mean is smaller.

7.A psychologist has prepared an "optimism Test" that is administered to graduating college seniors. The test measures how each class feels about its future--the higher the score, the more optimistic the class. Last year's class had a mean score of m = 15. A sample of n = 8 seniors from this year's class was selected and tested. The scores for these seniors are listed below. On the basis of this sample, can the psychologist conclude that this year's class has a different level of optimism than last year's class?

This year's class optimism scores:

7, 12, 11, 15, 7, 8, 15, 9

 

a. State the hypotheses and select an alpha level.

The null hypothesis is that the current class's optimism scores are no different than last year's scores.

The experimental hypothesis is that there is a significant difference between this year's class and last year's class in optimism about its future.

A two-tailed test with an alpha level of .05 will be used.

b. Determine the degrees of freedom, determine the critical values of the t statistic, and locate the critical rfegion.

df = n - 1

df = 8 - 1

df = 7

t = plus or minus 2.365

The observed t must be more extreme than either of these critical values of t to reject the hypothesis.

c. Compute the test statistic.

s = 3.295

s_M = s / square root of n

s_M = 3.295/2.83

s_M = 1.16

t = M - m /s_M

t = 10 - 15/1.16

t = -4.31

d. Make a decision about the null hypothesis and state your conclusion.

Since t = -4.39 is in the critical region, our sample data are unusual enough to reject the null hypothesis at the .05 level of significance. We can conclude thsat there is a significant difference in the level of optimism between this year's and last year's graduating class, t(8) = -4.39, p < .05, two-tailed.

8. A family therapist states that parents talk to their teenagers an average of 27 minutes per week. Surprised by that claim, a psychologist decided to collect some data on the amount of time parents spend in conversation with their teenage children. For n = 12 parents, the study revealed the following times (in minutes) devoted to conversation in a week. Do the psychologist's findings differ significantly from the therapist's claims? If so, is the family expert's claim an overestimate or underestimate of actual time spent talking to children? Use the .05 level of significance?

29, 22, 26, 22, 19, 27, 21, 24, 30, 22, 25, 28

 

a. State the hypothesis using symbols. Explain what they predict for this experiment.

H₀ : M = m

The parents in the sample would converse with their children an average of 27 minutes per week, and therefore, not differ from the time reported by the family therapist.

b. Sketch the distribution and locate the critical region.

 

 

c. Calculate the test statistic.

s = 3.48

s_M = s / square root of n

s_M = 3.48/3.464

s_M = 1.00

t = M - m /s_M

t = 24.58 - 27.00/1.00

t = - 2.42

d. Make a decision regarding the null hypothesis and state your conclusion.

 The family expert may have overestimated the actual time spent talking to children. The sample actually spent significantly less time per week talking than what the expert claimed. We there fore reject the null hypothesis and there was no difference.

9. For a standard set of discrimination problems that have been used for years in primate research, it is known that monkeys require an average of m = 20 trials before they can successfully reach the criterion (five consecutive correct solutions). A psychologist hypothesizes that the animals can learn the task vicariously, that is, simply by watching other animals perform the task. To test this hypothesis, the researcher selects a random sample of n = 4 monkeys. These animals are placed in neighboring enclosures from which they can watch another animal learn the task. After a week of viewing other animals, the four monkeys in the sample are tested on the problem. These animals require an average of 15 trials to solve the problem, with SS = 300. On the basis of these data, can the psychologist conclude that there is evidence that the animals perform significantly better after viewing others? Test the null hypothesis at the .01 level of significance.

a. State the hypothesis using symbols. Explain what they predict for this experiment.

H₁ : M < m

The monkeys in the sample will take significantly fewer trials to learn the task than monkeys on the average would take.

b. Sketch the distribution and locate the critical region.

 

c. Calculate the test statistic.

s = 10.00

s_M = s / square root of n

s_M = 10.00/2

s_M = 5.00

t = M - m /s_M

t = 15.00 - 20.00/5.00

t = - 1.00

d. Make a decision regarding the null hypothesis and state your conclusion.

 Fail to reject the null hypothesis. These data do not provide sufficient evidence to conclude that animals perform significantly better after viewing others on the task.

Psy 429 Main Page	Syllabus	Class Exercises	Homework Assign.	Experimental Dilemmas	Sample Tests;Answers	Statistical Tables
Learning Objectives	Statistical Formula	Answers to Class Ex.	Answers to Homework	Answers to Dilemmas	Web Links	Additional Exercises

To go back to Dr. Ryan's Home Page, click on kitty.

 

 

Celtic Clip Art - http://artinfo@aon-celtic.com