- Evolutionary theories often emphasize that humans have adapted to their physical environment. One such theory hypothesizes that people should spontaneously follow a 24-hour cycle of sleeping and waking—even if they are not exposed to the usual pattern of sunlight. To test this notion, eight paid volunteers were placed (individually) in a room in which there was no light from the outside and no clocks or other indications of time. They could turn the lights on and off as they wished. After a month in the room, each individual tended to develop a steady cycle. Their cycles at the end of the study were as follows: 25, 27, 25, 23, 24, 25, 26, and 25. Using the 5% level of significance, what should we conclude about the theory that 24 hours is the natural cycle? (That is, does the average cycle length under these conditions differ significantly from 24 hours?)
a. Use the steps of hypothesis testing.
i. Population 1: People who have no awareness of time or the day/night cycle.
ii. Population 2: People who are aware of time and the day/night cycle and follow a 24-hour cycle of sleeping and waking.
iii. Null Hypothesis: The sample scores will not vary significantly from a 24-hour cycle of sleep and waking, and therefore the knowledge of daylight/nighttime cannot be found to be statistically significant as to the 24-hour cycle.
iv. Alternative Hypothesis: The sample will vary enough from a 24-hour cycle of sleep and waking that the knowledge of daylight/nighttime can be found to be statistically significant as to the 24-hour cycle.
v. Comparison Distribution:
- Mean: 24
- SD: .42
- T Distribution: df=7
- T Score: 1.895
vii. Sample’s Score on Distribution:
- Mean: 25
- Standard Deviations: 2.38
viii. Conclusion: The sample mean score was statistically significant at p<.05, therefore we must reject the null hypothesis. This means that the effects of the knowledge of daylight/nighttime are significant to the 24-hour cycle.
b. Sketch the distributions involved.
c. Explain your answer to someone who has never taken a course in statistics. The point of the experiment is to find out if the knowledge of night and day/time affects a 24-hour sleep and wake cycle. If there is no difference between a person who has knowledge of day and night/time and someone who does not, then we can conclude that it is statistically significant as to the 24-hour cycle. Since we don’t know the full population’s mean and variance we have to hypothesize the distribution using a t distribution. There are eight people in the sample, so the degrees of freedom (df) is seven. In order to ascertain the variance of the population (comparison distribution) we have to divide the sum of squares of the scores by the df. Then, we need to further divide the quotient of that equation by the number of participants in the sample, in order to surmise the distribution of means’ variance. Lastly, we need to find the square root of the variance, which is the standard deviation. Once we know the characteristics we can find the cutoff t score for statistical significance. The t distribution chart indicates that at p<.05, with seven df, the t score cutoff is 1.895. The mean of the sample is 25, which when converted becomes a t score of 2.38 deviations from the comparison distributions mean. This means that we can reject the null hypothesis and embrace the alternative hypothesis as statistically significant.
Chapter 8, Problem 18
- Twenty students randomly assigned to an experimental group receive an instructional program; 30 in a control group do not. After 6 months, both groups are tested on their knowledge. The experimental group has a mean of 38 on the test (with an estimated population standard deviation of 3); the control group has a mean of 35 (with an estimated population standard deviation of 5). Using the .05 level, what should the experimenter conclude?
a. Use the steps of hypothesis testing,
b. sketch the distributions involved, and
c. explain your answer to someone who is familiar with the t test for a single sample, but not with the t test for independent means.
Chapter 9, Problem 17
- Do students at various colleges differ in how sociable they are? Twenty-five students were randomly selected from each of three colleges in a particular region and were asked to report on the amount of time they spent socializing each day with other students. The results for College X was a mean of 5 hours and an estimated population variance of 2 hours; for College Y, M = 4, S2 = 1.5; and for College Z, M = 6, S2 = 2.5. What should you conclude? Use the .05 level.
a. Use the steps of hypothesis testing,
i. Population 1: College students from College X.
ii. Population 2: College Students from College Y.
iii. Population 3: College Students from College Z.
iv. Null Hypothesis: That all three colleges have the same mean.
v. Alternative Hypothesis: That the three colleges have means that are not the same.
vi. Characteristics of Comparison Distribution: 2 numerator and 72 denominator degrees of freedom (df)
vii. Determine Cutoff Sample Score: Using the F table the f ratio cutoff is 3.13.
viii. Determine Sample’s score on comparison distribution:
- Estimate the variance of the distribution of means: 2
- Figure the estimated variance of the population of individuals: 50
- Divide the between-groups variance by the within-groups variance: 25
ix. Conclusion: We can reject the null hypothesis that there is no difference between the college populations and embrace the alternative hypothesis that the college populations are significantly different at a p-value of .05.
b. figure the effect size for the study; and
i. Take the variance between times the df between and divide it by the variance between times the df between plus the variance within times the df within.
ii. The effect size for this study is .99 or 99%
c. explain your answers to (a) and (b) to someone who has never had a course in statistics. First, assign a population to each group of scores. In this case we need to assign a population to each college. The null hypothesis is that there will be no significant difference between the means of scores and the variances thereof. The alternative hypothesis is that there will be significant difference between the means of scores and the variances thereof. Now we need to determine the characteristics of the comparison distribution. We use a f table when determining an f distribution, such as this. The between group df is determined by taking the number of groups and taking away one. In this case we have three groups, so the df of the between groups is 3-1=2. To determine the within-groups df we need to take the df from each group and add them together. In this case there are 25 participants in each group, which would be a df of 24, so 24+24+24=72. Now we need to locate the cutoff point on the comparison distribution by cross referencing the between groups df with the within groups df on a f table. This gives us a cutoff of 3.13. Now that we have the characteristics of the comparison distribution and the cutoff for the sample score, we need to calculate the f ratio. To calculate the f ratio we need to first estimate the variance of the distribution of means. You do this by calculating the sum of squares of the groups means and dividing by the df of the between groups. This figure comes out to 2. Then we need to calculate the variance of the population of individual scores. To do this we must multiply the number in each group by the estimated variance of the distribution of means. This figure comes out to 50. Now we need to divide the variance of the population of individual scores by the variance of the distribution of means, which comes out to 25. 25 is much larger than the 3.13 f score cutoff ratio at a p value of .05. Therefore we can reject the null hypothesis that the groups are not significantly different and embrace the alternative hypothesis that the groups are significantly different. Furthermore, we can calculate the effect size of the f ratio as compared to the comparison distribution by taking the variance between times the df between and divide it by the variance between times the df between plus the variance within times the df within. On this research analysis that comes out to .99 or 99%, which is very big effect size.
Chapter 11, Problems 11, 12
- Make up a scatter diagram with 10 dots for each of the following situations:
a. perfect positive linear correlation,
b. large but not perfect positive linear correlation,
c. small positive linear correlation,
d. large but not perfect negative linear correlation,
e. no correlation,
f. clear curvilinear correlation.
- Four research participants take a test of manual dexterity (high scores mean better dexterity) and an anxiety test (high scores mean more anxiety). The scores are as follows.
a. Make a scatter diagram of the scores;
b. describe in words the general pattern of correlation, if any; The trend line indicates a large, negative, linear correlation.
c. figure whether the correlation is statistically significant (use the .05 significance level, two-tailed);
i. Population 1: People like those in this study.
ii. Population 2: People for who there is no correlation between manual dexterity and anxiety.
iii. Null Hypothesis: That the two populations have the same correlation.
iv. Alternative Hypothesis: That the two populations do not have the same correlation at a statistically significant level of .05, two-tailed.
v. Characteristics of Comparison Distribution: df=2
vi. Determine cutoff sample score: At the .05 level for a two-tailed test the cutoff scores will be -4.303 and +4.303.
vii. Determine the sample’s score on the comparison distribution: By using the t test we can conclude that the sample score is -69 on the comparison distribution.
viii. Conclusion: We can reject the null hypothesis that there is no correlation and adopt the alternative hypothesis that there is a statistically significant correlation between manual dexterity and anxiety.
d. explain the logic of what you have done, writing as if you are speaking to someone who has never heard of correlation (but who does understand the mean, deviation scores, and hypothesis testing); Here we are trying to find out the statistical significance of a correlation. Population 1 represents the people in this study and population 2 represents people that have no correlation between manual dexterity and anxiety. The null hypothesis that the correlation will be the same for both populations. The alternative hypothesis is that the correlation in population 1 will be significantly greater. First, we need to determine the characteristics of the comparison distribution. We do this by taking the number of people in the sample and subtracting 2. This gives us our degrees of freedom to be used later. Next, we need to isolate the cutoff score at which the test will be considered statistically significant. Looking at the t table, with a two-tailed study at the .05 level this comes out to +4.303 or -4.303. By using the t test for a correlation coefficient we can determine that the sample score is at -69, which is far over the cutoff score. Therefore, we can reject the null hypothesis that the correlations for population 1 and two will be the same and embrace the alternative hypothesis that the correlation in population 1 is significantly greater than in population 2.
e. give three logically possible directions of causality, indicating for each direction whether it is a reasonable explanation for the correlation in light of the variables involved (and why).
i. X could be causing Y: That better manual dexterity could be causing the lower levels of anxiety, because the correlation is a negative correlation.
ii. Y could be causing X: That lower anxiety causes higher manual dexterity.
iii. Some third factor could be causing both X and Y: Job expectations or perceived job performance could possibly be affecting both manual dexterity and the anxiety level.
Aron, A., Aron, E., & Coups, E. (2006). Statistics for psychology (4th ed.). Upper Saddle River, NJ: Pearson/Allyn Bacon.