For all problems, unless otherwise indicated, assume a Type 1 error rate = .05.
An experiment is conducted in which behaviorally inhibited and behaviorally uninhibited children are exposed to an unfamiliar robot. The researcher quantifies heart rate responses in beats per minute (BPM) during exposure to the robot. The data are as follows:
Subject Number | Group | Average Heart Rate (in BPM) |
---|---|---|
1 | Uninhibited | 66 |
2 | Uninhibited | 61 |
3 | Uninhibited | 67 |
4 | Uninhibited | 60 |
5 | Uninhibited | 71 |
6 | Uninhibited | 63 |
7 | Uninhibited | 67 |
8 | Uninhibited | 66 |
9 | Inhibited | 89 |
10 | Inhibited | 78 |
11 | Inhibited | 67 |
12 | Inhibited | 86 |
13 | Inhibited | 82 |
14 | Inhibited | 66 |
15 | Inhibited | 85 |
16 | Inhibited | 79 |
17 | Inhibited | 88 |
18 | Inhibited | 79 |
19 | Inhibited | 90 |
20 | Inhibited | 64 |
The researcher is initially interested in testing the hypothesis that the inhibited group has more variable response times across subjects than the uninhibited group. (One could also assess variability on a within-subjects basis, but that is not the focus of the present study.) Using hand calculations test the hypothesis that the variability of heart rate within the behaviorally inhibited group is greater than the variability of heart rate within the behaviorally uninhibited group.
A reasonable test when comparing variances is the two-tailed F-test. The F-test will fail to reject a two-tailed H0 iff:
This particular F-test is one-tailed and so the measure is:
Subject | Group | Heart Rate | |||
---|---|---|---|---|---|
1 | Uninhibited | 66 | 65.125 | 0.766 | 12.982 |
2 | Uninhibited | 61 | 17.016 | ||
3 | Uninhibited | 67 | 3.516 | ||
4 | Uninhibited | 60 | 26.266 | ||
5 | Uninhibited | 71 | 5.641 | ||
6 | Uninhibited | 63 | 4.516 | ||
7 | Uninhibited | 67 | 3.516 | ||
8 | Uninhibited | 66 | 0.766 | ||
9 | Inhibited | 89 | 79.427 | 91.840 | 84.811 |
10 | Inhibited | 78 | 2.007 | ||
11 | Inhibited | 67 | 154.174 | ||
12 | Inhibited | 86 | 43.340 | ||
13 | Inhibited | 82 | 6.674 | ||
14 | Inhibited | 66 | 180.007 | ||
15 | Inhibited | 85 | 31.174 | ||
16 | Inhibited | 79 | 0.174 | ||
17 | Inhibited | 88 | 73.674 | ||
18 | Inhibited | 79 | 0.174 | ||
19 | Inhibited | 90 | 112.007 | ||
20 | Inhibited | 64 | 237.674 |
, the null hypothesis is rejected.
A study involves testing the effects of three medications on the ability of phobics to approach a feared object. 30 individuals who are phobic of spiders are randomly assigned to receive one of 3 medications (Zoloft, Naltrexone, and Valium) (10 subjects per group). After ingesting the medication, each participant attempts to approach a live spider. The researcher measures the distance in feet of each participant from the spider. The raw data are below:
Zoloft | 9 | 11 | 5 | 12 | 15 | 14 | 13 | 12 | 7 | 6 |
---|---|---|---|---|---|---|---|---|---|---|
Naltrexone | 15 | 16 | 12 | 12 | 18 | 19 | 23 | 20 | 13 | 17 |
Valium | 9 | 11 | 12 | 5 | 13 | 15 | 11 | 8 | 6 | 9 |
In order to assess whether medication condition has an effect on approach behavior, analyze these data using a one-way between subjects ANOVA. Please do the calculations by hand and show your work. Present the results in an ANOVA source table. Also be sure to indicate the statistical hypotheses (H0 & H1) that you are testing and whether or not you reject the null hypothesis.
Drug | Sample | Mean | SSW | MSW | SSB | MSB | |
---|---|---|---|---|---|---|---|
Zoloft | 9 | 10.40 | 1.960 | 11.622 | 34.844 | 135.033 | 11.619 |
11 | 0.360 | ||||||
5 | 29.160 | ||||||
12 | 2.560 | ||||||
15 | 21.160 | ||||||
14 | 12.960 | ||||||
13 | 6.760 | ||||||
12 | 2.560 | ||||||
7 | 11.560 | ||||||
6 | 19.360 | ||||||
Naltrexone | 15 | 16.50 | 2.250 | 179.211 | |||
16 | 0.250 | ||||||
12 | 20.250 | ||||||
12 | 20.250 | ||||||
18 | 2.250 | ||||||
19 | 6.250 | ||||||
23 | 42.250 | ||||||
20 | 12.250 | ||||||
13 | 12.250 | ||||||
17 | 0.250 | ||||||
Valium | 9 | 9.90 | 0.810 | 56.011 | |||
11 | 1.210 | ||||||
12 | 4.410 | ||||||
5 | 24.010 | ||||||
13 | 9.610 | ||||||
15 | 26.010 | ||||||
11 | 1.210 | ||||||
8 | 3.610 | ||||||
6 | 15.210 | ||||||
9 | 0.810 |
As an ANOVA source table, this is:
Source of Variation | Sum of Squares | df | Mean Square | F |
---|---|---|---|---|
Between | 270.067 | 2 | 135.033 | 11.619 |
Within | 313.80 | 27 | 11.622 | |
Total | 583.876 |
, the null hypothesis is rejected.
Replicate your results by doing an ANOVA using SAS. Turn in both the SAS program and your output.
spider_analysis.sas is a simple SAS program to compute ANOVA on spider_data.csv. More interestingly, homework_1.R is a R program to do the same thing.
The output from R is:
Response: Response Df Sum Sq Mean Sq F value Pr(>F) Drug 2 270.067 135.033 11.618 0.0002289 *** Residuals 27 313.800 11.622 Signif. codes: 0 '***' 0.001 '**' 0.01 '**' 0.05 '.' 0.1 ' ' 1
One of the undergraduate students in your lab left the results of an ANOVA in your mailbox. Due to some really odd printer problem only some of the cells are legible. After unsuccessfully trying to contact the student you realize that you actually have enough information left to figure out what the missing cells have to be.
Source of Variation | Sum of Squares | df | Mean Square | F |
---|---|---|---|---|
Between | 5 | 3.22 | ||
Within | 10.60 | |||
Total | 382.65 |
Fill in the blanks.
Source of Variation | Sum of Squares | df | Mean Square | F |
---|---|---|---|---|
Between | 170.66 | 5 | 34.132 | 3.22 |
Within | 211.99 | 20 | 10.60 | |
Total | 382.65 |
Indicate whether the results would lead to rejection of the null hypothesis.
This F-value would reject H0 at the 95% confidence level.
Assuming that the assumption of homogeneity of population variances is met here, present your best estimate of the variance within the populations.
When population variances are homogenous, the estimated value of the mean square within is an unbiased estimator of the population variance. For this data, the mean square within is 10.60.
In some statistical computations it's helpful to know the variance of the sample variance: that is . That is, consider the sampling distribution of the sample variance across a large number of random samples. The "variance of the sample variance" is the variance of this distribution and is an indicator of how variable the sample variance is likely to be from sample to sample. Prove:
Begin with a well-known distribution involving the sample variance:
In order for the two distributions to be the same, their variances must be the same. The distribution of the chi-square is known:
Constant factors within a variance may be factored out, but their value must be squared. Both the population variance and number of samples are constant in relation to the sample variance:
The rest is simply algebra: