PSY-304B: Quantitative Methods & Experimental Design

For all problems, unless otherwise indicated, assume a Type 1 error rate = .05.

1. 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.

   $$\begin{array}{ccccc}{H}_0& :& {\sigma }_1^2& <& {\sigma }_2^2\end{array}$$

   A reasonable test when comparing variances is the two-tailed F-test. The F-test will fail to reject a two-tailed H₀ iff:

   $$\begin{array}{ccccc}{}_{\frac{\alpha }{2}}F_{n_1-1,n_2-1}& \le & \frac{s_1^2}{s_2^2}& \le & {}_{1-\frac{\alpha }{2}}F_{n_1-1,n_2-1}\end{array}$$

   This particular F-test is one-tailed and so the measure is:

   $$\begin{array}{ccc}\frac{s_1^2}{s_2^2}& \le & {}_{1-\alpha }F_{n_1-1,n_2-1}\end{array}$$

   Subject	Group	Heart Rate	$\stackrel{\_}{Y}$	${\left(y-\stackrel{\_}{Y}\right)}^2$	$s_i^2=\frac{\sum {\left(y_i-\stackrel{\_}{Y_i}\right)}^2}{n_i-1}$
   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

   $$\begin{array}{ccccccccc}\frac{s_1^2}{s_2^2}& \approxeq & \frac{12.928}{84.811}& \approxeq & 0.153& \le & 0.21& \approxeq & {}_{.025}F_{7,11}\end{array}$$

   ${}_{\mathrm{0.01}}F_{7,11}\approxeq 0.153$, the null hypothesis is rejected.

2. 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

   1. 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.

      $$\begin{array}{cccccccccccccccc}{H}_0& :& \forall & {\mu }_i& ;& {\mu }_i& =& {\mu }_j& \Rightarrow & H_0& :& {\mu }_{\text{zoloft}}& =& {\mu }_{\text{naltrexone}}& =& {\mu }_{\text{valium}}\end{array}$$ $$\begin{array}{cccccccc}{H}_1& :& \exists & i,j& \ni & {\mu }_i& \ne & {\mu }_j\end{array}$$

      Drug	Sample	Mean	SSW	MSW	SSB	MSB	$\frac{\text{<html:acronym title="Mean Square Between">MSB</html:acronym>}}{\text{<html:acronym title="Mean Square Within">MSW</html:acronym>}}$
      		$\stackrel{\_}{Y}$	${\left(y-\stackrel{\_}{Y}\right)}^2$	$\frac{\sum {\left(y_i-\stackrel{\_}{Y_i}\right)}^2}{N-J}$	$n{\left(\stackrel{\_}{Y}-\stackrel{\_}{\stackrel{\_}{Y}}\right)}^2$	$\frac{\sum n_j{\left({\stackrel{\_}{Y}}_j-\stackrel{\_}{\stackrel{\_}{Y}}\right)}^2}{J-1}$	$\frac{\frac{\sum n_j{\left({\stackrel{\_}{Y}}_j-\stackrel{\_}{\stackrel{\_}{Y}}\right)}^2}{J-1}}{\frac{\sum {\left(y_i-\stackrel{\_}{Y_i}\right)}^2}{N-J}}$
      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

      $\begin{array}{ccccccc}{}_{\mathrm{0.999}}F_{2,27}& \approxeq & 11.619& \ge & 3.35& \approxeq & {}_{\mathrm{0.95}}F_{2,27}\end{array}$, the null hypothesis is rejected.

   2. 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

3. 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

   1. Fill in the blanks.

      $$\begin{array}{ccccccccccc}\frac{\text{<html:acronym title="Mean Square Between">MSB</html:acronym>}}{\text{<html:acronym title="Mean Square Within">MSW</html:acronym>}}& ~& F_{J-1,N-J}& \Rightarrow & \text{<html:acronym title="Mean Square Between">MSB</html:acronym>}& =& \left(F\right)\text{<html:acronym title="Mean Square Within">MSW</html:acronym>}& =& \left(3.22\right)\left(10.60\right)& =& 34.132\end{array}$$ $$\begin{array}{ccccccccccccc}\text{<html:acronym title="Mean Square Between">MSB</html:acronym>}& =& \frac{\text{<html:acronym title="Sum of Squares Between">SSB</html:acronym>}}{J-1}& =& \frac{\text{<html:acronym title="Sum of Squares Between">SSB</html:acronym>}}{{\mathrm{df}}_{\text{<html:acronym title="Sum of Squares Between">SSB</html:acronym>}}}& \Rightarrow & \text{<html:acronym title="Sum of Squares Between">SSB</html:acronym>}& =& \left(\text{<html:acronym title="Mean Square Between">MSB</html:acronym>}\right){\mathrm{df}}_{\text{<html:acronym title="Sum of Squares Between">SSB</html:acronym>}}& =& \left(34.132\right)\left(5\right)& =& 170.66\end{array}$$ $$\begin{array}{ccccccccccc}\text{<html:acronym title="Sum of Squares Total">SST</html:acronym>}& =& \text{<html:acronym title="Sum of Squares Between">SSB</html:acronym>}+\text{<html:acronym title="Sum of Squares Within">SSW</html:acronym>}& \Rightarrow & \text{<html:acronym title="Sum of Squares Within">SSW</html:acronym>}& =& \text{<html:acronym title="Sum of Squares Total">SST</html:acronym>}-\text{<html:acronym title="Sum of Squares Between">SSB</html:acronym>}& =& 382.65-170.66& =& 211.99\end{array}$$ $$\begin{array}{ccccccccccccc}\text{<html:acronym title="Mean Square Within">MSW</html:acronym>}& =& \frac{\text{<html:acronym title="Sum of Squares Within">SSW</html:acronym>}}{N-J}& =& \frac{\text{<html:acronym title="Sum of Squares Within">SSW</html:acronym>}}{{\mathrm{df}}_{\text{<html:acronym title="Sum of Squares Within">SSW</html:acronym>}}}& \Rightarrow & {\mathrm{df}}_{\text{<html:acronym title="Sum of Squares Within">SSW</html:acronym>}}& =& \frac{\text{<html:acronym title="Sum of Squares Within">SSW</html:acronym>}}{\text{<html:acronym title="Mean Square Within">MSW</html:acronym>}}& =& \frac{211.99}{10.60}& \approxeq & 20\end{array}$$

      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

   2. Indicate whether the results would lead to rejection of the null hypothesis.

      $$\begin{array}{ccccccccc}{}_{\alpha }F_{J-1,N-J}& =& {}_{\mathrm{0.97}}F_{5,20}& \approxeq & 3.22& \ge & 2.71& \approxeq & {}_{\mathrm{0.95}}F_{5,20}\end{array}$$

      This F-value would reject H₀ at the 95% confidence level.

   3. 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.

4. In some statistical computations it's helpful to know the variance of the sample variance: that is $Var\left({\hat{s}}^2\right)$. 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:

   $$\begin{array}{ccc}Var\left({\hat{s}}^2\right)& =& \frac{2{\sigma }^4}{n-1}\end{array}$$

   ---

   Begin with a well-known distribution involving the sample variance:

   $$\begin{array}{ccc}\frac{\left(n-1\right){\hat{s}}^2}{{\sigma }^2}& ~& {\chi }_{n-1}^2\end{array}$$

   In order for the two distributions to be the same, their variances must be the same. The distribution of the chi-square is known:

   $$\begin{array}{ccccc}Var\left(\frac{\left(n-1\right){\hat{s}}^2}{{\sigma }^2}\right)& =& Var\left({\chi }_{n-1}^2\right)& =& 2\left(n-1\right)\end{array}$$

   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:

   $$\begin{array}{ccc}{\left(\frac{n-1}{{\sigma }^2}\right)}^{2}Var\left({\hat{s}}^2\right)& =& 2\left(n-1\right)\end{array}$$

   The rest is simply algebra:

   $$\begin{array}{ccccc}Var\left({\hat{s}}^2\right)& =& \frac{2\left(n-1\right){\sigma }^4}{{\left(n-1\right)}^2}& =& \frac{2{\sigma }^4}{n-1}\end{array}$$