Project #18459 - Statistic unit 6

1. Six schizophrenics are randomly assigned to three drug treatment conditions. After three months, each patient is rated on normality of behavior in an interview situation. Using the scores below, determine whether there is a significant difference among the drug treatments using alpha level a of .05 and a two-tailed test
Drug 1 Drug 2 Drug 3
     3       4        6
     7       10      12
2. Nine subjects were randomly divided into three groups of three subjects each, and each group
received a different treatment for statophobia. The number of panic attacks experienced by each subject is given below. Test whether the type of treatment makes a difference in the number of panic attacks. Use alpha level of .05 and a two-tailed test.
    10            5            5
     8             6            7
     9             7            6
3. Determine which treatments were significantly different from each other, using the data from problem #2. Use Tukey HSD test.

4. Right-handers, left-handers, and ambidextrous individuals were tested on short-term memory after consuming a low dosage or high dosage of caffeine. Determine if there were significant main effects or an interaction. Use alpha level of .05 and a two-tailed test.
                                       RIGHT LEFT AMBI.
                                         4       5      9
LOW DOSAGE                      2       6      8
                                         3       4     10

                                          6       4      7
HIGH DOSAGE                       4      5       5
                                          2       6       6


This is how far I have gotten, I think I just need number 4.


1.       Using one way anova I computed the sum of squares between groups, then the sum withing groups, then df within(3),  then f critical value, critical value of f=9.55; f test=.5480 so fcrit is greater than f stat so test is insignificant.  27 % of variability is explained by drug treatments


2.       Using same formula as above I went through each step overall mean 7, SS total 24, 18=SS bw, 6=SS within, df total=8, df between=2, df within=6,  Mean square within=1, f=9 , f crit value =5.14 (9>5.14)=test is significant.  75% of variability is defined by treatment.


3.       Tukey:  df error=6 (4.34), 4.34 times square root of 1/3=4.34(.57)=2.5;


Placebo-cognitive 9-6=3

Placebo-gestalt   9-6=3

Cognitive-gestalt 6-6=0


Placebo vs cog  3>2.5=significant

Placebo vs gestalt 3>2.5= significant

Cog vs gestalt 0<2.5= not significant


Subject Mathematics
Due By (Pacific Time) 12/2/13 6 pm
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