full2.gif (11401 bytes)
Practice Test #3
Answers


1.  The national population mean for the SAT was 950 last year and the population standard deviation was 180.  The last year's graduating class  of 97 students from Albert Gore Technical High School (Gore Tech) had a mean score on the SAT of 967.  Please do the appropriate statistical test to tell if the Gore Tech mean was significantly greater than the population mean.

ans3 1.gif (1221 bytes)

Please include in your answer:
a)  Your hypotheses
Null hypothesis:  Gore Tech no different from the population
ans3 2.gif (965 bytes)
Research hypothesis:  Gore Tech higher than the population
ans3 4.gif (972 bytes)
b)  The value you computed for t
First compute the standard error of the mean

Hint:  Remember that when you have the population standard deviation you do not need to
         estimate the standard error or the mean. 

ans3 5.gif (1356 bytes)

Now compute t

ans3 18.gif (1433 bytes)
c)  The degrees of freedom = df = n-1 = 97-1= 96
d)  Wether your research hypothesis is 1-tail or 2-tail
        Research hypothesis is one-tail
e)  The critical value from Table T = 1.671 for one tail at 96 degrees of freedom

Hint:  If you enter Table T and you cannot find the exact degrees of freedom (df = 96 is not in the table)
            use the next lowest degree of freedom that is in the table.  In this case use I used df = 60.

f)  Your conclusion:  Because the computed value for t, .930, is less than the critical value, 1.671,
        we fail to reject the null hypothesis. 
        The students at Gore Tech did not score significantly higher than the national population mean.


2.  One group of 6 participants was shown a movie on the evils of alcohol use and the other group of 7 participants was shown a movie about dog nutrition.  Both groups were then given a test measuring their attitude toward drug use.  The scores for both groups are listed below (high scores indicate a resistance to the use of drugs).

Alcohol Movie Dog Movie
132
144
131
145
156
139
121
130
122
125
119
100
105

 

 

Given this data please see if there is a significant difference between the effect that the alcohol movie and the dog movie have on attitudes toward drug use. 

Before we begin we need to compute the mean, standard deviation, and n for each of the 2 samples.

Please include in your answer to this question:
a)  Your hypotheses
The null hypothesis:  The drug attitude scores for those watching the alcohol movie
            will not differ from those watching the dog movie.
ans3 8.gif (974 bytes)
The research hypothesis:  The drug attitude scores for those watching the alcohol movie
            will be higher than those watching the dog movie.
ans3 9.gif (977 bytes)
b)  The value you computed for t
First we compute the estimate of the standard error of the mean for each sample
            then use those values to compute the estimate of the standard error of the difference.

Hint:  Pay careful attention to each n, they may not be the same.

 

Finally we compute the t

ans3 11.gif (1624 bytes)
c)  The degrees of freedom = (n1-1) + (n2-1) = (6-1) + (7-1) = 11
d)  Determine if your research hypothesis is 1-tail or 2-tail
             The research hypothesis is one-tail
e)  The critical value from Table T = 1.796 for a one-tail test at 11 degrees of freedom
f)  Your conclusion: Because the computed value for t, 4.237, is greater than the critical value, 1.796,
        we reject the null hypothesis and accept the research hypothesis.
The people viewing the alcohol movie scored significantly higher than the people viewing the dog movie.
   


3.  A researcher has just finished a research project using 8 anorexia nervosa patients as participants.  She was testing the effectiveness of a particular type of psychotherapy on these patients.  Below is listed the weights of the 8 patients before and after the therapy.

Patient Weight Before Weight After D D2
M.R.
C.D.
D.R.
M.S.
M.A.
B.A.
B.S.
E.Q.
67
74
79
77
82
69
54
99
74
92
88
92
77
82
84
99
-7
-18
-9
-15
5
-13
-30
0
49
324
81
225
25
169
900
0
      ans3 14.gif (991 bytes) ans3 15.gif (1017 bytes)

Before we begin we need to compute the mean of the Ds and the sum of the D squares.

Please conduct a t-test on this data and include in your answer:
a)  Your hypotheses
The null hypothesis:  The therapy has no effect on the weight of the patient.
ans3 12.gif (995 bytes)
The research hypothesis:  The patients will gain weight after the therapy.
ans3 13.gif (997 bytes)
b)  The value you computed for t
First we need to compute the estimate of the standard error of the difference.
ans3 16.gif (2616 bytes)

Hint:  Remember that when you square a negative number it becomes positive.
            Also remember to take the square root at the end of the equation.

Finally we can compute the t.

ans3 17.gif (1320 bytes)
c)  The degrees of freedom = the number of pairs - 1 = 8-1 = 7
d)  Wether your research hypothesis is 1-tail or 2-tail
The research hypothesis is one-tail
e)  The critical value from Table T = 1.895 for a one-tail test at 7 degrees of freedom
f)  Your conclusion:  Because the absolute value of the computed value for t, |-2.830|, is greater
        than the critical value, 1.895, we reject the null hypothesis and accept the research hypothesis.
        The women weighed significantly more after the therapy than they did before the therapy.

[Top of Page]
[Course News] [Course Outline] [Chapter Outlines] [Homework] [Practice Tests] [Dr. Vernoy]

Copyright 2004 by Mark W. Vernoy