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.
Please include in your answer:
a) Your hypotheses
Null hypothesis: Gore Tech no different from the population
Research hypothesis: Gore Tech higher than the population
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.
Now compute t
c) The degrees of freedom = df = n1 = 971= 96
d) Wether your research hypothesis is 1tail or 2tail
Research hypothesis is onetail
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.
The research hypothesis: The drug attitude scores for those watching the alcohol
movie
will be higher than
those watching the dog movie.
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
c) The degrees of freedom = (n_{1}1) + (n_{2}1) = (61) +
(71) = 11
d) Determine if your research hypothesis is 1tail or 2tail
The research hypothesis is onetail
e) The critical value from Table T = 1.796 for a onetail 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  D^{2} 
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 
Before we begin we need to compute the mean of the Ds and the sum of the D squares.
Please conduct a ttest 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.
The research hypothesis: The patients will gain weight after the therapy.
b) The value you computed for t
First we need to compute the estimate of the standard error of the difference.
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.
c) The degrees of freedom = the number of pairs  1 = 81 = 7
d) Wether your research hypothesis is 1tail or 2tail
The research hypothesis is onetail
e) The critical value from Table T = 1.895 for a onetail 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.
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