In a study of the effects of rewards on learning in statistics students, an expe
ID: 3224922 • Letter: I
Question
In a study of the effects of rewards on learning in statistics students, an experimenter divided 20 students into 4 groups of equal sizes using random assignment to the groups. Each student was given a computer-generated statistics test. The students were rewarded with jelly beans as they took the test. The investigators observed the total number of correct answers on the test. The treatments were as follows:
Group A: The students were rewarded immediately for every correct answer.
Group B: The students were randomly rewarded 75% of the time whether their answer was correct or not.
Group C: The students were randomly rewarded 25% of the time whether their answer was correct or not.
Group D: The students were not rewarded at all.
The following table contains summary information for this experiment.
a) Construct the appropriate ANOVA table and test the hypothesis that there is no difference between the mean number of correct answers for the four treatments for the population of statistics students from which this sample was taken.
b) Is there evidence that the 25% random reward schedule and the 75% random reward schedule result in different achievement? Give statistical evidence to support your answer.
Treatment Sample size Mean 12 11 16 17 Standard deviation 1.707 2.345 1.707 2.702Explanation / Answer
Solution:-
SS
df
MS
F
p
Between:
130.000
3
43.333
9.305
0.001
Within:
74.510
16
4.657
Total:
204.510
19
a)
State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.
Null hypothesis: 1 = 2 = 3 = 4
Alternative hypothesis: At-least one of the is not equal.
Formulate an analysis plan. For this analysis, the significance level is 0.05.
Analyze sample data.
F statistics is given by:-
F = 9.305
The P-value = 0.001
Interpret results. Since the P-value (0.001) is less than the significance level (0.05), we have to reject the null hypothesis.
Conclusion:-
Reject H0, There is sufficient evidence for significant differences between the four treatments.
b)
State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.
Null hypothesis: B - C = 0
Alternative hypothesis: B - C 0
Note that these hypotheses constitute a two-tailed test. The null hypothesis will be rejected if the difference between sample means is too big or if it is too small.
Formulate an analysis plan. For this analysis, the significance level is 0.10. Using sample data, we will conduct a two-sample t-test of the null hypothesis.
Analyze sample data. Using sample data, we compute the standard error (SE), degrees of freedom (DF), and the t statistic test statistic (t).
SE = sqrt[(s12/n1) + (s22/n2)]
SE = 1.297
D.F = 8
t = [ (x1 - x2) - d ] / SE
t = 0.492
where s1 is the standard deviation of sample 1, s2 is the standard deviation of sample 2, n1 is the size of sample 1, n2 is the size of sample 2, x1 is the mean of sample 1, x2 is the mean of sample 2, d is the hypothesized difference between the population means, and SE is the standard error.
Since we have a two-tailed test, the P-value is the probability that a t statistic having 8 degrees of freedom is more extreme than - 0.492; that is, less than - 0.492 or greater than 0.492.
Thus, the P-value = 0.636
Interpret results. Since the P-value (0.636) is greater than the significance level (0.05), we have to accept the null hypothesis.
SS
df
MS
F
p
Between:
130.000
3
43.333
9.305
0.001
Within:
74.510
16
4.657
Total:
204.510
19
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.