A psychologist would like to examine the effects of different testing methods on the final performance of college students. One
group gets regular quizzes, one group gets three large exams, and the third group only gets a final exam. At the end of the course, the psychologist interviews each student to get a measure of the student’s overall knowledge of the material.
Quizzes Exams Final Only
4 1 0
6 4 2 G = 36
3 5 0 ΣX2= 164
7 2 2
T = 20 T = 12 T = 4
SS = 10 SS = 10 SS = 4
a. State the null hypothesis.
b. Compute the necessary hypothesis test using a two-tailed test with α= .05. Show your calculations.
c. Make a decision about your null hypothesis. (Include the critical value you used tobase your decision.)
d. Write an APA format statement summarizing your findings. (Include a measure ofeffect size if necessary.)
e. Explain what would need to be done next to follow up on any significant effects.
Quizzes Exams Final Only
4 1 0
6 4 2 G = 36
3 5 0 ΣX2= 164
7 2 2
T = 20 T = 12 T = 4
SS = 10 SS = 10 SS = 4
a. State the null hypothesis.
b. Compute the necessary hypothesis test using a two-tailed test with α= .05. Show your calculations.
c. Make a decision about your null hypothesis. (Include the critical value you used tobase your decision.)
d. Write an APA format statement summarizing your findings. (Include a measure ofeffect size if necessary.)
e. Explain what would need to be done next to follow up on any significant effects.
1 answer:
Answer:
Check the explanation
Step-by-step explanation:
One way ANOVA
The null and alternative hypothesis for this one way ANOVA is given as below:
Null hypothesis: H0: There is no significant difference in the averages of the scores for the quizzes, exams and final only.
Alternative hypothesis: There is a significance difference in the averages of the scores for the quizzes, exams and final only.
The ANOVA table with calculations can be seen in the attached images below:
In the attached image below, we get the p-value for this one way ANOVA test as 0.0221. We do not reject the null hypothesis if the p-value is greater than the given level of significance and we reject the null hypothesis if the p-value is less than the given level of significance or alpha value.
In the attached image below, we are given that the p-value = 0.0221 and level of significance or alpha value = 0.05, that is p-value is less than the given level of significance. So, we reject the null hypothesis that there is no significant difference in the averages of the scores for the quizzes, exams and final only. This means we conclude that there is a significance difference in the averages of the scores for the quizzes, exams and final only.
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Answer:
The value that is greater than 45% of the data values is approximately 137.84.
Step-by-step explanation:
The key is transforming values from this distribution to a z-score range and finding the corresponding value using a z-score table.
We are looking for a value x which attains a critical z-score that corresponds to the (100-45)%=55-th percentile:
The critical z value (from z-score table, online) is: -0.12, so:
The value that is greater than 45% of the data values is approximately 137.84.