Classic One-Way ANOVA
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Hi,
This is very confusing to me and I feel lost, can you help me find the answers to these questions:
1. A professor had students in a large marketing class rate his performance as excellent, good, fair, or poor. A graduate student collected the ratings and assumed the students that the professor would not receive them until after course grades had been sent to the records office. The rating (i.e. treatment) a student gave the professor was matched with their course grade, which could range from 0 to 100. The sample information is reported below. Is there a difference in the mean score of the students in each of the four rating categories? Use an los of 0.01.
COURSE GRADES
excellent good fair poor
94 75 70 68
90 68 73 70
85 77 76 72
80 83 78 65
88 80 74
68 65
65
How do I do the 5 step Hypothesis procedure?
This is the Hypothesis setup:
1. Set up the null and alternative hypothesis.
2. Pick the value of alpha and find the rejection region.
3. Calculate the test statistic.
4. decide whether or not to reject the null hypothesis.
5. Interpret the statistical decision in terms of the stated problem.
Problem 2
Chi-Square-Goodness of fit: Equal expected frequencies
The CV is 7.82 for this problem
A total of 171 students were asked their drinking habits in college. The following data was found in the survey.
CLASS OBSERVED FREQUENCY
freshman 86
sophomore 36
junior 30
senior 19
TOTAL 171
USING THE 0.05 LOS IS THE DISTRIBUTION OF STUDENTS UNIFORM OVER CLASSES?
How do I do the 5 step Hypothesis procedure??
Hypothesis setup:
1. Set up the null and alternative hypothesis.
2. Pick the value of alpha and find the rejection region
3.Calculate the test statistic.
4. Decide whether or not to reject the null hypothesis.
5. Interpret the statistical decision in terms of the stated problem
THANK YOU!
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Solution Summary
This solution provides a null and alternative hypothesis for each scenario and conducts a statistical test and decide whether or not to accept or reject the null hypothesis.
Education
- BSc , Wuhan Univ. China
- MA, Shandong Univ.
Recent Feedback
- "Your solution, looks excellent. I recognize things from previous chapters. I have seen the standard deviation formula you used to get 5.154. I do understand the Central Limit Theorem needs the sample size (n) to be greater than 30, we have 100. I do understand the sample mean(s) of the population will follow a normal distribution, and that CLT states the sample mean of population is the population (mean), we have 143.74. But when and WHY do we use the standard deviation formula where you got 5.154. WHEN & Why use standard deviation of the sample mean. I don't understand, why don't we simply use the "100" I understand that standard deviation is the square root of variance. I do understand that the variance is the square of the differences of each sample data value minus the mean. But somehow, why not use 100, why use standard deviation of sample mean? Please help explain."
- "excellent work"
- "Thank you so much for all of your help!!! I will be posting another assignment. Please let me know (once posted), if the credits I'm offering is enough or you ! Thanks again!"
- "Thank you"
- "Thank you very much for your valuable time and assistance!"
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