High School

1. A teacher was interested in the effect of class schedule on student performance. She randomly assigns students to three groups:

- Group 1 receives instruction in six 30-minute classes.
- Group 2 receives instruction in three 60-minute classes.
- Group 3 receives instruction in two 90-minute classes.

All students received the same total amount of instruction and completed the same quiz at the end of the experiment. Quiz scores for each group are listed below.

**Quiz Scores:**

- Group 1: 26, 18, 22, 19, 22, 24, 23, 30
- Group 2: 15, 20, 12, 15, 13, 20, 20, 5
- Group 3: 22, 15, 16, 14, 17, 10, 12, 22

Test the hypothesis that the duration of the class influenced quiz performance. Your answer should include an ANOVA table (see text and lecture for examples) and all of the usual steps for hypothesis testing:
- State the hypotheses.
- Identify the criterion for rejecting the null hypothesis.
- Compute the test statistic.
- Make a decision.

2. Write the results of the problem above in APA format. Include your best attempt at making a graph of the means and standard deviations for each group. The graph should be done using software (Excel would be my choice). Notice that I am looking for more than just a sentence stating whether or not the ANOVA was significant. Your results should be typed with the appropriate reporting of degrees of freedom and statistical significance or nonsignificance. Also, your results should report a measurement of effect size.

Answer :

Final answer:

The hypothesis test using ANOVA will determine if the duration of the class influenced quiz performance. The ANOVA table will provide the necessary statistics to make a decision. If the p-value is less than the significance level (usually 0.05), we reject the null hypothesis and conclude that there is a significant difference in the mean quiz scores between the three groups. If the p-value is greater than or equal to the significance level, we fail to reject the null hypothesis and conclude that there is no significant difference in the mean quiz scores between the three groups.

Explanation:

To test the hypothesis that the duration of the class influenced quiz performance, we will use Analysis of Variance (ANOVA). ANOVA is a statistical test used to compare the means of three or more groups and determine if there are any significant differences.

Hypotheses:

  • Null Hypothesis (H0): The duration of the class does not influence quiz performance. There is no significant difference in the mean quiz scores between the three groups.
  • Alternative Hypothesis (Ha): The duration of the class influences quiz performance. There is a significant difference in the mean quiz scores between the three groups.

Criterion for rejecting the null hypothesis:

If the p-value is less than the significance level (usually 0.05), we reject the null hypothesis and conclude that there is a significant difference in the mean quiz scores between the three groups.

ANOVA Table:

Steps for hypothesis testing:

  1. Calculate the sum of squares (SS) for between groups (SSB) and within groups (SSE).
  2. Calculate the degrees of freedom (df) for between groups (k-1) and within groups (N-k).
  3. Calculate the mean square (MS) for between groups (MSB = SSB / (k-1)) and within groups (MSE = SSE / (N-k)).
  4. Calculate the F-statistic (F = MSB / MSE).
  5. Find the p-value associated with the F-statistic.
  6. Compare the p-value with the significance level (usually 0.05) to make a decision.

Decision:

If the p-value is less than the significance level (usually 0.05), we reject the null hypothesis and conclude that there is a significant difference in the mean quiz scores between the three groups. If the p-value is greater than or equal to the significance level, we fail to reject the null hypothesis and conclude that there is no significant difference in the mean quiz scores between the three groups.

Learn more about testing the effect of class schedule on student performance using anova here:

https://brainly.com/question/29554661

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