a. Confidence Interval: To construct a 95% confidence interval for the difference between the average test scores of the two groups, use the formula CI = (mean1 - mean2) ± Z * sqrt((s1^2 / n1) + (s2^2 / n2)). The 95% confidence interval is approximately -9.769 to 0.529. b. Hypotheses: The competing hypotheses are H0: The average test scores of Trisha's and Oliver's groups are equal and Ha: The average test scores of Trisha's and Oliver's groups are different. c. Null Hypothesis and Rejection: The null hypothesis cannot be rejected at the 5% significance level, indicating no significant difference between the average test scores of Trisha's and Oliver's groups.
a. Confidence Interval:
To construct a 95% confidence interval for the difference between the average test scores of the two groups, we can use the formula:
CI = (mean1 - mean2) ± Z * sqrt((s1^2 / n1) + (s2^2 / n2))
Where:
mean1 and mean2 are the average test scores of Trisha's (Group A) and Oliver's (Group B) groups respectively.
s1 and s2 are the standard deviations of Trisha's (Group A) and Oliver's (Group B) groups respectively.
n1 and n2 are the number of students in Trisha's (Group A) and Oliver's (Group B) groups respectively.
Z is the z-score associated with the desired confidence level, which in this case is 95%.
Plugging in the given values, we get:
CI = (25.7 - 30.6) ± 1.96 * sqrt((98.2^2 / 20) + (87.4^2 / 25))
Simplifying, we find:
CI ≈ -9.769 to 0.529
Therefore, the 95% confidence interval for the difference between the average test scores of the two groups is approximately -9.769 to 0.529.
b. Hypotheses:
The competing hypotheses can be formulated as:
Null hypothesis (H0): The average test scores of Trisha's (Group A) and Oliver's (Group B) groups are equal.
Alternative hypothesis (Ha): The average test scores of Trisha's (Group A) and Oliver's (Group B) groups are different.
c. Null Hypothesis and Rejection:
From the 95% confidence interval obtained in part a, we can observe that the range of values includes zero. Therefore, we cannot reject the null hypothesis at the 5% significance level. This means that there is not sufficient evidence to conclude that there is a significant difference between the average test scores of Trisha's (Group A) and Oliver's (Group B) groups.
The probable question may be:
Trisha and Oliver are conducting a study on the average test scores of two groups of students. Trisha's group (Group A) has an average test score of 25.7 with a standard deviation of 98.2, while Oliver's group (Group B) has an average test score of 30.6 with a standard deviation of 87.4. Trisha has 20 students in her group, and Oliver has 25 students.
a. Confidence Interval:
Construct a 95% confidence interval for the difference between the average test scores of the two groups.
b. Hypotheses:
Formulate the competing hypotheses to determine whether there is a significant difference between the average test scores of the two groups.
c. Null Hypothesis and Rejection:
Based on the confidence interval from part a, assess whether the null hypothesis can be rejected at the 5% significance level. Provide an explanation for your conclusion.
Additional Information:
The study aims to understand if there's a meaningful difference in the average test scores between Trisha's and Oliver's groups. The confidence interval helps quantify the uncertainty in this difference, and the hypotheses provide a framework for statistical testing. Students, as budding statisticians, are tasked with interpreting the results in the context of educational research.