The midterm scores for samples of students from two sections of a large course h
ID: 3175580 • Letter: T
Question
The midterm scores for samples of students from two sections of a large course have the following summary statistics: # of Students Mean Standard Deviation Section A01 10 64.6 20.20 Section A02 8 71.0 9.27 We would like to conduct a hypothesis test to determine whether the true mean midterm score for Section A02 students is higher than the true mean midterm score for Section A01 students.
What are the hypotheses for the appropriate test of significance?
Question 12 options:
A) H0:XA02=XA01 vs. Ha: XA02XA01
B) H0:A02=A01 vs. Ha: A02>A01 '
C) H0:A02=A01 vs. Ha: A02A01
D) H0:XA02=XA01 vs. Ha: XA02>XA01
Explain what it would mean to make a Type I error in the context of this example.
What is the P-value for the appropriate test of significance? Between What is the conclusion of the hypothesis test at the 5% level of significance?
Explanation / Answer
The researcher wants to check whether mean midterm score for section A02 students is higher than mean midterm score for section A01 students, therefore, he postulates the null hypothesis as follows:
H0: mu A02=muA01 (there is no difference in mean midterm score for section A01 and section A02) and alternative hypothesis becomes:
H1:muA02>muA01 (mean midterm score for section A02 students is higher than mean midterm score for A01 students). Ans>B.
Type I error refer to incorrect rejection of true null hypothesis, that is if the reseracher falsely conlcude mean midterm score for section A02 students is higher than mean midterm score for A01 students, then he committs Type I error.
--- Assume the Independence assumption, Independent assumption, Randomization condition, and Nearly normal conditions are reasonable met. Therefore, for two independent groups, use 2-sample t test.
t=(xbarA02-xbarA01)/sqrt[s^2A01/nA01+s^2A02/nA02], where, xbar is sample mean, s is sample standard deviation, n is sample size.
=(64.6-71)/sqrt[20.2^2/10+9.27^2/8]
=-0.89
p value at 13 degrees of freedom is 0.806.
df=[{(s1^2/n1+s2^2/n2)^2}/{1/n1-1(s1^2/n1)^2+1/n2-1(s2^2/n2)^2}].
Conclusion: Rule states that reject H0, if p value is less than 0.05. Here, p value is not less than 0.05, therefore, fail to reject H0. Insufficient sample evidence to conclude that mean midterm score for section A02 students is significantly higher than mean midterm score for A01 students.
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