Answer:
The teacher randomly divides the class into two groups. One of the groups receives computer-based instruction and the other group receives traditional instruction without computers. After instruction, each student takes a test and the teacher wants to compare the test scores of the two groups.
Step-by-step explanation:
A two sample t-test is used to compared the means of two samples who were exposed to different scenarios and therefore, it helps us to see the difference between these two samples.
In this problem, the teacher is experimenting with a new computer-based instruction. To be able to use the two-sample t-test she would have to have two randomly assigned samples and expose each of these two samples to different scenarios (one would receive a traditional instruction and the other one would receive a computer instruction), and finally she would need to compare the scores in both samples.
Therefore, the correct answer would be b) The teacher randomly divides the class into two groups. One of the groups receives computer-based instruction and the other group receives traditional instruction without computers. After instruction, each student takes a test and the teacher wants to compare the test scores of the two groups.
<u>*Option A is incorrect</u> because the students are only exposed to the lesson using a computer program and therefore they aren't exposed to different scenarios.
<u>*Option C is incorrect</u> because although in this case the teacher uses both traditional and computer methods, there are not two different samples and she only gets the opinion of the students which is not an exact method of getting information on the effectiveness of the computer based method.
Answer:
The student's overall grade is 85.79%
Step-by-step explanation:
Given in the question that:
Exam = 50%; Quiz = 30%, Home work = 15% and Class participation = 5%
The total giving us 100%.
A particular student scored the following;
87.9% on exams, 77.8% quiz, 90% Home work and 100% on class participation
Calculating the percentages the student had for each;
Exam total = 50*0.879 = 43.95% earned
Quiz = 30*0.778 = 23.34% earned
Home work = 15*0.9 = 13.5% earned
Class participation = 5*1 = 5% earned
Total earned = 43.95+ 23.34 + 13.5+ 5 = 85.79%
What type of model are you talking about?
im sorry but did you ever get the answer?
