A teacher is experimenting with computer-based instruction. In which situation could the teacher use a hypothesis test for a dif
ference in two population means? (i) She gives each student a pretest. Then she teaches a lesson using a computer program. Afterwards, she gives each student a posttest. The teacher wants to see if the difference in scores will show an improvement. (ii) She randomly divides the class into two groups. One group receives computer-based instruction. The other group receives traditional instruction without computers. After instruction, each student has to solve a single problem. The teachers wants to compare the proportion of each group who can solve the problem. (iii) She gives each student a pretest. She then randomly divides the class into two groups. One group receives computer-based instruction. The other group receives traditional instruction without computers. After instruction, each student takes a post-test. The teacher compares the improvement in scores (post-test minus pretest) in the two groups. (iv) The teacher uses a combination of traditional methods and computer-based instruction. She asks students which they liked better. She wants to determine if the majority prefer the computer-based instruction.
2 answers:
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<u>Answer:</u>
The present value of 10,000 if interest is paid at a rate of 6.2% compounded weekly for 8 years is 6097.56
<u>Explanation:</u>
We know that compound interest is given by
Where ,
Where A = final amount (which is given to be = 10000)
P = Principal amount (which is the present amount which we have to find)
r = interest rate = 6.2 = 0.062
n = no. of times interest applied per time period = it is given that the interest is applied weekly, so in one year there are 52 weeks so n = 52
t = time period = 8 years
Substituting the given values, we get
P = 6097.5
We get, P = 6097.56 which is the present value of a sum of money