Ans:
16)
Paired t test:
let d=preparation -no preparation
preparation class | No preparation | d | |
1 | 1130 | 1040 | 90 |
2 | 1470 | 1240 | 230 |
3 | 1230 | 1270 | -40 |
4 | 630 | 1020 | -390 |
5 | 1030 | 1350 | -320 |
d-bar | -86 | ||
sd | 264.6318 |
Test statistic:
t=(-86-0)/(264.63/SQRT(5))
t=-0.727
df=5-1=4
p-value=tdist(0.727,4,1)=0.2538
Fail to reject the null hypothesis(as p-value is high)
There is not sufficient evidence to conclude that preparation course increase the SAT score.
16. It is believed that spending money on a professional preparation service for the SAT exams...
number 15 how do i find the sample size? it is said the p-value is 0.275 13.13 CHAPTER 13 EXERCISES 367 the Grape flavor was right on target for the last test with a variance of 2.30 grams based on a sample of size 16. The machine filling the Fruity flavor had a variance of 2.58 grams for a sample size of 16. Assuming the distribution of the fills for each machine is approximately normal, is there evidence to show...
A high school is examining the effectiveness of a new SAT prep program. The following scores are from practice tests taken by students before and after they took the prep program: SAT Score (before) SAT Score (after) 1000 1020 1250 1260 1300 1330 1070 1050 1100 1130 1050 1070 1100 1080 1350 1400 1250 1290 980 1020 Test the null hypothesis that the pre-program and post-program scores are equal (alpha=0.05).