A new SAT study course is tested on 12 individuals. Pre-course and post-course scores are recorded. Of interest is the average increase in SAT scores. The following data is collected. Conduct a hypothesis test at the 5% level. Pre-course score NOTE: If you are using a Student's t-distribution for the problem, including for paired data, you may assume that the underlying population is normally distributed. (In general, you must first prove that assumption, though.)
Pre-course score | Post-course score |
1230 | 1340 |
940 | 930 |
1090 | 1140 |
840 | 880 |
1100 | 1070 |
1250 | 1320 |
860 | 860 |
1330 | 1370 |
790 | 770 |
990 | 1040 |
1110 | 1200 |
740 | 850 |
PART A) State the null hypothesis. H_{0}: μ_{d} ≤ 0
PART B) State the alternative hypothesis. H_{a}: μ_{d} = 0
PART C) In words, state what your random variable X_{d} represents.-->The variable X_{d} represents the sample mean difference in SAT scores before the course and after the course.
PART D) State the distribution to use for the test. t={12-1}
PART E) What is the test statistic?
PART F) What is the p-value? If
H_{0} is true, then there is a chance equal to the p-value that the sample average difference between the post-course scores and pre-course scores is at least 41.67.
PART G) Sketch a picture of this situation. Label and scale the horizontal axis and shade the region(s) corresponding to the p-value
PART H) Indicate the correct decision ("reject" or "do not reject" the null hypothesis), the reason for it, and write an appropriate conclusion.
α = 0.05. reject the null hypothesis. Since p-value < α, we reject the null hypothesis.
There is sufficient evidence to show that the average post-course SAT score is larger than the average pre-course SAT score.
PART I) Explain how you determined which distribution to use.
A new SAT study course is tested on 12 individuals. Pre-course and post-course scores are recorded....
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