Question

A company that sponsors LSAT prep courses would like to be able to claim that their...

A company that sponsors LSAT prep courses would like to be able to claim that their courses improve scores by at least 3 percentage points. To test this, they take a sample of 8 people, have each take an initial diagnostic test, then take the prep course, and then take a post-test after the course. The test results are below (scores are out of 100%): Person12345678Initial Test7676807179786779Post-Test8382928575917678 Is there evidence, at an ?=0.06 level of significance, to conclude that the prep course improves scores by at least 3 percentage points? Carry out an appropriate hypothesis test, filling in the information requested. (Arrange your data so that the standardized test statistic is for the change from the initial test to the post-test.) A. The value of the standardized test statistic: Note: For the next part, your answer should use interval notation. An answer of the form (−∞,?) is expressed (-infty, a), an answer of the form (?,∞) is expressed (b, infty), and an answer of the form (−∞,?)∪(?,∞) is expressed (-infty, a)U(b, infty). B. The rejection region for the standardized test statistic: C. The p-value is D. Your decision for the hypothesis test: A. Reject ?0. B. Do Not Reject ?1. C. Reject ?1. D. Do Not Reject ?0.

Homework Answers

Answer #1

Null hypothesis H0: mean_pre = mean_post

I.e. there is no significant difference between the mean increase/improve in scores due to LSAT prep process

Alternative hypothesis H1: Mean_post - Mean_pre>= 3%

Pre test

X 76 76 80 71 79 78 67 79. X_bar = 606/8= 75.75

Sum (X- x_bar )= 0

Sum( x- x_bar)2  = 143.5

Y 83 82 92 85 75 91 76 78 Y_bar= 662/8= 82.75

Sum(Y- Y_bar) =0

Sum(Y - Y_bar)2 = 287.5

n1= 8 n2= 8

S​​​​​​2= 1/(n1+n2-2)*(sum( X - x-bar)2+sum(Y-Y_bar)2)

S​​​​​​2= 30.7857

Under the null hypothesis H0

t =( X_bar - Y_bar)/ √(S​​​​​​2(1/n1+1/n2) ~ t​​​​​​n1+n2-2

t = -7/2.7742

t= -2.5232

tabulate t0.05 for (8+8-2)= 14 d.f is 1.7613

conclusions: since calculated mod_t is greater than tabulated t so H0 is rejected at 5% l.o.s.

so we my conclude that prep course improve scores by at least 3% points

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