Question

Example 7: CAOS Comparisons (Paired Differences) The CAOS (Comprehensive Assessment of Outcomes in Statistics) exam is an online multiple choice test on concepts covered in a typical introductory statistics course. Students take one version before the start of the course and another version after the course ends. Before and After scores for a possible random sample of 10 students are shown in the table. (An actual random sample of scores are given in Exercise C.68 on page 455 of the text.) Student C D E F G H I Before 48 65 6048 43 After 45 5570 557152 35 55 After - Before 60 We are interested in determining whether taking the course increases students' understanding, as measured by this test. Why should we do a paired design rather than two separate groups? Calculate the differences (After - Before) for all 10 students. Then use your calculator to find the sample mean, standard deviation, and sample size for the differences. Draw a histogram (use 2 Stat Plot) of the distribution and evaluate the CLT conditions. Use the window:(X: -5, 25, 5 Y:-1, 5.1) Let d-after-before H:Hd <0 HA:Hd > 0 c. Use your calculator (STAT -> Test Test) to test to see if scores at the end of the course are higher, on average, than scores at the beginning of the course. Show all steps of this hypothesis test. Be sure to include defining the variable of interest and sample statistic, stating the hypotheses, Enter Ho, 1,5, n . and the results 0,4,p-value, X,s,n. Write your general conclusion and write your conclusion in context. d. What is the average increase on the exam after taking the course? Compute and interpret a 95% confidence interval for the improvement in mean CAOS scores between the Before and After scores. Use STAT TESTS T Interval. Write what is entered and the results.

Answer 1

Sol:

a).

Let d=after -before

$H_0:\mu _d\leq 0$

$H_a:\mu _d> 0$

b).

Student	A	B	C	D	E	F	G	H	I	J
Before	43	40	48	65	60	48	43	38	43	55
After	60	45	55	70	55	71	52	35	54	55
Difference(d)	17	5	7	5	15	23	9	-3	11	0

mean difference=10.9

std. deviation(sd)=9.219

c).

n=10

df=10-1=9

Test statistic:

t=(10.9-0)/(9.219/sqrt(10))

t=3.739

p-value=tdist(3.739,9,1)=0.0023

Asusme significance level=0.05

As,p-value<0.05,we reject null hypothesis.There is sufficient evidence to conclude that scores at the end of the course are higher, on average, than scores at the beginning of the course.  

d).

critical t value=tinv(0.05,9)=2.262

95% confidence interval for improvement in mean score

=10.9+2.262*9.219/sqrt(10)

=10.9+/-6.6

=(4.3, 17.5)

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