Do students reduce study time in classes where they achieve a higher midterm sco
ID: 3327347 • Letter: D
Question
Do students reduce study time in classes where they achieve a higher midterm score? In a Journal of Economic Education article (Winter 2005), Gregory Krohn and Catherine O’Connor studied student effort and performance in a class over a semester. In an intermediate macroeconomics course, they found that “students respond to higher midterm scores by reducing the number of hours they subsequently allocate to studying for the course.” Suppose that a random sample of n = 8 students who performed well on the midterm test was taken and weekly study times before and after the test were compared. The resulting data are given in Table 10.6. Assume that the population of all possible paired differences is normally distributed. Table 10.6 Weekly Study Time Data for Students Who Perform Well on the MidTerm Students 1 2 3 4 5 6 7 8 Before 11 12 18 17 13 12 18 11 After 4 10 6 9 2 10 9 9 Paired T-Test and CI: Study Before, Study After Paired T for Study Before - Study After N Mean StDev SE Mean StudyBefore 8 14.0000 3.1168 1.1019 StudyAfter 8 7.3750 3.0208 1.0680 Difference 8 6.62500 4.13824 1.46309 95% CI for mean difference: (3.16535, 10.08465) T-Test of mean difference = 0 (vs not = 0): T-Value = 4.53, P-Value = .0027 (a) Set up the null and alternative hypotheses to test whether there is a difference in the true mean study time before and after the midterm test. H0: µd = 0 versus Ha: µd ? 0 (b) Above we present the MINITAB output for the paired differences test. Use the output and critical values to test the hypotheses at the .10, .05, and .01 level of significance. Has the true mean study time changed? (Round your answer to 2 decimal places.) t = We have evidence. (c) Use the p-value to test the hypotheses at the .10, .05, and .01 level of significance. How much evidence is there against the null hypothesis? There is against the null hypothesis.
Explanation / Answer
Given that,
null, H0: Ud = 0
alternate, H1: Ud != 0
level of significance, = 0.05
from standard normal table,right tailed t /2 =1.895
since our test is right-tailed
reject Ho, if to > 1.895
we use Test Statistic
to= d/ (S/n)
where
value of S^2 = [ di^2 – ( di )^2 / n ] / ( n-1 ) )
d = ( Xi-Yi)/n) = 6.625
We have d = 6.625
pooled variance = calculate value of Sd= S^2 = sqrt [ 471-(53^2/8 ] / 7 = 4.1382
to = d/ (S/n) = 4.5281
critical Value
the value of |t | with n-1 = 7 d.f is 1.895
we got |t o| = 4.5281 & |t | =1.895
make Decision
hence Value of | to | > | t | and here we reject Ho
p-value :right tail - Ha : ( p > 4.5281 ) = 0.00135
hence value of p0.05 > 0.00135,here we reject Ho
ANSWERS
---------------
null, H0: Ud = 0
alternate, H1: Ud != 0
test statistic: 4.5281
critical value: reject Ho, if to > 1.895
decision: Reject Ho
p-value: 0.00135
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PART B.
MINITAB RESULT
WHEN 0.05
Paired T-Test and CI: X, Y
Descriptive Statistics
Sample N Mean StDev SE Mean
X 8 14.00 3.12 1.10
Y 8 7.38 3.02 1.07
Estimation for Paired Difference
Mean StDev SE Mean 95% CI for
_difference
6.63 4.14 1.46 (3.17, 10.08)
µ_difference: mean of (X - Y)
Test
Null hypothesis H: _difference = 0
Alternative hypothesis H: _difference 0
T-Value P-Value
4.53 0.003
Paired T-Test and CI: X, Y
Descriptive Statistics
Sample N Mean StDev SE Mean
X 8 14.00 3.12 1.10
Y 8 7.38 3.02 1.07
Estimation for Paired Difference
Mean StDev SE Mean 90% CI for
_difference
6.63 4.14 1.46 (3.85, 9.40)
µ_difference: mean of (X - Y)
Test
Null hypothesis H: _difference = 0
Alternative hypothesis H: _difference 0
T-Value P-Value
4.53 0.003
Paired T-Test and CI: X, Y
Descriptive Statistics
Sample N Mean StDev SE Mean
X 8 14.00 3.12 1.10
Y 8 7.38 3.02 1.07
Estimation for Paired Difference
Mean StDev SE Mean 99% CI for
_difference
6.63 4.14 1.46 (1.50, 11.75)
µ_difference: mean of (X - Y)
Test
Null hypothesis H: _difference = 0
Alternative hypothesis H: _difference 0
T-Value P-Value
4.53 0.003
X Y X-Y (X-Y)^2 11 4 7 49 12 10 2 4 18 6 12 144 17 9 8 64 13 2 11 121 12 10 2 4 18 9 9 81 11 9 2 4 53 471Related Questions
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