## Related questions with answers

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 exam was taken and weekly study times before and after the exam were compared. The resulting data are given in Table $10.5$. Assume that the population of all possible paired differences is normally distributed.

Table $10.5$ Weekly Study Time Data for Students Who Perform Well on the MidTerm

Students | $1$ | $2$ | $3$ | $4$ | $5$ | $6$ | $7$ | $8$ |
---|---|---|---|---|---|---|---|---|

Before | $15$ | $14$ | $17$ | $17$ | $19$ | $14$ | $13$ | $16$ |

After | $9$ | $9$ | $11$ | $10$ | $19$ | $10$ | $14$ | $10$ |

Use the p-value to test the hypotheses at the $.10$, $.05$, and $.01$ levels of significance. How much evidence is there against the null hypothesis?

Paired T-Test and CI: StudyBefore, StudyAfter

$\begin{array}{lrrrr} \text{Paired T for} & & \text{StudyBefore}&\text{- StudyAfter}\\ & \text { N } & \text { Mean } & \text { StDev } & \text { SE Mean } \\ \text { StudyBefore } & 8 & 15.6250 & 1.9955 & 0.7055 \\ \text { StudyAfter } & 8 & 11.5000 & 3.4226 & 1.2101 \\ \text { Difference } & 8 & 4.12500 & 2.99702 & 1.05961\end{array}$

$95 \%$ CI for mean difference: $(1.61943,6.63057)$ $T$-Test of mean difference $=0$ (vs not $=0$ ): $\mathrm{T}$-Value $=3.89 \quad \mathrm{P}$-Value $=\mathbf{0 . 0 0 6}$

Solution

VerifiedThe goal of this task is to test the null and the alternative hypothesis. The hypotheses that will help us try to determine if the population mean of the time that was spent studying for midterm exam has changed, is this a null hypothesis

$H_0: \mu_1-\mu_2 = 0,$

and this alternative hypothesis:

$H_a: \mu_1-\mu_2 \neq 0.$

Here $\mu_1$ represents the population mean of the time that was spent studying before the midterm exam and $\mu_2$ is the population mean of time that was spent studying after that exam.

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