91Ó°ÊÓ

Given data:

$\begin{array}{|llllllll|}\hline & \text{Sunday}& \text{Monday}& \text{Tuesday}& \text{Wednesday}& \text{Thursday}& \text{Friday}& \text{Saturday}\\ \begin{array}{l}\text{Number of}\\ \text{Students}\end{array}& 11& 8& 10& 7& 10& 5& 5\\ \hline\end{array}$

The null hypothesis is expressed as follows:

$H_0$: Students complete the majority of their assignments over the course of many weeks.

And the alternative hypothesis is expressed as follows:

$H_a$: The majority of homework is not completed evenly throughout the week by students.

If students complete their assignments evenly throughout the week, the predicted frequency for a class of $56$students should be $8$on Sunday, $8$on Monday, localid="1653557837693" $8$on Tuesday, localid="1653557844797" $8$on Wednesday, localid="1653557854908" $8$on Thursday, localid="1653557863907" $8$on Friday, and 8 on Saturday.

The Chi-square test can be used with the predicted frequencies as follows: localid="1653557870164" $(8,8,8,8,8,8,and8)$and observed frequencies as localid="1653557879024" $(11,8,10,7,10,5and5)$,

Let's see how to do the calculation in Excel:

In Excel, use the $CHIDIST\left(\right)$formula to calculate the p-value:

The alpha value is $0.05$. Given that $p-$value $>\mathrm{Î\pm }$, the conclusion will be to reject the null hypothesis, $H_0$. As a result, the null hypothesis is not rejected because it is accepted. As a result, there is enough evidence to ensure that students do the majority of their homework evenly across the weeks.