91Ó°ÊÓ

We need to find out the evidence for all the students completing their assigned work last week.

We know that

The sample proportion is a division of the number of successes and sample size i.e. $\hat{p}=\frac{x}{n}=\frac{45}{100}=0.45=45\%$

We should notice that the sample proportion of $45\%$is lower than the advertised $60\%$, implying that less than$60\%$ of all students at the school completed their assigned homework last week.

We need to find out the explanations for the evidence described in part (a).

Because the population proportion is $60\%$and we obtained a sample with a sample proportion of only $45\%$by chance, the sample proportion could be less than $60\%$.

However, it's also feasible that the sample proportion is lower than $60\%$because the population proportion is likewise lower than $60\%$.

We need to find out the representation of value $0.73$.

Each dot in the dotplot indicates a sample proportion $p$for a simple random sample (SRS) of $100$students, where the sample proportion is the proportion of students in the sample who completed their assigned homework the previous week.

The dot at $0.73$indicates a basic random sample of $100$pupils, with $73$of them having done their assigned homework the previous week (which then results in a sample proportion of $0.73$).

We need to find whether it is surprising to get smaller SRS or not.

We can see that there are no dots above $0.45$and no dots to the left of $0.45$in the following dotplot. As a result, obtaining a sample with a sample percentage of $p=0.45$is highly implausible when the population proportion is actually $p=0.60$, and we would be shocked if we obtained a sample with a sample proportion of $p=0.45$.

We need to find the convincing answer for part (d).

We can see that there are no dots above $0.45$and no dots to the left of $0.45$in the following dotplot. This means that the sample proportion is very unlikely to be $p=0.45$when the population proportion is actually $p=0.60$, and we would be shocked if we got a sample with sample proportion $p=0.45$.

This suggests that less than $60\%$of all pupils at the school completed their given homework last week, according to credible information.