Question

Suppose that the sample proportion of students who did all their assigned homework last week is $\hat{p}=\frac{57}{100}=0.57$. Would this sample proportion provide convincing evidence that less than $60\%$of all students at the school completed all their assigned homework last week? Explain your reasoning.

Short answer

A sample proportion of $\hat{p}\le 0.57$happened $\frac{78}{250}=31.27\%$of the time, so it would not be surprising to get $\hat{p}\le 0.57$. Since it is not surprising, it does not provide convincing evidence.

Step-by-step solution

Given information

We have been given that the sample proportion of students who did all their assigned homework last week is $\hat{p}=\frac{57}{100}=0.57$

Explanation

The sample proportion ($\hat{p}$) refers to the percentage of people in a sample who have a particular attribute or trait. The sample proportion is the percentage of successful samples; to calculate it, divide the number of people (or objects) who have the desired feature by the total number of persons (or items) in the sample.

Because $\frac{78}{250}=31.27\%$, of the values of $\hat{p}$in the simulation are less than or equal to $0.57$, a sample fraction of $\hat{p}=0.57$or less in an SRS of size $100$from a population with p =$0.60$ would not be surprising. A sample proportion of $\hat{p}\le 0.57$does not provide convincing evidence that the population proportion of students who completed all of their assigned homework is less than $\hat{p}=60\%$.

The disparity between the observed proportion ($\hat{p}=0.57$) and the proportion stated (p =$0.60$) could be explained by sampling variability.