Question

"The Craven," by Mark Salangsang

Once upon a morning dreary

In stats class I was weak and weary.

Pondering over last night’s homework

Whose answers were now on the board

This I did and nothing more.

While I nodded nearly napping

Suddenly, there came a tapping.

As someone gently rapping,

Rapping my head as I snore.

Quoth the teacher, “Sleep no more.”

“In every class you fall asleep,”

The teacher said, his voice was deep.

“So a tally I’ve begun to keep

Of every class you nap and snore.

The percentage being forty-four.”

“My dear teacher I must confess,

While sleeping is what I do best.

The percentage, I think, must be less,

A percentage less than forty-four.”

This I said and nothing more.

“We’ll see,” he said and walked away,

And fifty classes from that day

He counted till the month of May

The classes in which I napped and snored.

The number he found was twenty-four.

At a significance level of $0.05$,

Please tell me am I still alive?

Or did my grade just take a dive

Plunging down beneath the floor?

Upon thee I hereby implore.

Short answer

The null hypothesis is not rejected and the proportion of children sleeping at class is $44\%$.

Step-by-step solution

Given information

Sample size, $n=50$

According to the teacher, the percentage of students who slept in the class is$44\%$

Explanation

Hypothesis: The null hypothesis states that $44\%$of children sleep in class, while the alternate hypothesis states that less than $44\%$of children sleep in class.

$$\begin{gathered} H_0:p=0.44 \\ {H}_0:p<0.44 \end{gathered}$$

The normal distribution is:

$N\left(0.44,\sqrt{\frac{\left(0.44\right)\left(0.56\right)}{50}}\right)$

The Z test statistic is,

Where n is the sample size of $50$respondents:

$$\begin{gathered} z=\frac{p^{\wedge }-p}{\sqrt{{\displaystyle \frac{p\left(1-p\right)}{n}}}} \\ {P}^{\wedge }=\frac{x}{n} \\ =\frac{24}{50} \\ =0.48 \end{gathered}$$

Substitute the $P^{\wedge }$value in the Z test statistic

$$\begin{gathered} z=\frac{p^{\wedge }p}{\sqrt{{\displaystyle \frac{p\left(1-p\right)}{n}}}} \\ =\frac{0.48-0.44}{\sqrt{{\displaystyle \frac{p\left(1-p\right)}{n}}}} \\ =\frac{0.04}{0.0701} \\ =0.571 \end{gathered}$$