Question

"Phillip’s Wish," by Suzanne Osorio

My nephew likes to play

Chasing the girls makes his day.

He asked his mother

If it is okay

To get his ear pierced.

She said, “No way!”

To poke a hole through your ear,

Is not what I want for you, dear.

He argued his point quite well,

Says even my macho pal, Mel,

Has gotten this done.

It’s all just for fun.

C’mon please, mom, please, what the hell.

Again Phillip complained to his mother,

Saying half his friends (including their brothers)

Are piercing their ears

And they have no fears

He wants to be like the others.

She said, “I think it’s much less.

We must do a hypothesis test.

And if you are right,

I won’t put up a fight.

But, if not, then my case will rest.”

We proceeded to call fifty guys

To see whose prediction would fly.

Nineteen of the fifty

Said piercing was nifty

And earrings they’d occasionally buy.

Then there’s the other thirty-one,

Who said they’d never have this done.

So now this poem’s finished.

Will his hopes be diminished,

Or will my nephew have his fun?

Short answer

The null hypothesis is rejected and the alternate hypothesis states that less than half of the friends of Philip have pierced their ears remains true.

Step-by-step solution

Given information

Philip told his mother that half of his friends have pierced their ears, but his mother replied that it is less than half. Out of$50$friends$19$have pierced their ears.

Explanation

State the hypothesis:

The null hypothesis states that half of the friends of Philip have pierced their ears and the alternate hypothesis states that less than half of the friends of Philip have pierced their ears.

$$\begin{gathered} H_0:p=0.50 \\ {H}_0:p<0.50 \end{gathered}$$

The normal distribution is:

$N\left(0.50,\sqrt{\frac{\left(0.50\right)\left(0.50\right)}{50}}\right)$

The Z test statistic is calculated as:

$z=\frac{p^{\wedge }-p}{\sqrt{{\displaystyle \frac{p\left(1-p\right)}{n}}}}$

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

$$\begin{gathered} p^{\wedge }=\frac{x}{n} \\ =\frac{19}{50} \\ =0.38 \end{gathered}$$

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

$$\begin{gathered} z=\frac{p^{\wedge }-p}{\sqrt{{\displaystyle \frac{p\left(1-p\right)}{n}}}} \\ =\frac{0.38-0.50}{\sqrt{{\displaystyle \frac{p\left(1-p\right)}{n}}}} \\ =\frac{-0.12}{0.0707} \\ =-1.697 \end{gathered}$$