Question

"Asian Family Reunion," by Chau Nguyen every two years it comes around. We all get together from different towns. In my honest opinion, It's not a typical family reunion. Not forty, or fifty, or sixty, But how about seventy companions! The kids would play, scream, and shout One minute they're happy, another they'll pout. The teenagers would look, stare, and compare from how they look to what they wear. The men would chat about

their business . That they make more, but never less. Money is always their subject and there's always talk of more new projects. The women get tired from all of the chats. They head to the kitchen to set out the mats. Some would sit and some would stand eating and talking with plates in their hands. Then come the games and the songs and suddenly, everyone gets along! With all that laughter, it's sad to say that it always ends in

the same old way. They hug and kiss and say "good-bye" and then they all begin to cry! I say that 60 percent shed their tears but my mom counted 35 people this year. She said

that boys and men will always have their pride, so we won't ever see them cry. I myself don't think she's correct, so could you please try this problem to see if you object?

Short answer

The alternate hypothesis states that the proportion of people who shed tears is not more than$60\%$remains true.

Step-by-step solution

Given information

The phrases from "Asian Family Reunion" is given. It is mentioned that $60\%$of relatives shed tears.

Explanation

Hypothesis:

The hypothesis indicates that the percentage of people who cry is more than or equal to $60\%$, while the second hypothesis states that the percentage of people who cry is less than $60\%$.

$$\begin{gathered} H_0:p\ge 0.60 \\ {H}_0:p<0.60 \end{gathered}$$

The normal distribution for this test is,

$N\left(0.60,\sqrt{\frac{\left(0.60\right)\left(0.40\right)}{70}}\right)$

Where n is the sample size of $50$respondents. The $P^{\wedge }$is calculated as,

$$\begin{gathered} P^{\wedge }=\frac{x}{n} \\ =\frac{35}{70} \\ =0.5 \end{gathered}$$

The Z test statistic is:

$$\begin{gathered} z=\frac{p^{\wedge }-p}{\sqrt{{\displaystyle \frac{p\left(1-p\right)}{n}}}} \\ =\frac{0.5-0.60}{\sqrt{{\displaystyle \frac{p\left(1-p\right)}{n}}}} \\ =\frac{-0.1}{0.0586} \\ =-1.708 \end{gathered}$$