"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

Expert verified

Confidence Interval: (0.245, 0.515)

Step by step solution

01

Step 1: Find H0 and Ha: We want to test if guys take less than 50% of the piercing ears, on average.

H0:p=0.50;Ha:p<0.50

02

Determine the distribution needed: In words, CLEARLY state what your random variable P′ represents.Let P’ = the proportion of friends that has a pierced ear

Normal:N0.5,(0.5)(1-0.5)31

Test Statistic:role="math" localid="1649422725054" z=1.70

03

Calculate the p-value using the normal distribution for proportions:

p-value=0.0448

In one to two complete sentences, explain what the p-value means for this problem. If the null hypothesis is true (the proportion is 0.50), then there is a 0.04408 probability that the sample (estimated) proportion is 0.38 or more.

04

Compare α and the p-value:Indicate the correct decision (“reject” or “do not reject” the null hypothesis), the reason for it, and write an appropriate conclusion, using complete sentences.

alphadecisionreason for decision
0.05Reject the null hypothesis
p-value<0.05

Conclusion: There is sufficient evidence to support the claim that less than 50% of his friends have pierced ears.

05

Confidence Interval

Confidence Interval: (0.245, 0.515): The “plus-4s” confidence interval is (0.259, 0.519).

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Most popular questions from this chapter

"Blowing Bubbles," by Sondra Prull

Studying stats just made me tense,

I had to find some sane defense.

Some light and lifting simple play

To float my math anxiety away.

Blowing bubbles lifts me high

Takes my troubles to the sky.

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Bubble therapy is the best.

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20per blow became the mean

They deviated by 6, and not16.

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Was22a reasonable guess?

Find the answer and pass this test!

"Japanese Girls’ Names"

by Kumi Furuichi

It used to be very typical for Japanese girls’ names to end with “ko.” (The trend might have started around my

grandmothers’ generation and its peak might have been around my mother’s generation.) “Ko” means “child” in Chinese characters. Parents would name their daughters with “ko” attaching to other Chinese characters which have meanings that they want their daughters to become, such as Sachiko—happy child, Yoshiko—a good child, Yasuko—a healthy child, and so on.

However, I noticed recently that only two out of nine of my Japanese girlfriends at this school have names which end with “ko.” More and more, parents seem to have become creative, modernized, and, sometimes, westernized in naming their children.

I have a feeling that, while 70 percent or more of my mother’s generation would have names with “ko” at the end,

the proportion has dropped among my peers. I wrote down all my Japanese friends’, ex-classmates’, co-workers, and

acquaintances’ names that I could remember. Following are the names. (Some are repeats.) Test to see if the proportion has

dropped for this generation.

Ai, Akemi, Akiko, Ayumi, Chiaki, Chie, Eiko, Eri, Eriko, Fumiko, Harumi, Hitomi, Hiroko, Hiroko, Hidemi, Hisako,

Hinako, Izumi, Izumi, Junko, Junko, Kana, Kanako, Kanayo, Kayo, Kayoko, Kazumi, Keiko, Keiko, Kei, Kumi, Kumiko,

Kyoko, Kyoko, Madoka, Maho, Mai, Maiko, Maki, Miki, Miki, Mikiko, Mina, Minako, Miyako, Momoko, Nana, Naoko,

Naoko, Naoko, Noriko, Rieko, Rika, Rika, Rumiko, Rei, Reiko, Reiko, Sachiko, Sachiko, Sachiyo, Saki, Sayaka, Sayoko,

Sayuri, Seiko, Shiho, Shizuka, Sumiko, Takako, Takako, Tomoe, Tomoe, Tomoko, Touko, Yasuko, Yasuko, Yasuyo, Yoko, Yoko, Yoko, Yoshiko, Yoshiko, Yoshiko, Yuka, Yuki, Yuki, Yukiko, Yuko, Yuko.

Previously, an organization reported that teenagers spent 4.5hours per week, on average, on the phone. The organization thinks that, currently, the mean is higher. Fifteen randomly chosen teenagers were asked how many hours per week they spend on the phone. The sample mean was 4.75hours with a sample standard deviation of 2.0. Conduct a hypothesis test.

At a significance level of a=0.05, what is the correct conclusion?

a. There is enough evidence to conclude that the mean number of hours is more than4.75

b. There is enough evidence to conclude that the mean number of hours is more than 4.5

c. There is not enough evidence to conclude that the mean number of hours is more than 4.5

d. There is not enough evidence to conclude that the mean number of hours is more than 4.75

Instructions: For the following ten exercises,

Hypothesis testing: For the following ten exercises, answer each question.

a. State the null and alternate hypothesis.

b. State the p-value.

c. State alpha.

d. What is your decision?

e. Write a conclusion.

f. Answer any other questions asked in the problem

A teacher believes that 85%of students in the class will want to go on a field trip to the local zoo. She performs a hypothesis test to determine if the percentage is the same or different from 85%. The teacher samples 50students and 39reply that they would want to go to the zoo. For the hypothesis test, use a 1%level of significance.

First, determine what type of test this is, set up the hypothesis test, find the p-value, sketch the graph, and state your conclusion.

The cost of a daily newspaper varies from city to city. However, the variation among prices remains steady with a

standard deviation of 20¢. A study was done to test the claim that the mean cost of a daily newspaper is $1.00. Twelve costs

yield a mean cost of 95¢ with a standard deviation of 18¢. Do the data support the claim at the 1% level?

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