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.

Short Answer

Expert verified

a. The test is two-tailed test .

b. The value of p is 0.166

c. We conclude that there is insufficient information to establish that the percent of pupils in the class who will desire to attend on a field trip to the local zoo is the same or different from 85%at the 5%significance level.

d. Graph is,

Step by step solution

01

Introduction

A two-tailed test is required when the degree of uncertainty cannot be identified on the higher or lower side.

02

Explanation Part a

Because the problem is the percentage of pupils in the class who will desire to go on a field trip to the local zoo, this is a check of a singular sample size.

We must put your ideas to the reality.

H0:p=0.85vsHa:p0.85

We have, α=1%=0.01

As this is a two-tailed test, the phrase "is the same or different from" indicate this is a two-tailed test.

03

Explanation Part b

The teacher polls n=50students, x=39

say they'd want to visit the zoo.

The random variable Prepresents the percentage of children in the school who desire to travel to the nearby zoo on a school trip. The test's distribution is normal, i.e

P:Np,p(1p)n=N0.85,0.85×0.1550.

Now we determine the p-value for a mean using the data is normally distributed:

P-value=Pp<0.78orp>0.78=2Pp<0.780.166

where its data from getting is given as in the issue

p=xnp=3950p=0.78

04

Explanation Part c

The conclusion is

α=0.01<0.166=P-value.

As a result, we do not rule out H0:p=0.85In other words, the sample proportion phas a 0.166chance of being 0.79or0.92or less.

The sample data do not reveal sufficient evidence that the percentage of kids who will desire to go on a field trip to a local zoo is different from85%at1%level of significance.

05

Explanation Part d

The following is the diagram for the problem

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Assume the null hypothesis states that the mean is at least 18. Is this a left-tailed, right-tailed, or two-tailed test?

La Leche League International reports that the mean age of weaning a child from breastfeeding is age four to five worldwide. In America, most nursing mothers wean their children much earlier. Suppose a random survey is conducted of 21 U.S. mothers who recently weaned their children. The mean weaning age was nine months (3/4 year) with a standard deviation of 4 months. Conduct a hypothesis test to determine if the mean weaning age in the U.S. is less than four years old.

A normal distribution has a standard deviation of 1.We want to verify a claim that the mean is greater than12.

A sample of 36is taken with a sample mean of12.5.

H0:μ12

Ha:μ>12

Thep-value is0.0013.

Draw a graph that shows thep-value.

"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.

Sixty-eight percent of online courses taught at community colleges nationwide were taught by full-time faculty. To test if 68% also represents California’s percent for full-time faculty teaching the online classes, Long Beach City College (LBCC) in California, was randomly selected for comparison. In the same year, 34 of the 44 online courses LBCC offered were taught by full-time faculty. Conduct a hypothesis test to determine if 68% represents California. NOTE: For more accurate results, use more California community colleges and this past year's data.

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free