/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Q.79 In 1955, Life Magazine reported ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

In 1955, Life Magazine reported that the 25year-old mother of three worked, on average, an 80 hour week. Recently,

many groups have been studying whether or not the women's movement has, in fact, resulted in an increase in the average

work week for women (combining employment and at-home work). Suppose a study was done to determine if the mean

work week has increased. 81women were surveyed with the following results. The sample mean was83the sample

standard deviation was ten. Does it appear that the mean work week has increased for women at the role="math" localid="1650381098713" 5%level?

Short Answer

Expert verified

Yes, the average work hours for women have 5%grown.

Step by step solution

01

 Given Information

Letμ=populationmeanworkweekforwomen

So, Null Hypothesis , H0: μ≤80hour week

( the mean work week has not increased for women )

Alternate Hypothesis , H0:μ>80

( the mean work week has increased for women )

02

Explanation

The test statistics that will be used here is One-sample t test statistics because we do not know about population standard deviation ;

T.S=X¯-μsn~t=n-1

Where,

X¯=samplemeanworkweek=83

s = sample standard deviation =10

n= sample of women =75

So,t=83-801075~t=74

=2.598

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 91Ó°ÊÓ!

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

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?

According to an article in Bloomberg Businessweek, New York City's most recent adult smoking rate is 14%. Suppose that a survey is conducted to determine this year’s rate. Nine out of 70 randomly chosen N.Y. City residents reply that they smoke. Conduct a hypothesis test to determine if the rate is still 14% or if it has decreased.

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:μ≤12

Ha:μ>12

Thep-value is0.0013.

Draw a graph that shows thep-value.

Assume H0:μ=9 and Ha:μ<9. Is this a left-tailed, right-tailed, or two-tailed test?

Previously, an organization reported that teenagers spent 4.5 hours 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 4.75 hours with a sample standard deviation of 2.0. Conduct a hypothesis test.

The null and alternative hypotheses are:

a.Ho:x¯=4.5,Ha:x¯>4.5b.Ho:μ≥4.5,Ha:μ<4.5c.Ho:μ=4.75,Ha:μ>4.75d.Ho:μ=4.5,Ha:μ>4.5

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.