Losing weight A Gallup poll found that 59% of the people in its sample said “Yes” when asked, “Would you like to lose weight?” Gallup announced: “For results based on the total sample of national adults, one can say with 95% confidence that the margin of (sampling) error is ±3 ±3percentage points.”12 Based on the confidence interval, is there convincing evidence that the true proportion of U.S. adults who would say they want to lose weight differs from 0.55? Explain your reasoning

Short Answer

Expert verified

The required answer is :

Yes, there are sufficient evidence

Step by step solution

01

Given information

Sample proportion (p^)=59%=0.59

Margin of error(E)=±3%=±0.03

02

The objective is to find out the confidence interval 

The formula for calculating the confidence interval for a population proportion is as follows:

Confidence interval =p^±E

The 95%confidence interval can be calculated as follows:

Confidence interval :

=p^±E=0.59±0.03=(0.56,0.62)

Therefore, the confidence interval is(0.56,0.62)

03

The objective is to find whether there are sufficient evidence to conclude that the proportion of U.S adults who are saying that they are interesting in losing the weight is different from 0.55 . 

Because 0.55does not fall within the above-mentioned confidence interval. Thus, there is sufficient evidence to conclude that the proportion of US adults who say they want to lose weight is greater than 0.55

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

Reality TVTelevision networks rely heavily on ratings of TV shows when deciding

whether to renew a show for another season. Suppose a network has decided that

“Miniature Golf with the Stars” will only be renewed if it can be established that more than 12%of U.S. adults watch the show. A polling company asks a random sample of 2000U.S. adults if they watch “Miniature Golf with the Stars.” The network uses the data to perform a test of

H0:p=0.12

Ha:p>0.12

where pis the true proportion of all U.S. adults who watch the show. Describe a Type Ierror and a TypeIIerror in this setting, and give a possible consequence of each.

Attitudes The Survey of Study Habits and Attitudes (SSHA) is a

psychological test with scores that range from 0to200.. The mean score for U.S. college students is115. A teacher suspects that older students have better attitudes toward school. She gives the SSHA to an SRS of 45students from the more than 1000students at her college who are at least 30years of age. The teacher wants to perform a test at the α=0.05significance level of

H0:μ=115Ha:μ>115

where μ=the mean SSHA score in the population of students at her college who are at least 30years old. Check if the conditions for performing the test are met.

Side effects A drug manufacturer claims that less than 10%of patients who take its new drug for treating Alzheimer’s disease will experience nausea. To test this claim, researchers conduct an experiment. They give the new drug to a random sample of 300out of 5000Alzheimer’s patients whose families have given informed consent for the patients to participate in the study. In all, 25of the subjects experience nausea.

a. Describe a Type I error and a Type II error in this setting, and give a possible

consequence of each.

b. Do these data provide convincing evidence for the drug manufacturer’s claim?

pg559¯

No homework Refer to Exercises 1 and 9. What conclusion would you make at theα=0.05α=0.05level?

A company that manufactures classroom chairs for high school students

claims that the mean breaking strength of the chairs is 300 pounds. One of the chairs collapsed beneath a 220-pound student last week. You suspect that the manufacturer is exaggerating the breaking strength of the chairs, so you would like to perform a test of H0:μ=300Ha:μ<300where μ is the true mean breaking strength of this company’s classroom chairs.

a. The power of the test to detect that μ=294 based on a random sample of 30

chairs and a significance level of α=0.05 is 0.71. Interpret this value.

b. Find the probability of a Type I error and the probability of a Type II error for the test in part (a).

c. Describe two ways to increase the power of the test in part (a).

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