/*! 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. 9. No homework? Refer to Exercise 1... [FREE SOLUTION] | 91影视

91影视

No homework? Refer to Exercise 1. The math teachers inspect the

homework assignments from a random sample of 50 students at the school. Only 68% of the students completed their math homework. A significance test yields a P-value of 0.1265.

a. Explain what it would mean for the null hypothesis to be true in this setting.

b. Interpret the P-value.

Short Answer

Expert verified

Part a) H0:p=75%=0.75

Part b)

Step by step solution

01

Part a) Step 1: Given information

The claim is proportion is less than 75%

02

Part a) Step 2: The objective is to explain the mean for the null hypothesis to be true in this setting.

The null hypothesis statement states that the population value is equal to the claim value:

H0:p=75%=0.75

If the null hypothesis H0:p=75%=0.75is correct,

then 75%of all students at the researcher's school finished their homework assignments last night.

03

part b) Step 1: Given information

P-value=0.1265=12.65%

p^=68%=0.68

04

Part b) Step 2: The objective is to explain the p value

The claim is proportion is less than75%

The null hypothesis or the alternative hypothesis is the claim. The null hypothesis statement states that the population is equal to the claim value. If the claim is the null hypothesis, then the alternative hypothesis statement is the inverse of the null hypothesis.

H0:p=75%=0.75Ha:p<0.75

When the null hypothesis is true, the P-value is the probability of receiving the value of the test statistic or a more extreme value.

When the population proportion of all students at the school who finished their homework assignments last night is0.75, there is a12.65%chance that the sample proportion of students in the sample who finished their homework assignments last night is0.68or less.

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 91影视!

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

Walking to school Refer to Exercise 36.

a. Explain why the sample result gives some evidence for the alternative hypothesis.

b. Calculate the standardized test statistic and P-value.

c. What conclusion would you make?

Losing weight A Gallup poll found that 59% of the people in its sample said 鈥淵es鈥 when asked, 鈥淲ould you like to lose weight?鈥 Gallup announced: 鈥淔or results based on the total sample of national adults, one can say with 95% confidence that the margin of (sampling) error is 卤3 3percentage points.鈥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

Side effects A drug manufacturer claims that less than 10%of patients who take its new drug for treating Alzheimer鈥檚 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鈥檚 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鈥檚 claim?

After once again losing a football game to the archrival, a college鈥檚 alumni association conducted a survey to see if alumni were in favor of firing the coach. An SRS of 100 alumni from the population of all living alumni was taken, and 64 of the alumni in the sample were in favor of firing the coach. Suppose you wish to see if a majority of all living alumni is in favor of firing the coach. The appropriate standardized test statistic is

(a)z=0.64-0.50.64(0.36)100z=0.64-0.50.64(0.36)100

role="math" localid="1654432946823" (b)t=0.64-0.50.64(0.36)100t=0.64-0.50.64(0.36)100

(c)z=0.64-0.50.5(0.5)100z=0.64-0.50.5(0.5)100

(d)z=0.64-0.50.64(0.36)64z=0.64-0.50.64(0.36)64

(e)z=0.5-0.640.5(0.5)100z=0.5-0.640.5(0.5)100

Home computers Jason reads a report that says 80%of U.S. high school

students have a computer at home. He believes the proportion is smaller than 0.80at his large rural high school. Jason chooses an SRS of 60students and finds that 41have a computer at home. He would like to carry out a test at the =0.05significance level of H0:p=0.80versus Ha:p<0.80, where p= the true

proportion of all students at Jason鈥檚 high school who have a computer at home. Check if the conditions for performing the significance test are met.

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.