/*! 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 R9.2. Checking conditions Refer to Exe... [FREE SOLUTION] | 91影视

91影视

Checking conditions Refer to Exercise R9.1. Identify the appropriate test to

perform in each setting and show that the conditions for carrying out the test are met.

Short Answer

Expert verified

a. t-test for the mean for the one sample

b. ztest for the proportion for one sample

Step by step solution

01

Part (a) Step 1: Given information

When evaluating a claim about a proportion, use a one-sample z test, and when testing a claim about a mean, use a one-sample t-test.

02

Part (a) Step2: Calculation

Claim about a mean and therefore use a one-sample t-test for the mean.

Conditions

Random: The reason for my satisfaction is that the sample is a simple random sample.

Independent: The reason for this is that the sample of 48female graduates represents less than 10%of the total number of female graduates in the local high school.

Large sample: The reason for this is that the sample size of 48is more than 30and hence the sample size is enormous.

Even though all of the prerequisites are met, a hypothesis test for the population mean is appropriate.

03

Part (b) Step 1: Calculation

As a result of the claim concerning the proportion, a one-sample z-test for the proportion was used.

p=0.25n=80

Conditions

Random: The reason for my satisfaction is that the sample is a random sample.

Independent: The reason for this is that the sample of 80pupils at the school represents less than 10%of the total student population.

Normal: satisfied, the reason is that

np=80(0.25)=20n(1p)=80(10.25)=60

Both are at least 10

Despite the fact that all of the preceding conditions are met, it is appropriate to calculate the confidence interval for the population percentage p

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

Which of the following has the greatest probability?

a.P(t>2)if t has 5 degrees of freedom.

b. P(t>2) if t has 2 degrees of freedom.

c. P(z>2) if z is a standard Normal random variable.

d.P(t<2)if t has 5 degrees of freedom.

e.P(z<2) if z is a standard Normal random variable.

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

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.

Proposition XA political organization wants to determine if there is convincing evidence that a majority of registered voters in a large city favor Proposition X. In an SRS of 1000registered voters, 482favor the proposition. Explain why it isn鈥檛 necessary to carry out a significance test in this setting.

Watching grass grow The germination rate of seeds is defined as the proportion of seeds that sprout and grow when properly planted and watered. A certain variety of grass seed usually has a germination rate of 0.80. A company wants to see if spraying the seeds with a chemical that is known to increase germination rates in other species will increase the germination rate of this variety of grass. The company researchers spray a random sample of 400grass seeds with the chemical, and 339of the seeds germinate. Do these data provide convincing evidence at the =0.05 significance level that the chemical is

effective for this variety of grass?

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.