/*! 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. 68 There are two common methods for... [FREE SOLUTION] | 91影视

91影视

There are two common methods for measuring the concentration of a pollutant in fish tissue. Do the two methods differ on average? You apply both methods to a random sample of 18carp and use

(a) the paired t-test for d.

(b) the one-sample z test for p.

(c) the two-sample t-test for1-2

(d) the two-sample z test for p1-p2

(e) none of these.

Short Answer

Expert verified

Apply both methods to a random sample of 18 carp and use option (a) the paired t-test for d.

Step by step solution

01

Given information

The two common methods for measuring the concentration of a pollutant in fish tissue.

Random sample=18carp

02

Explanation

One proportion: one-sample z test/interval

Two proportion: two-sample z test/interval

One mean: one-sample t-test/interval

Mean difference: two-sample t-test/interval

Two means or mean differences with paired data in the two samples: paired t-test/interval

Use a test if you want to test for a difference, equality, increase or decrease. Use an interval if you want to estimate an interval in which the true value lies.

Paired t-test ford.

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

Refer to Exercise16.

(a) Carry out a significance test at the =0.05level.

(b) Construct and interpret a 95%confidence interval for the difference between the population proportions. Explain how the confidence interval is consistent with the results of the test in part (a).

The Environmental Protection Agency is charged with monitoring industrial emissions that pollute the atmosphere and water. So long as emission levels stay within specified guidelines, the EPA does not take action against the polluter. If the polluter is in violation of the regulations, the offender can be fined, forced to clean up the problem, or possibly closed. Suppose that for a particular industry the acceptable emission level has been set at no more than 5parts per million (5ppm). The null and alternative hypotheses are H0role="math" localid="1650298159260" :=5versus Ha=>5. Which of the following describes a Type II error?

(a) The EPA fails to find evidence that emissions exceed acceptable limits when, in fact, they are within acceptable limits.

(b) The EPA concludes that emissions exceed acceptable limits when, in fact, they are within acceptable limits.

(c) The EPA concludes that emissions exceed acceptable limits when, in fact, they do exceed acceptable limits.

(d) The EPA takes more samples to ensure that they make the correct decision.

(e) The EPA fails to find evidence that emissions exceed acceptable limits when, in fact, they do exceed acceptable limits.

Pat wants to compare the cost of one- and two-bedroom apartments in the area of her college campus. She collects data for a random sample of 10 advertisements of each type. The table below shows the rents (in dollars per month) for the selected apartments.

Pat wonders if two-bedroom apartments rent for significantly more, on average than one-bedroom apartments. She decides to perform a test of H0:1=2 versus Ha:1<2, where 1 and 2 are the true mean rents for all one-bedroom and two-bedroom aparaments, respectively, near the campus.

(a) Name the appropriate test and show that the conditions for carrying out this test are met.

(b) The appropriate test from part (a) yields a P-value of 0.058. Interpret this P-value in context.

(c) What conclusion should Pat draw at the=0.05 significance level? Explain.

In an experiment to learn whether Substance M can help restore memory, the brains of 20rats were treated to damage their memories. The rats were trained to run a maze. After a day, 10rats (determined at random) were given M and 7of them succeeded in the maze. Only 2of the 10control rats were successful. The two-sample z test for 鈥渘o difference鈥 against 鈥渁 significantly higher proportion of the M group succeeds鈥

(a) gives z=2.25,P<0.02

(b) gives z=2.60,P<0.005

(c) gives z=2.25,P<0.04but not <0.02

(d) should not be used because the Random condition is violated

(e) should not be used because the Normal condition is violated.

T10.11. Researchers wondered whether maintaining a patient's body temperature close to normal by heating the patient during surgery would affect wound infection rates. Patients were assigned at random to two groups: the normothermic group (patients' core temperatures were maintained at near normal, 36.5C, with heating blankets) and the hypothermic group (patients" core temperatures were allowed to decrease to about34.5C). If keeping patients warm during surgery alters the chance of infection, patients in the two groups should have hospital stays of very different lengths. Here are summary statistics on hospital stay (in number of days) for the two groups:

GroupnxsxNormothemic10412.14.4Hypothermic9614.76.5

(a) Construct and interpret a 95%confidence interval for the difference in the true mean length of hospital stay for normothermic and hypothermic patients.

(b) Does your interval in part (a) suggest that keeping patients warm during surgery affects the average length of patients' hospital staves? Justify your answer.

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.