/*! 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} Q76 - Exercises Well water Trace metals found in... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 10: Q76 - Exercises (page 686)

Well water Trace metals found in wells affect the taste of drinking water, and high concentrations can pose a health risk. Researchers measured the concentration of zinc (in milligrams/liter) near the top and the bottom of 10 randomly selected wells in a large region. The data are provided in the following table.

a. Make a dot plot of the difference (Bottom鈥揟op) in the zinc concentrations for these 10 wells.

  1. Describe what the graph reveals about whether the amount of zinc at the top and bottom of the wells is the same, on average.

  2. Calculate the mean difference and the standard deviation of the differences. Interpret the mean difference.

Short Answer

Expert verified

Dot Plot graphically represents differences between tire and groove estimates, Maximum positive differences denote that major weight based estimates are higher than groove based estimates. Mean difference =0.0804,meanstandarddeviation=0.05227

Step by step solution

01

Dot Plot 

02

Interpretation 

The differences of zinc concentration between top and bottom levels of top wells

is positive in most of the sample units.

So, it states - zinc levels at the bottom of the well is more than zinc levels at the top of the well.

03

Mean and standard Deviation of Differences

MeanDifference=Sumofdifferences/Numberofdifferences=0.015+鈥�0.028+鈥�.....0.058+鈥�0.111=0.804/10=0.0804

StandardDeviation=鈭�(xi-x')2/(n-1)鈥�

where xi=eachdifference,x'=meandifference,n=numberofunits

  • 1st well : (xi-x')2=(0.015-0.0804)2=0.00427
  • 2nd well : (xi-x')2=0.028-0.0804=0.6021

Similarly finding all (xi-x')2and summing them,

[(0.015-0.0804)2+...+(0.111-0.0804)2]梅[10-1]=0.0245924/9=0.05227

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

A 96% confidence interval for the proportion of the labor force that is unemployed in a certain city is (0.07,0.10). Which of the following statements is true?

a. The probability is 0.96 that between 7%and10% of the labor force is unemployed.

b. About 96% of the intervals constructed by this method will contain the true proportion of the labor force that is unemployed in the city.

c. In repeated samples of the same size, there is a 96% chance that the sample proportion will fall between 0.07and0.10.

d. The true rate of unemployment in the labor force lies within this interval 96% of the time.

e. Between 7%and10%of the labor force is unemployed 96% of the time.

American-made cars Nathan and Kyle both work for the Department of Motor Vehicles (DMV), but they live in different states. In Nathan鈥檚 state, 80%of the registered cars are made by American manufacturers. In Kyle鈥檚 state, only 60%of the registered cars are made by American manufacturers. Nathan selects a random sample of 100cars in his state and Kyle selects a random sample of 70cars in his state. Let pn-pkbe the difference (Nathan鈥檚 state 鈥� Kyle鈥檚 state) in the sample proportion of cars made by American manufacturers.

a. What is the shape of the sampling distribution of pn-pk? Why?

b. Find the mean of the sampling distribution.

c. Calculate and interpret the standard deviation of the sampling distribution.

Starting in the 1970s, medical technology has enabled babies with very low birth weight (VLBW, less than1500grams, or about 3.3 pounds) to survive without major handicaps. It was noticed that these children nonetheless had difficulties in school and as adults. A long-term study has followed 242 randomly selected VLBW babies to age 20years, along with a control group of 233 randomly selected babies from the same population who had normal birth weight. 50

a. Is this an experiment or an observational study? Why?

b. At age20,179 of the VLBW group and 193 of the control group had graduated from high school. Do these data provide convincing evidence at the 伪=0.05 significance level that the graduation rate among VLBW babies is less than for normal-birth-weight babies?

Music and memory Refer to Exercise 87.

a. Construct and interpret a 99%confidence interval for the true mean difference. If you already defined the parameter and checked conditions in Exercise87, you don鈥檛 need to do them again here.

b. Explain how the confidence interval provides more information than the test in Exercise .

Ban junk food! A CBS News poll asked 606 randomly selected women and 442

randomly selected men, 鈥淒o you think putting a special tax on junk food would encourage more people to lose weight?鈥� 170 of the women and 102 of the men said 鈥淵es.鈥� A 99% confidence interval for the difference (Women 鈥� Men) in the true proportion of people in each population who would say 鈥淵es鈥� is 鈭�0.020to0.120. Does the confidence interval provide convincing evidence that the two population proportions are equal? Explain your answer.
See all solutions

Recommended explanations on Math Textbooks

Mechanics Maths

Read Explanation

Pure Maths

Read Explanation

Discrete Mathematics

Read Explanation

Applied Mathematics

Read Explanation

Decision Maths

Read Explanation

Logic and Functions

Read Explanation
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.