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Chapter 10: Q 75 - Exercises (page 686)

Groovy tires Researchers were interested in comparing two methods for

estimating tire wear. The first method used the amount of weight lost by a tire. The second method used the amount of wear in the grooves of the tire. A random sample of 16 tires was obtained. Both methods were used to estimate the total distance traveled by each tire. The table provides the two estimates (in thousands of miles) for each tire.

a. Make a dot plot of the difference (Weight–Groove) in the estimate of wear for each tire using the two methods.

b. Describe what the graph reveals about whether the two methods give similar estimates of tire wear, on average.

c. 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 = 4.556, Mean standard deviation=3.225

Step by step solution

01

Dot Plot 

02

Interpretation 

The differences of distance travelled as per weight and groove are positive in most of the sample units.

So, it states - that the distance travelled as per wear and tear, according to weight base ; are mostly higher than the distance travelled as per wear and tear according to grooves basis.

03

Mean and standard Deviation of Differences 

MeanDifference=Sumofdifferences/numberofdifferences

[10.2+2.7+6.4+...(otherdifferences)+...+3.7+2.2+0.2]÷16=72.9/16=4.556
role="math" localid="1655229348625" StandardDeviationofdifferences=(xi-x')2/(n-1)

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

  • 1st tire : (xi-x')2=(10.2-4.556)2=31.9676
  • 2nd tire : (xi-x')2=(2.7-4.556)2=3.45

Similarly finding all (xi-x')2and summing them, [(10.2-4.55625)2+...+(0.2-4.55625)2]÷[16-1]

= 56.059375/15

10.4039=3.225

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Most popular questions from this chapter

Coaching and SAT scores What we really want to know is whether coached students improve more than uncoached students, on average, and whether any advantage is large enough to be worth paying for. Use the information above to answer these questions:

a. How much more do coached students gain, on average, compared to uncoached students? Construct and interpret a 99%confidence interval.

b. Does the interval in part (a) give convincing evidence that coached students gain more, on average, than uncoached students? Explain your answer.

c. Based on your work, what is your opinion: Do you think coaching courses are worth paying for?

Children make choices Refer to Exercise 15.

a. Explain why the sample results give some evidence for the alternative hypothesis.

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

c. What conclusion would you make?

Have a ball! Can students throw a baseball farther than a softball? To find out, researchers conducted a study involving 24randomly selected students from a large high school. After warming up, each student threw a baseball as far as he or she could and threw a softball as far as he she could, in a random order. The distance in yards for each throw was recorded. Here are the data, along with the difference (Baseball – Softball) in distance thrown, for each student:

a. Explain why these are paired data.

b. A boxplot of the differences is shown. Explain how the graph gives some evidence that students like these can throw a baseball farther than a softball.

c. State appropriate hypotheses for performing a test about the true mean difference. Be sure to define any parameter(s) you use.

d. Explain why the Normal/Large Sample condition is not met in this case. The mean difference (Baseball−Softball) in distance thrown for these 24students is xdiff = 6.54yards. Is this a surprisingly large result if the null hypothesis is true? To find out, we can perform a simulation assuming that students have the same ability to throw a baseball and a softball. For each student, write the two distances thrown on different note cards. Shuffle the two cards and designate one distance to baseball and one distance to softball. Then subtract the two distances (Baseball−Softball) . Do this for all the students and find the simulated mean difference. Repeat many times. Here are the results of 100trials of this simulation

e. Use the results of the simulation to estimate the P-value. What conclusion would you draw ?

Two samples or paired data? In each of the following settings, decide whether you should use two-sample t procedures to perform inference about a difference in means or paired t procedures to perform inference about a mean difference. Explain your choice.

a. To compare the average weight gain of pigs fed two different diets, nine pairs of pigs were used. The pigs in each pair were littermates. A coin toss was used to decide which pig in each pair got Diet A and which got Diet B.

b. Separate random samples of male and female college professors are taken. We wish to compare the average salaries of male and female teachers.

c. To test the effects of a new fertilizer, 100 plots are treated with the new fertilizer, and 100 plots are treated with another fertilizer. A computer’s random number generator is used to determine which plots get which fertilizer.

A scatterplot and a least-squares regression line are shown in the figure. What effect does point P have on the slope of the regression line and the correlation?

a. Point P increases the slope and increases the correlation.

b. Point P increases the slope and decreases the correlation.

c. Point P decreases the slope and decreases the correlation.

d. Point P decreases the slope and increases the correlation .

e. No conclusion can be drawn because the other coordinates are unknown.

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