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Chapter 12: Q. 53 (page 818)

Students in Mr. Handford’s class dropped a kickball beneath a motion detector. The detector recorded the height of the ball (in feet) as it bounced up and down several times. Here is a computer output from a linear regression analysis of the transformed data of log(height) versus bounce number. Predict the highest point the ball reaches on its seventh bounce.

a. 0.35feet

b. 2.26feet

c. 0.37feet

d. 2.32feet

e. 0.43feet

Short Answer

Expert verified

The correct option is (e).

Step by step solution

01

Given information

The given data is

02

Explanation

The general equation is y^=b0+b1x

The computation of the constant b0is noted in the computer output's row "Constant" and column "Coef":

b0=0.45374

The computation of the slope b1is mentioned in the computer output's row "Bounce" and column "Coef":

b1=-0.117160

Substitute the values in the equation

role="math" localid="1654189474140" =0.45374-0.117160x

logy=0.45374-0.117160x

Put the value of x

logy=0.45374-0.117160(7)

=-0.36638

Then take the exponential

y^=10logyy^=10-0.36638y^=0.43

The option is (e).

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Most popular questions from this chapter

Sam has determined that the weights of unpeeled bananas from his local store have a mean of116grams with a standard deviation of 9grams. Assuming that the distribution of weight is approximately Normal, to the nearest gram, the heaviest 30%of these bananas weigh at least how much?

a.107g

b.121g

C.111g

d.125g

e.116g

Western lowland gorillas, whose main habitat is in central Africa, have a mean weight of 275pounds with a standard deviation of 40pounds. Capuchin monkeys, whose main habitat is Brazil and other parts of Latin America, have a mean weight of 6pounds with a standard deviation of 1.1pounds. Both distributions of weight are approximately Normally distributed. If a particular western lowland gorilla is known to weigh 345pounds, approximately how much would a capuchin monkey have to weigh, in pounds, to have the same standardized weight as the gorilla?

a. 4.08

b. 7.27

c. 7.93

d.8.20

e. There is not enough information to determine the weight of a capuchin monkey.

Multiple Choice Select the best answer for Exercises 23-28. Exercises 23-28 refer to the following setting. To see if students with longer feet tend to be taller, a random sample of 25students was selected from a large high school. For each student, x=footlength&y=heightwere recorded. We checked that the conditions for inference about the slope of the population regression line are met. Here is a portion of the computer output from a least-squares regression analysis using these data:

Which of the following would have resulted in a violation of the conditions for inference?

a. If the entire sample was selected from one classroom

b. If the sample size was 15instead of 25

c. If the scatterplot of x=footlength&y=heightdid not show a perfect linear relationship

d. If the histogram of heights had an outlier

e. If the standard deviation of foot length was different from the standard deviation of height

Beer and BAC How well does the number of beers a person drinks predict his or her blood alcohol content (BAC)? Sixteen volunteers aged 21or older with an initial BAC of 0took part in a study to find out. Each volunteer drank a randomly assigned number of cans of beer. Thirty minutes later, a police officer measured their BAC. A least-squares regression analysis was performed on the data using x=number of beers and y=BAC. Here is a residual plot and a histogram of the residuals. Check whether the conditions for performing inference about the regression model are met.

The students in Mr. Shenk’s class measured the arm spans and heights (in inches) of a random sample of 18students from their large high school. Here is computer output from a least-squares regression analysis of these data. Construct and interpret a 90%confidence interval for the slope of the population regression line. Assume that the conditions for performing inference are met.

PredictorCoefStdevt-ratioPConstant11.5475.6002.060.056Armspan0.840420.0809110.390.000S=1.613R-Sq=87.1%R-Sq(adj)=86.3%
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