Ideal proportions Refer to Exercise 10.

(a) What height would you predict for a student with an arm span of 76 inches? Show your work.

(b) About how far off do you expect the prediction in part (a) to be from the student's actual height? Justify your answer.

Short Answer

Expert verified

a) 75.41892 inches.

b)1.613

Step by step solution

01

Part(a) Step 1: Given Information

Need to find What height would you predict for a student with an arm span of 76 inches.

02

Part(a) Step 2: Explanation

The estimates a and b are given in the column of "Coef":

y^=11.547+0.84042x

Replace x by 5 and evaluate:

y=11.547+0.84042(76)=75.41892

This means that we predict that the student will be 75.41892inches tall.

03

Part(a) Step 1: Given Information

Given:

s=1.613

04

Part(b) Step 2: Explanation

Given:

s=1.613

The expected error made when making predictions is given by σ, which is estimated by s. Thus we expect that the prediction will be off by 1.613 on average.

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Most popular questions from this chapter

An old saying in golf is “You drive for show and you putt for dough.” The point is that good putting is more important than long driving for shooting low scores and hence winning money. To see if this is the case, data from a random sample of 69of the nearly 1000players on the PGA Tour’s world money list are examined. The average number of putts per hole and the player’s total winnings for the previous season is recorded. A least-squares regression line was fitted to the data. The following results were obtained from statistical software.

The correlation between total winnings and the average number of putts per hole for these players is

(a)-0.285

(b)-0.081

(c)0.007

(d)0.081

(e) 0.285

A company has been running television commercials for a new children’s product on five different

family programs during the evening hours in a large city over a one-month period. A random sample of families is taken, and they are asked to indicate the one program they viewed most often and their rating of the advertised product. The results are summarized in the following table.

The advertiser decided to use a chi-square test to see if there is a relationship between the family program viewed and the product’s rating. What would be the degrees of freedom for this test?

(a) 3 (c) 12 (e) 19

(b) 4 (d) 18

SAT Math scores In Chapter 3, we examined data on the percent of high school graduates in each state who took the SAT and the state's mean SAT Math score in a recent year. The figure below shows a residual plot for the least-squares regression line based on these data. Are the conditions for performing inference about the slope βof the population regression line met? Justify your answer.

In the casting of metal parts, molten metal flows through a “gate” into a die that shapes the part. The gate velocity (the speed at which metal is forced through the gate) plays a critical role in die casting. A firm that casts cylindrical aluminium pistons examined a random sample of 12pistons formed from the same alloy of metal. What is the relationship between the cylinder wall thickness (inches) and the gate velocity (feet per second) chosen by the skilled workers who do the casting? If there is a clear pattern, it can be used to direct new workers or to automate the process. A scatterplot of the data is shown below

A least-squares regression analysis was performed on the data. Some computer output and a residual plot are shown below. A Normal probability plot of the residuals (not shown) is roughly linear.

Do these data provide convincing evidence of a straight-line relationship between thickness and gate velocity in the population of pistons formed from this alloy of metal? Carry out an appropriate significance test at the α=0.05level.

The P-value for the test in Question 5 is 0.0087. A correct interpretation of this result is that

(a) the probability that there is no linear relationship between an average number of putts per hole and total winnings for these 69players is 0.0087.

(b) the probability that there is no linear relationship between the average number of putts per hole and total winnings for all players on the PGA Tour’s world money list is 0.0087.

(c) if there is no linear relationship between an average number of putts per hole and total winnings for the players in the sample, the probability of getting a random sample of 69players yields a least-squares regression line with a slope of -4139198or less is 0.0087.

(d) if there is no linear relationship between an average number of putts per hole and total winnings for the players on the PGA Tour’s world money list, the probability of getting a random sample of 69 players yields a least-squares regression line with a slope of -4139198or less is 0.0087.

(e) the probability of making a Type II error is0.0087.

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