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Question: Impact of race on football card values. Refer to the Electronic Journal of Sociology (2007) study of the Impact of race on the value of professional football players’ “rookie” cards, Exercise 12.72 (p. 756). Recall that the sample consisted of 148 rookie cards of NFL players who were inducted into the Football Hall of Fame (HOF). The researchers modelled the natural logarithm of card price (y) as a function of the following independent variables:

Race:x1=1ifblack,0ifwhiteCardavailability:x2=1ifhigh,0iflowCardvintage:x3=yearcardprintedFinalist:x4=naturallogarithmofnumberoftimesplayeronfinalHOFballotPosition-QB::x5=1ifquarterback,0ifnotPosition-RB:x7=1ifrunningback,0ifnotPosition-WR:x8=1ifwidereceiver,0ifnotPosition-TEx9=1iftightend,0ifnotPosition-DL:x10=1ifdefensivelineman,0ifnotPosition-LB:x11=1iflinebacker,0ifnotPosition-DB:x12=1ifdefensiveback,0ifnot

[Note: For position, offensive lineman is the base level.]

  1. The model E(y)=β0+β1x1+β2x2+β3x3+β4x4+β5x5+β6x6+β7x7+β8x8+β9x9+β10x10+β11x11+β12x12 was fit to the data with the following results:R2=0.705,Ra2=0.681,F=26.9.Interpret the results, practically. Make an inference about the overall adequacy of the model.
  2. Refer to part a. Statistics for the race variable were reported as follows:β^1=-0.147,sβ^1=-0.145,t=-1.014,p-value=0.312 .Use this information to make an inference about the impact of race on the value of professional football players’ rookie cards.
  3. Refer to part a. Statistics for the card vintage variable were reported as follows:β^3=-0.074,sβ^3=0.007,t=-10.92,p-value=.000.Use this information to make an inference about the impact of card vintage on the value of professional football players’ rookie cards.
  4. Write a first-order model for E(y) as a function of card vintage x3and position x5-x12that allows for the relationship between price and vintage to vary depending on position.

Short Answer

Expert verified
  1. The value of R2 is 0.705 indicating that nearly 70% of variation in the data is explained by the model. The value of 68.1% for Ra2 indicates that the variables are explaining the model to a higher degree. At 95% significance level, it can be concluded thatthe model is not a good fit for the data.
  2. At95% significance level,β1=0 . Hence it can be concluded with enough evidence that x1 is a significance variable.
  3. At 95% significance level, β3=0. Hence it can be concluded with enough evidence that x2 is a significance variable.
  4. The model is:
  5. Ey=β0+β1x3+β2x5+β3x6+β4x7+β5x8+β6x9+β7x10+β8x11+β9x12+β10x3x5+β11x3x6+β12x3x7+β13x3x8+β14x3x9+β15x3x10+β16x3x11+β17x3x12

.

Step by step solution

01

Given Information

The model is given as-Ey=β0+β1x1+β2x2+β3x3+β4x4+β5x5+β6x6+β7x7+β8x8+β9x9+β10x10+β11x11+β12x12

Where , R2=0.705andRa2=0.681 .

The statistics for the race variable are given asβ^1=-0.147,sβ^1=-0.145,t=-1.014,p-value=0.312 whereas the statistics for the card vintage variable are given as β^3=-0.074,sβ^3=0.007,t=-10.92,p-value=.000 .

02

Interpretation of results

The results got from the model were:R2=0.705,Ra2=0.681,F=26.9. .

Here, the value of R2 is 0.705 indicating that nearly 70% of variation in the data is explained by the model which is very good number. This denotes that the model is a good fit for the data.

Value of ,Ra2=0.681, which adjusts for the added variables and checks if the added variables is explaining the variation in the model or not. The value of 68.1% indicates that the variables are explaining the model to a higher degree.

F-test statistic value is 26.9, to check the overall adequacy of the model, we conduct the f-test.

H0:β1=β2=β3=0

Ha :At least one of the parameters β1,β2,β3 is non zero.

H0:is rejected if F statistic >F0.025,147,147.

For α=0.025,F0.025,147,147=1.311. Since , there is sufficient evidence to reject

Hence the model is not a good fit for the data.

03

 Step 3: Significance of  β1

b.

Here,

H0:β1=0Ha:β10

t-test statistic=β^1sβ^1=-0.1470.145=-1.0137

Value of t0.025,147is 1.98

Ho: is rejected if t statistic>t0.05,24,24 .

For α=0.025, since t<t0.025,147

Not sufficient evidence to reject at 95% confidence interval.

Therefore, β1=0. Hence it can be concluded with enough evidence that x1 is a significant variable.

04

Significance of  β2 

c.

H0:β3=0Ha:β30

Here,

t-teststatistic=β^3sβ^3=-0.0740.007=-10.57

Value of t0.025,147 is 1.98

H0 is rejected if t statistic >t0.05,24,24. For α=0.025, since t<t0.025,147

Not sufficient evidence to reject H0 at 95% confidence interval.

Therefore,β3=0. Hence it can be concluded with enough evidence that x3 is a significant variable.

05

Model equation

d.

To write a first-order model for E(y) as a function of card vintage x3 and position x5-x12 that allows the relationship between price and vintage to vary depending on the position can be expressed by an interaction model.

The equation can be written as:

.Ey=β0+β1x3+β2x5+β3x6+β4x7+β5x8+β6x9+β7x10+β8x11+β9x12+β10x3x5+β11x3x6+β12x3x7+β13x3x8+β14x3x9+β15x3x10+β16x3x11+β17x3x12

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Most popular questions from this chapter

Forensic analysis of JFK assassination bullets. Following theassassination of President John F. Kennedy (JFK) in 1963, the House Select Committee on Assassinations (HSCA) conducted an official government investigation. The HSCA concluded that although there was a probable conspiracy involving at least one shooter in addition to Lee Harvey Oswald, the additional shooter missed all limousine occupants. A recent analysis of assassination bullet fragments, reported in the Annals of Applied Statistics(Vol. 1, 2007), contradicted these findings, concluding that the evidence used by the HSCA to rule out a second assassin is fundamentally flawed. It is well documented that at least two different bullets were the source of bullet fragments found after the assassination. Let E= {bullet evidence used by the HSCA}, T= {two bullets used in the assassination}, and= {more than two bullets used in the assassination}. Given the evidence (E), which is more likely to have occurred— two bullets used (T) or more than two bullets used ?

a. The researchers demonstrated that the ratio,P(T\E)/P(Tc\E), is less than 1. Explain why this result supports the theory of more than two bullets used in the assassination of JFK.

b. To obtain the result, part a, the researchers first showed that P(T\E)P(Tc\E)=[PE\T.PT][PE\Tc.PTc]Demonstrate this equality using Bayes’s Rule.

A paired difference experiment produced the following results:

nd=38,x¯1=92,x¯2=95.5,d¯=-3.5,sd2=21

a. Determine the values zfor which the null hypothesis μ1μ2=0would be rejected in favor of the alternative hypothesis μ1μ2<0 Use .role="math" localid="1652704322912" α=.10

b. Conduct the paired difference test described in part a. Draw the appropriate conclusions.

c. What assumptions are necessary so that the paired difference test will be valid?

d. Find a90% confidence interval for the mean difference μd.

e. Which of the two inferential procedures, the confidence interval of part d or the test of the hypothesis of part b, provides more information about the differences between the population means?

Non-destructive evaluation. Non-destructive evaluation(NDE) describes methods that quantitatively characterize materials, tissues, and structures by non-invasive means, such as X-ray computed tomography, ultrasonic, and acoustic emission. Recently, NDE was used to detect defects in steel castings (JOM,May 2005). Assume that the probability that NDE detects a “hit” (i.e., predicts a defect in a steel casting) when, in fact, a defect exists is .97. (This is often called the probability of detection.) Also assume that the probability that NDE detects a hit when, in fact, no defect exists is .005. (This is called the probability of a false call.) Past experience has shown a defect occurs once in every 100 steel castings. If NDE detects a hit for a particular steel casting, what is the probability that an actual defect exists?

Question: Refer to the Bulletin of Marine Science (April 2010) study of lobster trap placement, Exercise 6.29 (p. 348). Recall that the variable of interest was the average distance separating traps—called trap-spacing—deployed by teams of fishermen. The trap-spacing measurements (in meters) for a sample of seven teams from the Bahia Tortugas (BT) fishing cooperative are repeated in the table. In addition, trap-spacing measurements for eight teams from the Punta Abreojos (PA) fishing cooperative are listed. For this problem, we are interested in comparing the mean trap-spacing measurements of the two fishing cooperatives.

BT Cooperative

93

99

105

94

82

70

86

PA Cooperative

118

94

106

72

90

66

98


Source: Based on G. G. Chester, “Explaining Catch Variation Among Baja California Lobster Fishers Through Spatial Analysis of Trap-Placement Decisions,” Bulletin of Marine Science, Vol. 86, No. 2, April 2010 (Table 1).

a. Identify the target parameter for this study.b. Compute a point estimate of the target parameter.c. What is the problem with using the normal (z) statistic to find a confidence interval for the target parameter?d. Find aconfidence interval for the target parameter.e. Use the interval, part d, to make a statement about the difference in mean trap-spacing measurements of the two fishing cooperatives.f. What conditions must be satisfied for the inference, part e, to be valid?

Question: Independent random samples from approximately normal populations produced the results shown below.

Sample 1

Sample 2

52 33 42 4441 50 44 5145 38 37 4044 50 43

52 43 47 5662 53 61 5056 52 53 6050 48 60 55

a. Do the data provide sufficient evidence to conclude that (μ1-μ2)>10? Test usingα=0.1.

b. Construct a confidence interval for (μ1-μ2). Interpret your result.

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