WebAssign Printed Access Card for Brase/Brase's Understandable Statistics: Concepts and Methods, 12th Edition, Single-Term
WebAssign Printed Access Card for Brase/Brase's Understandable Statistics: Concepts and Methods, 12th Edition, Single-Term
12th Edition
ISBN: 9781337652551
Author: Charles Henry Brase, Corrinne Pellillo Brase
Publisher: Cengage Learning
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Chapter 9.3, Problem 7P

(a)

To determine

Verify the values of x, x2, y, y2, xy and r.

(a)

Expert Solution
Check Mark

Explanation of Solution

Calculation:

The variable x denotes a random variable that represents the percentage of successful free throws a professional basketball player makes in a season and y denotes a random variable that represents the percentage of successful field goals a professional basketball player makes in a season.

The formula for correlation coefficient is,

r=nxy(x)(y)nx2(x)2ny2(y)2

In the formula, n is the sample size.

The values are verified in the table below,

xyx2y2xy
6744448919362948
6542422517642730
7548562523043600
8651739626014386
7344532919363212
7351532926013723
x=439y=280x2=32,393y2=13,142xy=20,599

Hence, the values are verified.

The number of data pairs are n=6. The value of correlation r is,

r=6(20599)(439×280)6(32393)(439)26(13142)(280)2=1235941229201637452=674860.18835=0.784

Hence, the value of r is verified as approximately 0.784.

(b)

To determine

Check whether the claim ρ>0 is significant or not.

(b)

Expert Solution
Check Mark

Answer to Problem 7P

The claim ρ>0 is significant.

Explanation of Solution

Calculation:

Null hypothesis:

H0:ρ=0

Alternative hypothesis:

H1:ρ>0

Test statistic:

The test statistic formula for test correlation r is,

t=rn21r2

Where r is the sample correlation coefficient, n is the sample size with degrees of freedom d.f.=n2.

Substitute r as 0.784, and n as 6 in the test statistic formula.

t=0.784621(0.784)2=1.5680.6208=2.526

The test statistic value is 2.526.

The degrees of freedom is,

d.f.=62=4

Step by step procedure to obtain P-value using MINITAB software is given below:

  • Choose Graph > Probability Distribution Plot choose View Probability > OK.
  • From Distribution, choose ‘t’ distribution.
  • Enter the Degrees of freedom as 4.
  • Click the Shaded Area tab.
  • Choose X Value and Right Tail, for the region of the curve to shade.
  • Enter the X value as 2.526.
  • Click OK.

Output using MINITAB software is given below:

WebAssign Printed Access Card for Brase/Brase's Understandable Statistics: Concepts and Methods, 12th Edition, Single-Term, Chapter 9.3, Problem 7P , additional homework tip  1

From Minitab output, the P-value is 0.0325.

Rejection rule:

  • If the P-value is less than or equal to α, then reject the null hypothesis and the test is statistically significant. That is, P-valueα.

Conclusion:

The P-value is 0.0325 and the level of significance is 0.05.

The P-value is less than the level of significance.

That is, 0.0325(=P-value)<0.05(=α).

By the rejection rule, the null hypothesis is rejected.

Hence, the claim ρ>0 is significant at level 0.05. That is, there is a positive correlation between x and y.

(c)

To determine

Verify the values of Se, a, b, and x¯.

(c)

Expert Solution
Check Mark

Explanation of Solution

Calculation:

The value of x¯ is,

x¯=xn=4396=73.167

The value of x¯ is 73.167.

The value of y¯ is,

y¯=yn=2806=46.667

The value of y¯ is 46.667.

The value of b is,

b=nxy(x)(y)nx2(x)2=6(20599)(439×280)6(32393)(439)2=6741637=0.4117

The value of b is 0.4117.

The value of a is,

a=y¯bx¯=46.667(0.41173)73.167=46.66730.125=16.542

The value of a is 16.542.

The value of Se is,

Se=y2aybxyn2=13142(16.542×280)(0.41173×20599)62=29.013734

     =7.2706=2.6964

The value of Se is 2.6964.

The equation of the least-squares line is,

y^=a+bx=16.542+0.4117x

The regression equation is y^=16.542+0.4117x.

(d)

To determine

Find the predicted percentage y^ of successful field goals for a player with x=70% successful free throws.

(d)

Expert Solution
Check Mark

Answer to Problem 7P

The predicted percentage y^ of successful field goals for a player with x=70% successful free throws is 45.36%.

Explanation of Solution

Calculation:

The regression equation is y^=16.542+0.4117x.

Substitute x as 70 in the regression equation

y^=16.542+0.4117(70)=16.542+28.819=45.36145.36%

Hence, the predicted percentage y^ of successful field goals for a player with x=70% successful free throws is 45.36%.

(e)

To determine

Find the 90% confidence interval for y when x=70%.

(e)

Expert Solution
Check Mark

Answer to Problem 7P

The 90% confidence interval for y when x=70% is 39.06<y<51.67.

Explanation of Solution

Calculation:

Step by step procedure to obtain confidence interval using MINITAB software is given below:

  • Choose Stat > Regression > Regression.
  • In Response, enter the column containing the response as y.
  • In Predictors, enter the columns containing the predictor as x.
  • Choose Options.
  • In Prediction intervals for new observations, enter the value as 70.
  • In Confidence level, enter value as 90.
  • Click OK.

Output using MINITAB software is given below:

WebAssign Printed Access Card for Brase/Brase's Understandable Statistics: Concepts and Methods, 12th Edition, Single-Term, Chapter 9.3, Problem 7P , additional homework tip  2

From Minitab output, the confidence interval is 39.06<y<51.67.

Hence, the 90% confidence interval for y when x=70% is 39.06<y<51.67.

(f)

To determine

Check whether the claim β>0 is significant or not.

(f)

Expert Solution
Check Mark

Answer to Problem 7P

The claim β>0 is significant.

Explanation of Solution

Calculation:

Null hypothesis:

H0:β=0

Alternative hypothesis:

H1:β>0

Test statistic:

The test statistic formula for slope β is,

t=bβSex21n(x)2

Where Se is the standard error of estimate computed from the sample, n is the sample size with degrees of freedom d.f.=n2.

Substitute Se as 2.6964, x2 as 32393, x as 439 and n as 6 in the test statistic formula

t=0.41172.69643239316(439)2=6.80030.6208=2.522

The test statistic value is 2.522.

The degrees of freedom is,

d.f.=62=4

Step by step procedure to obtain P-value using MINITAB software is given below:

  • Choose Graph > Probability Distribution Plot choose View Probability > OK.
  • From Distribution, choose ‘t’ distribution.
  • Enter the Degrees of freedom as 4.
  • Click the Shaded Area tab.
  • Choose X Value and Right Tail, for the region of the curve to shade.
  • Enter the X value as 2.522.
  • Click OK.

Output using MINITAB software is given below:

WebAssign Printed Access Card for Brase/Brase's Understandable Statistics: Concepts and Methods, 12th Edition, Single-Term, Chapter 9.3, Problem 7P , additional homework tip  3

From Minitab output, the P-value is 0.0326.

Rejection rule:

  • If the P-value is less than or equal to α, then reject the null hypothesis and the test is statistically significant. That is, P-valueα.

Conclusion:

The P-value is 0.0326 and the level of significance is 0.05.

The P-value is less than the level of significance.

That is, 0.0326(=P-value)<0.05(=α).

By the rejection rule, the null hypothesis is rejected.

Hence, the claim β>0 is significant at level 0.05. That is, there is a positive correlation between x and y.

(g)

To determine

Find a 90% confidence interval for β.

Interpret the confidence interval.

(g)

Expert Solution
Check Mark

Answer to Problem 7P

The 90% confidence interval for β is 0.064<β<0.760.

Explanation of Solution

Calculation:

Confidence interval for slope:

The confidence interval formula for slope β is,

bE<β<b+E

Where E=tcSex21n(x)2, tc is the critical value, Se is the standard error of estimate computed from the sample, n is the sample size with degrees of freedom d.f.=n2.

Critical value:

Use the Appendix II: Tables, Table 6: Critical Values for Student’s t Distribution:

  • In d.f. column locate the value 4.
  • In the row of two-tail area locate the level of significance α=0.10.
  • The intersecting value of row and columns is 2.132.

The critical value is ±2.132.

The margin of error is,

E=2.132×2.69643239316(439)2=5.748716.5177=0.3480

The 90% confidence interval for β is,

0.41170.3480<β<0.4117+0.34800.064<β<0.760

Hence, the 90% confidence interval for β is 0.064<β<0.760.

The percentage of successful field goals for a basketball player increases by an amount that ranges between 0.064 and 0.760, if percentage of successful free throws increases by one unit.

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Chapter 9 Solutions

WebAssign Printed Access Card for Brase/Brase's Understandable Statistics: Concepts and Methods, 12th Edition, Single-Term

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