Practice of Statistics in the Life Sciences
Practice of Statistics in the Life Sciences
4th Edition
ISBN: 9781319013370
Author: Brigitte Baldi, David S. Moore
Publisher: W. H. Freeman
Question
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Chapter 23, Problem 23.5AYK

a)

To determine

To find t statistic and p-value.

a)

Expert Solution
Check Mark

Answer to Problem 23.5AYK

Test statistic is -4.989 and p-value is 0.0038

Explanation of Solution

Given:

    Number of D.magna grazersNet growth rate of G. semen
    1-1.9
    2-2.5
    3-2.2
    4-3.9
    5-4.1
    6-4.3
    Regression Analysis    
      
     0.862 n 6  
     r -0.928 k 1  
     Std. Error 0.443 Dep. Var. Y 
      
    ANOVA table 
    Source SS df MS F p-value
    Regression 4.8893 1 4.8893 24.89.0075
    Residual 0.7857 4 0.1964  
    Total 5.6750 5    
      
      
    Regression output  
    variables coefficients std. error t (df=4) p-value  
    Intercept-1.30 0.41 -3.151 .0345 
    X-0.53 0.11 -4.989 .0075 

Null and alternative hypotheses:

  H0:β=0

  H0:β0

Test statistic is,

t = -4.989

Therefore, P-value of one tailed test is half of the two-tailed test.

  Pvalue=0.00752=0.0038

Decision: P-value< 0.05, reject H0.

Conclusion: There is sufficient evidence to conclude that there is linear relationship between two variables.

b)

To determine

To find correlation coefficient and find p-value for this one-tailed test.

b)

Expert Solution
Check Mark

Answer to Problem 23.5AYK

Test statistic is -4.981 and p-value is 0.0038

Explanation of Solution

Given:

    Number of D.magna grazersNet growth rate of G. semen
    1-1.9
    2-2.5
    3-2.2
    4-3.9
    5-4.1
    6-4.3
    Regression Analysis    
      
     0.862 n 6  
     r -0.928 k 1  
     Std. Error 0.443 Dep. Var. Y 
      
    ANOVA table 
    Source SS df MS F p-value
    Regression 4.8893 1 4.8893 24.89.0075
    Residual 0.7857 4 0.1964  
    Total 5.6750 5    
      
      
    Regression output  
    variables coefficients std. error t (df=4) p-value  
    Intercept-1.30 0.41 -3.151 .0345 
    X-0.53 0.11 -4.989 .0075 

Formula:

  t=rn21r2

Correlation coefficient is r = -0.928

Null and alternative hypotheses:

  H0:ρ=0

  H0:ρ<0

Test statistic is

  t=0.928621(0.928)2t=4.981

Therefore, P-value of one tailed test is.

  Pvalue=0.0038

Decision: P-value< 0.05, reject H0.

Conclusion: There is sufficient evidence to conclude that there is linear relationship between two variables.

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