a)
To find t statistic and p-value.
a)
Answer to Problem 23.5AYK
Test statistic is -4.989 and p-value is 0.0038
Explanation of Solution
Given:
Number of D.magna grazers | Net growth rate of G. semen |
1 | -1.9 |
2 | -2.5 |
3 | -2.2 |
4 | -3.9 |
5 | -4.1 |
6 | -4.3 |
r² | 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:
Test statistic is,
t = -4.989
Therefore, P-value of one tailed test is half of the two-tailed test.
Decision: P-value< 0.05, reject H0.
Conclusion: There is sufficient evidence to conclude that there is linear relationship between two variables.
b)
To find
b)
Answer to Problem 23.5AYK
Test statistic is -4.981 and p-value is 0.0038
Explanation of Solution
Given:
Number of D.magna grazers | Net 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 | |||||
r² | 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:
Null and alternative hypotheses:
Test statistic is
Therefore, P-value of one tailed test is.
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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Chapter 23 Solutions
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