Question

6. (4 points) The data and sample regression equation on plant weight x (in grams) and quantity of volatile emissions y in hundreds of nanograms) are given below. Presuming that the assumptions for regression inferences are met, use SPSS to decide at significance level 0.10 whether the data provide sufficient evidence to conclude that the predictor variable is useful for predicting the response variable. Perform the five steps of hypothesis testing using a p-value test. 57 83 57 63 565 62 78 73 5268 a) What are the null and alternative hypothesis? b) What is the test statistic? c) What is the p-value? d) State and interpret your hypothesis test conclusion

Answer 1

Null hypothesis: H:B=0 There is no linear relationship between x and y. Alternative hypothesis: HAB=0

There is a linear relationship between x and y. (6) SPSS output:

Model Summary Adjusted R Square - 036 Model R R Square 260 .068 a. Predictors: (Constant), X Std. Error of the Estimate 5.4536 Change Statistics R Square Change F Change 0f1 .0686541 Sig. F Change d r2 9 .439 ANOVA df Sig Mean Square 19.456 29.742 .654 439 Sum of Model Squares Regression 19.456 Residual 267.680 Total 287.136 a Dependent Variable:y b. Predictors: (Constant), 10 Coefficients Standardized Coefficients Beta Sig 611 Unstandardized Coefficients Model Std. Error (Constant) 5.885 11.167 138 171 a. Dependent Variable:y .527 260 439 The value of the test statistic is t=0.809

p-value = 0.439. (d) Conclusion: Use a significance level, a = 0.10. Here, p-value is 0.439, which is greater than the value of a = 0.10. That is, p-value (=0.439) > a{=0.10).

Therefore, by the rejection rule, do not reject the null hypothesis H.- Interpretation: There is insufficient evidence at level of significance a = 0.10. Therefore, we cannot conclude that there is a linear relationship between x and y. The data does not provide sufficient evidence to conclude that the predictor variable is useful for predicting the response variable.