Analyzing the Relationship Between Weight and WingLength in Sparrows

1. a) The variables are Treatment, Weight, and WingLength b) c) Weight = β^0 x β^1 WingLength Weight=1.36549+0.46740×WingLength d) For each additional unit in WingLength, the expected Weight of a sparrow is estimated to increase by 0.46740.

2. a) b) The coefficient correlation r is .7835 - this suggests that there is a positive linear correlation between the variables of Weight and WingLength. This means that as WingLength increases, Weight should increase as well. R squared is .6139 - This suggests that 61.39% of linear regression relates to the variability on the model. There is a chance other factors and variables may be affecting this as well but there is a reasonable fit of the data.

3. a) Weight = 1.3655 + 0.4674*Winglength Weight = 1.3655 + 0.4674*16 Weight = 8.8439 b) Residual = Observed - Predicted Residual = 5.3 - 8.8439 Residual = -3.54 c) Yes it seems unusual but other factors may be affecting this number. d) As we saw the R squared only explained 61% of the variability of the data so this is consistent with that. This number does seem unusual as the weight is significantly lower than expected. The residual seems to be the minimum so it does seem very odd. The median for residuals is .0809 while the 45th residual is -3.54 which is a significant difference. Residuals: Min 1Q Median 3Q Max -3.5440 -0.9935 0.0809 1.0559 3.4168

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4. a) ● Linear relationship - There is a constant spread around the residual vs fitted plot and clear random scatter showing a linear relationship. ● Normality - The normal Q-Q plot is good for the most part as there are a few outliers such as 39, 12, and 53. But other than that, the Q-Q plot shows normality and we can proceed with caution. ● No/ little multicollinearity - The p-value is .00132 which is < .05 and see that the critical score is very low and assume little multicollinearity. ● Independence - The assumption of independence is met as there is no clear pattern to the residual plot. ● Homoscedasticity - Equal Variance - The variance of residuals is constant around the fitted line. The example given is Homoscedastic b) 1) Linearity - relationship between Audience score and Critic score is linear Independence - the residuals are independent of each other Homoscedasticity - the variance of the residuals are constant in the graph Normality - the residuals are distributed randomly and normal 2) Null hypothesis - there is no linear relationship between the 2 variables as B1 = 0 Alternative hypothesis - there is is a linear relationship between the variables as B1 =/ 0 3) T-test value is 3.341 4) P value is .001325

5) With the p value of the t-test being .001325 being less than .05, there is strong evidence to reject the null hypothesis and assume that there is a linear relationship between the variables of audience score and critic score. c) d) 1) Independence: each observation in sample is independent from one another Normality: the residuals are normally distributed Homoscedasticity: there is equal variance in the sample group 2) Null Hypothesis - there is no significant difference between Audience Scores and Critic Scores Alternative Hypothesis - there is a significant difference between Audience Scores and Critic Scores 3) The F statistic from the output given is 11.164 4) The output is given the P value of 0.00132 5)

As the p value of 0.00132 is less than the alpha of 0.05, we reject the null hypothesis. This suggests that there is a significant difference between audience and critic scores and that they are related/ affect eachother.

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STAT 3022 HW 1

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