Review for final exam with answers

Review for final exam with answers - In class Review for...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
In class Review for the Final #1 24.36 Sparrowhawk colonies. One of nature’s patterns connects the percent of adult birds in a colony that return from the previous year and the number of new adults that join the colony. Can the percent of returning adult birds be use to predict the number of new adults that join the colony? a. Is this an experiment or an observational study? Can we conclude causation? Observational study—cannot conclude causation b. What are the explanatory and response variables? What type of variable are they? Explanatory variable: Percent of adult birds returning Response variable: Number of new adults that join the colony c. What statistical procedure should be performed to answer the research question? Regression analysis since the data are bivariate quantitative—two different measurements on each individual. d. State the hypotheses in words and in symbols for testing whether a linear relationship exists between the two variables “Percent return” and “new birds”. H 0 : percent of returning adult birds cannot be used to predict number of new adult birds. H a : percent of returning adult birds can be used to predict number of new adult birds. H 0 : β = 0 versus H a : 0 e. Is the condition of equal variance met? Why or why not? Yes—no megaphone in residual plot Is the condition of linear relationship met? Why or why not? Yes—no smile or frown in residual plot Is the condition of Normality of residuals met? Why or why not? Yes—no outliers or strong skewness in dotplot of residuals so t distribution is robust with respect to Normality Is it reasonable to assume that the 13 colonies are independent of each other? Why? Yes—no reason to suspect that birds in one colony interact with birds in another colony. f. Using the regression printout, what are the values of the t test statistic and P- value for testing H 0 : β = 0 (slope)? On second row of computer output: t = -3.7432015 and P-value = 0.0032 g. At α = 0.05 what can you conclusions can you make about whether we can use the percentage of returning birds to predict the number of new adult birds? Since P-value < α , we reject H 0 and conclude that the percentage of returning birds can be used to predict the number of new birds. h. Interpret slope in context. For every one percent increase in percent of returning birds, there is an average decrease of 0.3 new adult birds. i. What is a 95% confidence interval for β (slope)? df = n – 2 = 13 – 2 = 11 t * = 2.201 for 95% confidence b = -0.304 SE b = 0.081 95% confidence interval for b is: -0.304 ± (2.201) ( 0.081) or (-0.482, -0.126) Note: values for b and SE b are found on the bottom row of the regression output. j. Interpret r 2 in context.
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
56% of the variation in new adult birds joining the colony can be explained by the percent of returning adult birds. k. What is the predicted number of new adult birds in a colony when 60% of the birds return?
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

Page1 / 9

Review for final exam with answers - In class Review for...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online