Excercise 7.29 present regression output from a model for predicting annual murders per million from percentage living in poverty based on a random sample of 20 metropolitan areas. The model output is also provided below
knitr::include_graphics("C:\\Users\\sergioor\\Documents\\images\\capture.png")
a)What are the hypotheses for evaluating wether poverty percentage is a significant predictor of murder rate?
\({ H }_{ 0 }\): The poverty percentage is not significant predictor of murder rate. \(\beta _{ 1 }=0\)
\({ H }_{ A }\): The poverty percentage is a significant predictor of murder rate. \(\beta _{ 1 }\neq 0\)
b)State the conlcusion of the hypothesis test from part (a) in context of the Data.
Based on the summary chart above, the null hypothesis should be rejected. A Pr value of 0 indicates that its highly likely the poverty predictor is chance
c)Calculate a 95% confidence interval for the slope of poverty percentage, and interpret in in the context of the data
The 95% confidence interval is calculated using a T Distribution, with degrees of freedom of 18 (as there are 20 metropolitan areas, less 2 sided test), and the equation (using the t-value from the t probability table) is essentially: poverty pct ± (2.10 * SE), giving (1.74, 3.38). The more exact "R" calculation is below. This interval basically means that each increase of one percent in poverty , will, within 95 confidence, increase the number of murders per million in the range of 1.74 to 3.38.
The confidence interval means that for each increase in one percent in poverty, the increase in murders per million will be within the range or 1.74 and 3.38, with 95% confidence.
ci_u <- 2.559 + (qt(.975, df=18)*.39)
ci_l <- 2.559 - (qt(.975, df=18)*.39)
list(ci_u, ci_l)
## [[1]]
## [1] 3.37836
##
## [[2]]
## [1] 1.73964
d)Do your results from the hypothesis test and the confidence interval agree? Explain.
The results do agree with the hypothesis test. The CI does not cross zero which also means that poverty is a predictor of murder rate