Correlation and Regression

Q1. Describe the Grafton and Coos Counties using ALL variables in the data set.

In the data set Grafton & Coos Counties have an ed_avg of 18.52291 and an income median of 30,000 and the region is 0 which means it is not in southeastern region.

Q2. Create a scatterplot to examine the relationship between ed_avg and income_median.

Q3. Compute the correlation coefficient between the two variables and interpret them.

Hint: Make sure to interpret the direction and the magnitude of the relationship. In addition, keep in mind that correlation (or regression) coefficients do not show causation but only association.

## [1] 0.8622811

There is a positive cooralation between varibales median income and education average on the scatterplot.

Q4. Build a regression model to predict income_median using ed_avg, save the regression result in mod_1, and show the summary result.

## 
## Call:
## lm(formula = income_median ~ ed_avg, data = residents_25to65)
## 
## Residuals:
##     Min      1Q  Median      3Q     Max 
## -3643.9 -2548.6   655.8  1730.7  4150.6 
## 
## Coefficients:
##             Estimate Std. Error t value Pr(>|t|)   
## (Intercept)  -201503      49675  -4.056  0.00365 **
## ed_avg         12695       2636   4.816  0.00133 **
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 2891 on 8 degrees of freedom
## Multiple R-squared:  0.7435, Adjusted R-squared:  0.7115 
## F-statistic: 23.19 on 1 and 8 DF,  p-value: 0.001328

Q5. Is the coefficient of ed_avg statistically significant at 5%? How do you know?

Hint: Discuss your answer in terms of the number of stars in the summary result. Refer to the interpretation section in quiz4_a.

** The end of the intercept line shows that the coefficiant is significant at .1% level. It shows we are 99% confident the intercept is true.

Q6. Further develop the regression model above by adding another variable, region, save the regression result in mod_2, and show the summary result.

## 
## Call:
## lm(formula = income_median ~ ed_avg + region, data = residents_25to65)
## 
## Residuals:
##     Min      1Q  Median      3Q     Max 
## -2016.2  -778.4  -373.5   353.4  2780.7 
## 
## Coefficients:
##             Estimate Std. Error t value Pr(>|t|)    
## (Intercept)  -166192      30700  -5.413 0.000994 ***
## ed_avg         10701       1638   6.532 0.000324 ***
## region          4524       1136   3.981 0.005314 ** 
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 1711 on 7 degrees of freedom
## Multiple R-squared:  0.9214, Adjusted R-squared:  0.899 
## F-statistic: 41.05 on 2 and 7 DF,  p-value: 0.0001359

Q7. Compare mod_1 and mod_2. Which of the two models better fits the data?

Hint: Discuss your answer by comparing the residual standard error and the adjusted R squared between the two models.

It looks like model 2 is a better mod because the residual standard error is lower. Also r-squared value is slightly higher, indicates stronger predictor.

Q8. How much median income does the second model predict for the Grafton and Coos Counties?

Hint: Note that the second model has two predictors. Use both predictors to compute the predicted income.

income=y coefficiant + intercept 166192+1070150=701242+166192=867434 166192+4524 50=392392+166192=558584

Q9. According to the result of the second regression model, are residents of southeastern regions of the State likely to make more income? Why or why not?

Hint: Discuss your answer based on the coefficient of region. You may refer to the interpretation section in quiz4_a.

According to the second regression model, residents of southeastern regions will make less income because the intercept is 166192+4524 *50=392392+166192=558584

Q10.a. Hide the code but display the results of the code on the webpage.

Q10.b. Display the title and your name correctly at the top of the webpage.

Q10.c. Use the correct slug.