Part I

Please put the answers for Part I next to the question number:

    1. daysDrive
    1. mean = 3.3, median = 3.5
    1. Randomly assign Ebola patients to one of two groups, either the treatment or placebo group, and then compare the fever of the two groups.
    1. eye color and natural hair color are independent
    1. 17.8 and 69.0
    1. median and interquartile range; mean and standard deviation

7a. Describe the two distributions.

Distribution A is unimodal and right-skewed. Distribution B is unimodal, symmetrical and nearly normal.

7b. Explain why the means of these two distributions are similar but the standard deviations are not.

Means are similar because when the the mean from 500 random samples of size 30 from the Observations distribution was performed it followed a normal distribution since it satisfies the randomness and minimum number of samples. Standart deviations are not similar because the Observations distribution has observations farther from the mean than in the Sampling Distribution.

7c. What is the statistical principal that describes this phenomenon (2 pts)?

It’s described by Central Limit Theorem.

Part II

Consider the four datasets, each with two columns (x and y), provided below.

options(digits=2)
data1 <- data.frame(x=c(10,8,13,9,11,14,6,4,12,7,5),
                    y=c(8.04,6.95,7.58,8.81,8.33,9.96,7.24,4.26,10.84,4.82,5.68))
data2 <- data.frame(x=c(10,8,13,9,11,14,6,4,12,7,5),
                    y=c(9.14,8.14,8.74,8.77,9.26,8.1,6.13,3.1,9.13,7.26,4.74))
data3 <- data.frame(x=c(10,8,13,9,11,14,6,4,12,7,5),
                    y=c(7.46,6.77,12.74,7.11,7.81,8.84,6.08,5.39,8.15,6.42,5.73))
data4 <- data.frame(x=c(8,8,8,8,8,8,8,19,8,8,8),
                    y=c(6.58,5.76,7.71,8.84,8.47,7.04,5.25,12.5,5.56,7.91,6.89))

For each column, calculate (to two decimal places):

a. The mean (for x and y separately; 1 pt).

mean_x1
## [1] 9
mean_y1 
## [1] 7.5
mean_x2 
## [1] 9
mean_y2 
## [1] 7.5
mean_x3 
## [1] 9
mean_y3 
## [1] 7.5
mean_x4 
## [1] 9
mean_y4 
## [1] 7.5

b. The median (for x and y separately; 1 pt).

median_x1
## [1] 9
median_y1 
## [1] 7.6
median_x2 
## [1] 9
median_y2 
## [1] 8.1
median_x3 
## [1] 9
median_y3 
## [1] 7.1
median_x4 
## [1] 8
median_y4 
## [1] 7

c. The standard deviation (for x and y separately; 1 pt).

sd_x1
## [1] 3.3
sd_y1 
## [1] 2
sd_x2 
## [1] 3.3
sd_y2 
## [1] 2
sd_x3 
## [1] 3.3
sd_y3 
## [1] 2
sd_x4 
## [1] 3.3
sd_y4 
## [1] 2

For each x and y pair, calculate (also to two decimal places; 1 pt):

d. The correlation (1 pt).

cor(data1$x, data1$y)
## [1] 0.82
cor(data2$x, data2$y)
## [1] 0.82
cor(data3$x, data3$y)
## [1] 0.82
cor(data4$x, data4$y)
## [1] 0.82

e. Linear regression equation (2 pts).

lm_1 <- lm(y ~ x, data=data1)
summary(lm_1)
## 
## Call:
## lm(formula = y ~ x, data = data1)
## 
## Residuals:
##     Min      1Q  Median      3Q     Max 
## -1.9213 -0.4558 -0.0414  0.7094  1.8388 
## 
## Coefficients:
##             Estimate Std. Error t value Pr(>|t|)   
## (Intercept)    3.000      1.125    2.67   0.0257 * 
## x              0.500      0.118    4.24   0.0022 **
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 1.2 on 9 degrees of freedom
## Multiple R-squared:  0.667,  Adjusted R-squared:  0.629 
## F-statistic:   18 on 1 and 9 DF,  p-value: 0.00217
lm_2 <- lm(y ~ x, data=data2)
summary(lm_2)
## 
## Call:
## lm(formula = y ~ x, data = data2)
## 
## Residuals:
##    Min     1Q Median     3Q    Max 
## -1.901 -0.761  0.129  0.949  1.269 
## 
## Coefficients:
##             Estimate Std. Error t value Pr(>|t|)   
## (Intercept)    3.001      1.125    2.67   0.0258 * 
## x              0.500      0.118    4.24   0.0022 **
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 1.2 on 9 degrees of freedom
## Multiple R-squared:  0.666,  Adjusted R-squared:  0.629 
## F-statistic:   18 on 1 and 9 DF,  p-value: 0.00218
lm_3 <- lm(y ~ x, data=data3)
summary(lm_3)
## 
## Call:
## lm(formula = y ~ x, data = data3)
## 
## Residuals:
##    Min     1Q Median     3Q    Max 
## -1.159 -0.615 -0.230  0.154  3.241 
## 
## Coefficients:
##             Estimate Std. Error t value Pr(>|t|)   
## (Intercept)    3.002      1.124    2.67   0.0256 * 
## x              0.500      0.118    4.24   0.0022 **
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 1.2 on 9 degrees of freedom
## Multiple R-squared:  0.666,  Adjusted R-squared:  0.629 
## F-statistic:   18 on 1 and 9 DF,  p-value: 0.00218
lm_4 <- lm(y ~ x, data=data4)
summary(lm_4)
## 
## Call:
## lm(formula = y ~ x, data = data4)
## 
## Residuals:
##    Min     1Q Median     3Q    Max 
## -1.751 -0.831  0.000  0.809  1.839 
## 
## Coefficients:
##             Estimate Std. Error t value Pr(>|t|)   
## (Intercept)    3.002      1.124    2.67   0.0256 * 
## x              0.500      0.118    4.24   0.0022 **
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 1.2 on 9 degrees of freedom
## Multiple R-squared:  0.667,  Adjusted R-squared:  0.63 
## F-statistic:   18 on 1 and 9 DF,  p-value: 0.00216

Linear regression equations is the same for all data sets: y=3+0.5x

f. R-Squared (2 pts).

Data1 R-Squared = 0.667

Data2 R-Squared= 0.666

Data3 R-Squared = 0.666

Data4 R-Squared = 0.667

For each pair, is it appropriate to estimate a linear regression model? Why or why not? Be specific as to why for each pair and include appropriate plots! (4 pts)

Data1

Conditions:

1-Linearity CHECK

2-Nearly normal residuals X

3-Constant variability CHECK

4-Independent observations UNKNOWN

Linear regression model is not appropriate.

par(mfrow=c(2,2))
plot(data1)
hist(lm_1$residuals)
qqnorm(lm_1$residuals)
qqline(lm_1$residuals)

Data2

Conditions:

1-Linearity X

2-Nearly normal residuals X

3-Constant variability X

4-Independent observations UNKNOWN

Linear regression model is not appropriate.

par(mfrow=c(2,2))
plot(data2)
hist(lm_2$residuals)
qqnorm(lm_2$residuals)
qqline(lm_2$residuals)

Data3

Conditions:

1-Linearity CHECK

2-Nearly normal residuals CHECK

3-Constant variability X

4-Independent observations UNKNOWN

Linear regression model is not appropriate.

par(mfrow=c(2,2))
plot(data3)
hist(lm_3$residuals)
qqnorm(lm_3$residuals)
qqline(lm_3$residuals)

Data4

Conditions:

1-Linearity X (very extreme outlier)

2-Nearly normal residuals X

3-Constant variability X

4-Independent observations UNKNOWN

Linear regression model is not appropriate.

par(mfrow=c(2,2))
plot(data4)
hist(lm_4$residuals)
qqnorm(lm_4$residuals)
qqline(lm_4$residuals)

Explain why it is important to include appropriate visualizations when analyzing data. Include any visualization(s) you create. (2 pts)

It is critical to visualize the data and check all conditions when creating a model. These data sets have very similar means, standard deviations, R-squared and linear regression equations, however using visualization methods we can conclude that it’s completely inappropriate for some data sets to use linear regression equation.