In: Statistics and Probability
Consider the following Data:
Year |
Tea |
Coffee |
---|---|---|
1994 |
42.4 |
95.85 |
1995 |
42.12 |
97.28 |
1996 |
47.61 |
87.62 |
1997 |
60.86 |
92.04 |
1998 |
55.58 |
99.21 |
1999 |
50.61 |
95.63 |
2000 |
49.89 |
97.42 |
2001 |
56.77 |
93.93 |
2002 |
62.53 |
95.67 |
2003 |
68.31 |
99.25 |
2004 |
69.88 |
101.31 |
2005 |
72.99 |
101.68 |
2006 |
71.36 |
104.02 |
2007 |
90.78 |
106.09 |
2008 |
74.7 |
105.8 |
2009 |
67.15 |
102.15 |
2010 |
67.03 |
101.15 |
2011 |
87.83 |
104.05 |
2012 |
93.4 |
102.7 |
2013 |
78.9 |
105.28 |
2014 |
111.32 |
106.3 |
2015 |
98.39 |
104.96 |
2016 |
105.25 |
103.57 |
By using the definition and discussing what is relevant to the situation, interpret each of the following for both the coffee and tea data. Also, compare each for coffee and tea. Be sure to include the relevant information (state the value of or, in the case of the distribution, include the graphs) with each component.
Here I attach the R code with output
tea=c(42.4,42.12,47.61,60.86,55.58,50.61,49.89,56.77,62.53,68.31,69.88,72.99,71.36,90.78,74.7,67.15,67.03,87.83,93.4,78.9,111.32,98.39,105.25)
coffee=c(95.85,97.28,87.62,92.04,99.21,95.63,97.42,93.93,95.67,99.27,101.31,101.68,104.02,106.09,105.8,102.15,101.15,104.05,102.7,105.28,106.3,104.96,103.57)
year=c(1994:2016)
mean(tea)
median(tea)
mfv1(tea)
range(tea)
mean(coffee)
median(coffee)
mfv1(coffee)
summary(tea)
summary(coffee)
IQRtea=83.36-56.18=27.18
IQRcoffee=104.00-96.56=7.44
sd(tea)
sd(coffee)
range(coffee)
hist(coffee)
hist(tea)
boxplot(coffee)
boxplot(tea)
plot(coffee,tea)
plot(year,coffee)
plot(year,tea)
m=lm(coffee~tea)
summary(m)
cor(coffee,tea)
The measure of central tendency are mean, median , mode(mfv-most frequent observation) etc..
Measure of spread is range
range(tea)=111.32-42.12=69.2
range(coffee)=106.30-87.62=18.68
Here the model interval length is 5 units.
Here the model interval length is taken as 10 units.
Boxplot of coffee
From the plot it is clear that the distribution is negatively skewed and there is no outliers in the data.
Boxplot of tea
From the plot we can infer that the distribution is positively skewed and there is no outliers in the model.
The correlation between Coffee and tea is 0.769, which implies there is positive linear relatioship exist between the variable tea and coffee.
The obtained model is:
coffee=86.3200+0.1954*tea