A ____ document identifies the purpose of the program being developed, the application title, the procedures to be followed when
1 answer:
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The intercept of a graph is the point where the graph crosses the axes of the graph.
The interpretation of the c-intercept is: <em>in 1996, the global wind power capacity was approximately 7 gigawatts.</em>
On the graph (see attachment), we have the following observations
- The vertical axis represents the global wind power capacity (i.e. the c-axis)
- The horizontal axis represents the years since 1996 (i.e. the t-axis)
So, the c-intercept is the point that crosses the vertical axis.
From the graph, the curve crosses the c-axis at c = 7, and t = 0 (i.e. years = 1996)
Hence, the interpretation of the c-intercept is: <em>in 1996, the global wind power capacity was approximately 7 gigawatts.</em>
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Read more about intercepts at:
Answer:
library(moments)
library(readxl)
Exam1Q1 <- read_excel("Exam1Q1.xlsx")
View(Exam1Q1)
## Warning in system2("/usr/bin/otool", c("-L", shQuote(DSO)), stdout = TRUE):
## running command ''/usr/bin/otool' -L '/Library/Frameworks/R.framework/
## Resources/modules/R_de.so'' had status 1
plot(density(Exam1Q1$Highschool))
gostino.test(Exam1Q1$Highschool)
##
## D'Agostino skewness test
##
## data: Exam1Q1$Highschool
## skew = -0.24041, z = -0.69269, p-value = 0.4885
## alternative hypothesis: data have a skewness
agostino.test(Exam1Q1$BS)
##
## D'Agostino skewness test
##
## data: Exam1Q1$BS
## skew = 0.53467, z = 1.49330, p-value = 0.1354
## alternative hypothesis: data have a skewness
anscombe.test(Exam1Q1$Highschool)
##
## Anscombe-Glynn kurtosis test
##
## data: Exam1Q1$Highschool
## kurt = 2.56950, z = -0.29677, p-value = 0.7666
## alternative hypothesis: kurtosis is not equal to 3
anscombe.test(Exam1Q1$BS)
##
## Anscombe-Glynn kurtosis test
##
## data: Exam1Q1$BS
## kurt = 3.6449, z = 1.2599, p-value = 0.2077
## alternative hypothesis: kurtosis is not equal to 3
t.test(Exam1Q1$Highschool,Exam1Q1$BS, var.equal = T)
##
## Two Sample t-test
##
## data: Exam1Q1$Highschool and Exam1Q1$BS
## t = 0.16929, df = 76, p-value = 0.866
## alternative hypothesis: true difference in means is not equal to 0
## 95 percent confidence interval:
## -1.131669 1.341926
## sample estimates:
## mean of x mean of y
## 39.51282 39.40769
Explanation:
Summary:
The data set include two dependent variables which states High School annual earnings and BS annual earnings in the same small firm. Firstly, I used density plot function to view the distribution of two sets of variables. Secondly, I tested the skewness and kurtosis.Skewness of High School is -0.24041 which is left tail. Skewness of BS is 0.53476 which is right tail.Kurtosis of High School is 2.5695 which lower than 3 is platykurtic. Kurtosis of BS is 3.6449 which larger than 3 is leptokurtic. Thirdly, the sample population is 39 for each and has unknow standard deviation, so that I’m going to use t-test to perform the comparing analysis. The result shows p-value which equal to 0.866 is larger than significant value(alpha = 0.05). To the concludsion, there is no difference between yearly earnings of High School and of BS at a small firm.
