Your group has invented a solar cell that works by applying photosynthesis chemi

Question

Your group has invented a solar cell that works by applying photosynthesis chemistry to generating hydrogen which is stored for use in a fuel cell. In 10 different regions you set up solar-city installations (using solar cells and lead acid batteries for storage) and your technology comparing the kilowatt hours generated in a month The results are. a. Test whether there is a significant difference between the amounts generated assuming a normal distribution, use alpha = 0.05 b. Test whether there is a significant difference assuming that the data is not normally distributed use alpha = 0.05

Explanation / Answer

I am using R software to solve this problem. At first we can load the data in R environment as below:

SolarCity <- c(1000,   1700,   1200,   700,   620,   740,   920,   640,   1500,   500)
PhotoFuel <- c(1200,   1500,   100,   900,   780,   860,   1200,   800,   2000,   650)

a) If the data is normally distributed, ew can use t test to see if there is significant difference. T tests can be done in R using t.test() function as below:

t.test(SolarCity,PhotoFuel,alternative = "two.sided",var.equal = T,conf.level = 0.95)

   Two Sample t-test

data: SolarCity and PhotoFuel
t = -0.22797, df = 18, p-value = 0.8222
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
-480.1417 386.1417
sample estimates:
mean of x mean of y
952 999

Here we can see that p value is 0.8222 which is far bigger than 0.05. So we can say that there is no significant difference.

b) When normality assumption is not met, we can use the non parametric approach. In this case we can use Wilcoxon rank sum test, also called as Mann-Whitney U test. In R, we can do this test using wilcox.test() function as below:

wilcox.test(SolarCity,PhotoFuel,alternative = "two.sided",conf.level = 0.95)

   Wilcoxon rank sum test with continuity correction

data: SolarCity and PhotoFuel
W = 43.5, p-value = 0.6495
alternative hypothesis: true location shift is not equal to 0

Here also we can see that p value is 0.6495, which is greater than 0.05. So we can conclude that there is no significant difference.

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