Statistic study to define resistance supplier

Standard Deviaton from sample

r<- c(3.14, 3.22, 3.30, 3.52, 3.05, 3.10, 3.54, 3.39, 3.19, 2.87, 3.23, 2.87, 2.91, 3.07, 3.29)

sd(r)
## [1] 0.2107967

Effect

0.1/sd(r)
## [1] 0.4743908

Power

  • Determined by us: 0.8

Sample size

pwr.t.test(d=(0.1/sd(r)),power=0.8,sig.level=0.05, type="one.sample", alternative="two.sided")
## 
##      One-sample t test power calculation 
## 
##               n = 36.84348
##               d = 0.4743908
##       sig.level = 0.05
##           power = 0.8
##     alternative = two.sided

New Sample

n<- c(3.14,3.40,3.23,3.09,3.38,3.42,3.21,3.17,3.37,3.25,3.23,3.11,2.98,3.29,3.26,3.23,3.28,3.11,3.25,3.01,3.14,3.09,3.13,2.99,3.29,3.10,3.25,3.16,3.15,3.24,3.08,3.30,3.28,3.01,3.17,3.07,3.24)

Determine new test to approve or reject Ho

t.test(n, mu=3.2, alternative="two.sided")
## 
##  One Sample t-test
## 
## data:  n
## t = -0.43267, df = 36, p-value = 0.6678
## alternative hypothesis: true mean is not equal to 3.2
## 95 percent confidence interval:
##  3.153886 3.229898
## sample estimates:
## mean of x 
##  3.191892

In this case we are getting a p-value of 0.6678, which is much more bigger than our alpha (0.05). We have enough evidence to fail to reject Ho and we will continue to work with our supplier. That means that we can say to the managers that we are 80% confident that the resistances given by our supplier are good enough for our manufacturing processes.