Assignment - HW Week 2 - 2.24

Given Data on Machine 1 & 2

Machine1 <- c(16.03,16.04,16.05,16.05,16.02,16.01,15.96,15.98,16.02,15.99)
Machine2 <- c(16.02,15.97,15.96,16.01,15.99,16.03,16.04,16.02,16.01,16.00)
M <- cbind(Machine1,Machine2)
print(M)

##       Machine1 Machine2
##  [1,]    16.03    16.02
##  [2,]    16.04    15.97
##  [3,]    16.05    15.96
##  [4,]    16.05    16.01
##  [5,]    16.02    15.99
##  [6,]    16.01    16.03
##  [7,]    15.96    16.04
##  [8,]    15.98    16.02
##  [9,]    16.02    16.01
## [10,]    15.99    16.00

Descriptive Statistics for Machine 1

summary(Machine1)

##    Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
##   15.96   15.99   16.02   16.02   16.04   16.05

Descriptive Statistics for Machine 2

summary(Machine2)

##    Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
##   15.96   15.99   16.01   16.00   16.02   16.04

Two sample t-Test

t.test(Machine1, Machine2, alternative = "two.sided", var.equal = TRUE, conf.level = .95)

## 
##  Two Sample t-test
## 
## data:  Machine1 and Machine2
## t = 0.79894, df = 18, p-value = 0.4347
## alternative hypothesis: true difference in means is not equal to 0
## 95 percent confidence interval:
##  -0.01629652  0.03629652
## sample estimates:
## mean of x mean of y 
##    16.015    16.005

Answers:-

(a) Stating Hypothesis:

Null hypothesis: Ho : u1=u2 : u1-u2=0

There is no significant difference in the net volume of filling bottle in both Machines.

Alternative hypothesis: Ha : u1/=u2 : u1-u2 /= 0

There is significant difference in net volume of filling bottle in the Machines.

(b) Here, we use the two sample t-test because we know the population standard deviation. Comparing the p value with the significance level alpha=0.05 we see that P value > alpha, so we can conclude that we have enough evidence to fail to reject the null hypothesis, and the difference between the two groups is not significantly different.

(c) The P-value for the test for [t = 0.79894] is 0.4347

(d) The 95 percent confidence interval on the difference in the mean fill volume for the two machines is -0.01629652 <= u1-u2 <= 0.03629652

Source Code

# All R code used in document

library(dplyr)
Machine1 <- c(16.03,16.04,16.05,16.05,16.02,16.01,15.96,15.98,16.02,15.99)
Machine2 <- c(16.02,15.97,15.96,16.01,15.99,16.03,16.04,16.02,16.01,16.00)
M <- cbind(Machine1,Machine2)
print(M)
summary(Machine1)
summary(Machine2)
t.test(Machine1, Machine2, alternative = "two.sided", var.equal = TRUE, conf.level = .95)