METHODS FOR HYPOTHESIS TESTING
- z-Test
 - Used when Sample Size is Large
 - Used for single Population or single Sample
 - Used for Proportion
 - t-Test
 - Used when Sample Size is small
 - Used when Standard Deviation of the population is not known
 - Types:
 - One Sample
 - Two Sample
 - Paired t Test
 - Chi-Square-Test
 
One Sample t-Test Example
A typical American mean mobile internet usage is 144 minutes. You select a sample size of 30.
- Is there a evidence that the population mean time spent accessing the internet via mobile is different than 144 minutes? Use P-Value approach and level of significance of 0.05
 
| Minutes | 
| 72 | 
| 144 | 
| 48 | 
| 72 | 
| 36 | 
| 360 | 
| 44 | 
| 30 | 
| 432 | 
| 24 | 
| 288 | 
| 144 | 
| 144 | 
| 240 | 
| 432 | 
| 144 | 
| 144 | 
| 144 | 
| 576 | 
| 216 | 
| 72 | 
| 72 | 
| 144 | 
| 288 | 
| 144 | 
| 36 | 
| 288 | 
| 48 | 
| 288 | 
| 144 | 
Load this data to some file and read it into R variable mydata
attach(mydata)
#Hypothesis testing => H0: Mu = 144, Ha: Mu < 144
#Sample Mean
Xbar = mean(Minutes); Xbar
#Sample Standard Deviation
Sigma = sd(Minutes); Sigma
#Population Mean
Mu = 144; Mu
#Sample Size
n = 30; n
#Degrees of Freedom
dF = n - 1; dF
#Function sqrt or ^0.5 i.e. to power 0.5
tval = (Xbar - Mu)/(Sigma/sqrt(n)); tval
#P-Value
Pval = pt(tval, dF); Pval
OUTPUT
CONCLUSION
Since, P-value = 0.885 or 88.5% > Alpha = 0.05 or 5%, we do not reject the NULL hypothesis i.e. Average time spent is 144 minutes
NOTE:
- 1 - P-value => This is the lowest value for a One-Tail test. Excel function - T.DIST
 - (1 - P-value) * 2 => This is the lowest value for a Two-Tail test. Excel function - T.DIST.2T
 
Comments
Post a Comment