Hypothesis Testing - continued

SCENARIO
A product manager wants to launch a product in the market where mean household income is greater than $ 29,000.
Sample mean of random sample of 400 households is found to be $30,000 with standard deviation of $8,000.
With this amount of data available, should the product manager proceed ahead with the launch?

SOLUTION
We will use One-Tailed Test to come to a conclusion with:

Population Mean = Mu
H0: Mu <= $29,000 - NULL hypothesis
Ha: Mu > $29,000 - Alternate hypothesis
NOTE: The Alternate Hypothesis will never have a equal(=) sign

Given:
Sample mean X-bar = $30,000
Sample size n = 400 households
Population mean(Income) Mu = $29,000
Standard Deviation Sigma = $8,000

  • Using, z = (30000 - 29000)/(8000/SQRT(400)) = 2.5
NOTE: The data distribution is not known but since sample size(= 400) is greater than 30, we use Central Limit Theorem for Normal Data Distribution

Critical Value for Rejecting the NULL hypothesis 

  • Critical/Rejection Region - Defined by Alpha in the normal distribution plot. The default value is 0.05 or 5%
  • The Critical Values are calculated with Probability = 1 - Alpha = 1 - 0.05 = 0.95 or 95%
  • To calculate Z-Critical(Zc) values in Excel:
    • If Alpha is not specified and it is a positive tail, use NORM.S.INV(0.95) = 1.645 to calculate Zc
    • If Alpha is not specified and it is a negative tail, use NORM.S.INV(0.05) = - 1.645 to calculate Zc
  • Using formula for Z and substituting Zc for Z and to calculate Critical Mean = X-bar-Critical(X-bar-c):
    • Sample mean X-bar = $30,000
    • Sample size n = 400 households
    • Population mean(Income) Mu = $29,000
    • Standard Deviation Sigma = $8,000
    • Using Zc = (X-bar-c - Mu)/(Sigma/SQRT(n))
    • X-bar-c = (Sigma/SQRT(n)) * Zc + Mu = $29,658
CONCLUSION
Now since, X-bar-c > $29,000, we reject the NULL hypothesis(H0). This means, the product manager can launch the product in the market




P-Value

  • Also called as Actual Significance Level. This is the level at which the NULL hypothesis gets rejected.
  • Sample Z value is calculated as 2.5. Using EXCEL, 
    • P-value = 1 - NORM.S.DIST(2.5, 1) = 0.0062 or 0.62%
  • P-Value of 0.0062 means you are taking a real risk of only 0.6%. Whereas you are allowed a risk of 5%. Signifies that we can be 99.4% confident of the output.
QUICK OBSERVATIONS
  • When P-Value < Alpha, reject NULL Hypothesis
  • When Z-Value falls in the rejection region, reject NULL Hypothesis
REFERENCES

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