Case Problem: Z test for mean, single sample, population standard deviation is known.
A manufacturer of a certain brand of 9-volt batteries claims that the average life of the battery is 40
hours when used in a radio, with a standard deviation of 5 hours. To test the manufacturer’s claim, a
random sample of 100 batteries was tested and it showed an average life of 38 hours. What can you
conclude about the manufacturer’s claim at a level of significance α = 0.05? Calculate p – value.
Solution:
Random variable X = Life of battery (in hours)
Sample Size (n) = 100
Sample mean ( X ¿ = 38 hours
Population average as claimed by manufacturer = 40 hrs
Population standard deviation as claimed by manufacturer = 5 hours
Step 1: Setting up Null and Alternative hypothesis
Let μ be the average life of the battery produced by the manufacturer.
Null Hypothesis
H0 : μ = 40 (Manufacturer claim or average life of battery produced by company is 40 hours)
Alternative Hypothesis
H1: μ ≠ 40 (Consumer claim or average life of battery produced by company is not equal to 40 hours)
Step 2: Level of significance of the test
Here, level of significance = prob (type I error) = 5 % = 0.05
Step 3: Test Statistic
The appropriate test statistic for this test is given by,
Where,
= sample mean
= Hypothesized value of mean
= Known population standard deviation
n = Sample size
The test statistic follows standard normal distribution with mean = 0 and standard deviation = 1
Step 4: Critical value or Tabulated value of Z
The test is two sided, hence there are two rejection regions.
The level of significance of test (α) = 5 % (given)
The critical value of Z from the table is given by,
Zc = Zα/2 = 1.96
AR: -1.96 < Z < + 1.96
RR: Either Z ≥ 1.96 Or Z ≤ - 1.96
Decision Rule: Accept H0 if cal Z falls in the acceptance region (-1.96, +1.96) otherwise reject H0
Step 5 : Calculated or Observed value of Z
The value of Z from sample is given by,
= =-4
Hence, cal Z = - 4
Step 6: Statistical decision
Since, calculated Z falls in the lower rejection region (Z ≤ - 1.96), we reject null hypothesis at 5 % level of
significance.
Step 7: Conclusion
The average life of batteries produced by certain manufacturer is not 40 hrs. i.e. it is significantly lower
than 40 hours. Hence manufacturer claim is not valid.
Step 8: p-value
two-tailed test p-value = 2 x P(Z ≤ - 4) = 2 x 0.00003 = 0.00006
left sided test p-value = P (Z ≤ - 4) = 0.00003
Decision
If p-value (0.00006) ≤ α (0.05) value reject H0
If p-value ≥ α value accept
