Answer:
Null hypotheses = H₀ = σ₁² ≤ σ₂²
Alternative hypotheses = Ha = σ₁² > σ₂²
Test statistic = 1.9
p-value = 0.206
Since the p-value is greater than α therefore, we cannot reject the null hypothesis.
So we can conclude that the night shift workers don't show more variability in their output levels than day workers.
Explanation:
Let σ₁² denotes the variance of night shift-workers
Let σ₂² denotes the variance of day shift-workers
State the null and alternative hypotheses:
The null hypothesis assumes that the variance of night shift-workers is equal to or less than day-shift workers.
Null hypotheses = H₀ = σ₁² ≤ σ₂²
The alternate hypothesis assumes that the variance of night shift-workers is more than day-shift workers.
Alternative hypotheses = Ha = σ₁² > σ₂²
Test statistic:
The test statistic or also called F-value is calculated using
Test statistic = Larger sample variance/Smaller sample variance
The larger sample variance is σ₁² = 38
The smaller sample variance is σ₂² = 20
Test statistic = σ₁²/σ₂²
Test statistic = 38/20
Test statistic = 1.9
p-value:
The degree of freedom corresponding to night shift workers is given by
df₁ = n - 1
df₁ = 9 - 1
df₁ = 8
The degree of freedom corresponding to day shift workers is given by
df₂ = n - 1
df₂ = 8 - 1
df₂ = 7
We can find out the p-value using F-table or by using Excel.
Using Excel to find out the p-value,
p-value = FDIST(F-value, df₁, df₂)
p-value = FDIST(1.9, 8, 7)
p-value = 0.206
Conclusion:
p-value > α
0.206 > 0.05 ( α = 1 - 0.95 = 0.05)
Since the p-value is greater than α therefore, we cannot reject the null hypothesis corresponding to a confidence level of 95%
So we can conclude that the night shift workers don't show more variability in their output levels than day workers.