Answer:
<em>Calculated |t |= 7.72 > 2.1314 at 0.05 level of significance </em>
<em>Null hypothesis is rejected </em>
<em> The night shift-workers show more variability in their output levels are greater than day workers </em>
Step-by-step explanation:
<u>Step(i):</u>-
Given first sample size n₁ = 9
Given mean of the first sample x₁⁻ = 520
Given sample variance of the first sample'S²₁' = = 38
Standard deviation of the first sample 'S₁' = √38 =6.16
Given second sample size n₂ = 8
Given mean of the second sample x₂⁻ = 540
Given sample variance of the second sample ( 'S²₂' = 20
Standard deviation of the second sample 'S₂' = √20 = 4.4712
<u><em>Step(ii):-</em></u>
<em>Null hypothesis :H₀: σ²₁ =σ²₂</em>
<em>Alternative Hypothesis:H₁: σ²₁ > σ²₂</em>
<em>Test statistic </em>
<em> </em>
<em></em>
<em> </em>
![t = \frac{520 - 540 }{\sqrt{\frac{38 }{9 }+\frac{20 }{8 } } }](https://tex.z-dn.net/?f=t%20%3D%20%5Cfrac%7B520%20-%20540%20%7D%7B%5Csqrt%7B%5Cfrac%7B38%20%7D%7B9%20%7D%2B%5Cfrac%7B20%20%7D%7B8%20%7D%20%20%7D%20%7D)
t = - 7.72
|t| = |- 7.72| = 7.72
Degrees of freedom
<em> ν = n₁ + n₂ -2 = 9 +8 -2 = 15</em>
<em>t₀.₀₅ = 2.1314</em>
<u><em>Final answer</em></u><em>:-</em>
<em>Calculated |t |= 7.72 > 2.1314 at 0.05 level of significance </em>
<em>Null hypothesis is rejected </em>
<em> The night shift-workers show more variability in their output levels are greater than day workers </em>