Question

It has been suggested that night shift-workers show more variability in their output levels than day workers (σ2N > σ2D). Below, you are given the results of two independent random samples.

Night-shift Day-shift
Sample-size 9 8
Sample-mean 520 540
sample-variance 38 20

Required:
a. State the null and alternative hypotheses to be tested.
b. Compute the test statistic.
c. Determine the p-value.
d. At 95% confidence, what do you conclude?

Answer 1

Answer:

Calculated |t |= 7.72 > 2.1314 at 0.05 level of significance

Null hypothesis is rejected

The night shift-workers show more variability in their output levels are greater than day workers

Step-by-step explanation:

Step(i):-

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

Step(ii):-

Null hypothesis :H₀:  σ²₁ =σ²₂

Alternative Hypothesis:H₁: σ²₁ > σ²₂

Test statistic

              $t = \frac{x^{-} _{1} - x^{-} _{2} }{\sqrt{\frac{S^2_{1} }{n_{1} }+\frac{S^2_{2} }{n_{2} } } }$

             

            $t = \frac{520 - 540 }{\sqrt{\frac{38 }{9 }+\frac{20 }{8 } } }$

           t =    - 7.72

          |t| = |- 7.72| = 7.72

Degrees of freedom

ν = n₁ + n₂ -2 = 9 +8 -2 = 15

t₀.₀₅ = 2.1314

Final answer:-

Calculated |t |= 7.72 > 2.1314 at 0.05 level of significance

Null hypothesis is rejected

The night shift-workers show more variability in their output levels are greater than day workers